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<MASK>
Those who don’t know how to find number of factors of large numbers can try problem given in Related Problems.
### EDITORIALIST’S SOLUTIONS:
Editorialist solution can be found here.
<MASK>
<UNMASK>
<MASK>
Author: Vineet Shah
Editorialist: Vaibhav Jain
<MASK>
Find the number of ordered pairs (b,c) such that it satisfies the following equation:
a+(b×c)=a×b+a×c
<MASK>
### EXPLANATION:
<MASK>
Now we can see that number of ordered pairs (b,c) shall be equal to the factors of a×(a-1) i.e. fact(a)×fact(a-1). This is because fact(a) is multiplicative function and gcd(a,a-1)=1. Note here that if (b,c) satisfies equation so does (-b,-c) and similarly when (b,-c) satisfies equation so does (-b,c). Hence, actual answer will be 2×fact(a)×fact(a-1). But finding factors of a will give TLE. Hence we will use concept here.
<MASK>
If primality test comes out to be true that means a has left only one factor otherwise it has two factors. For large numbers we will use one of the probabilistic methods of primality test whose complexity is O(klogn) where k is number of iterations.
<MASK>
Those who don’t know how to find number of factors of large numbers can try problem given in Related Problems.
### EDITORIALIST’S SOLUTIONS:
Editorialist solution can be found here.
### RELATED PROBLEMS:
<MASK>
1 Like
Mention this as well -
Let fact(x) = no of factors of x.
And fact(x) is a multiplicative function.
Because gcd(a,a-1) is 1.
Therefore, fact(a*(a-1)) = fact(a)*fact(a-1).
2 Likes
|
<MASK>
Author: Vineet Shah
Editorialist: Vaibhav Jain
<MASK>
Find the number of ordered pairs (b,c) such that it satisfies the following equation:
a+(b×c)=a×b+a×c
<MASK>
### EXPLANATION:
<MASK>
Now we can see that number of ordered pairs (b,c) shall be equal to the factors of a×(a-1) i.e. fact(a)×fact(a-1). This is because fact(a) is multiplicative function and gcd(a,a-1)=1. Note here that if (b,c) satisfies equation so does (-b,-c) and similarly when (b,-c) satisfies equation so does (-b,c). Hence, actual answer will be 2×fact(a)×fact(a-1). But finding factors of a will give TLE. Hence we will use concept here.
<MASK>
If primality test comes out to be true that means a has left only one factor otherwise it has two factors. For large numbers we will use one of the probabilistic methods of primality test whose complexity is O(klogn) where k is number of iterations.
<MASK>
Those who don’t know how to find number of factors of large numbers can try problem given in Related Problems.
### EDITORIALIST’S SOLUTIONS:
Editorialist solution can be found here.
### RELATED PROBLEMS:
<MASK>
1 Like
Mention this as well -
Let fact(x) = no of factors of x.
And fact(x) is a multiplicative function.
Because gcd(a,a-1) is 1.
Therefore, fact(a*(a-1)) = fact(a)*fact(a-1).
2 Likes
<UNMASK>
<MASK>
Author: Vineet Shah
Editorialist: Vaibhav Jain
<MASK>
### PROBLEM:
Find the number of ordered pairs (b,c) such that it satisfies the following equation:
a+(b×c)=a×b+a×c
<MASK>
### EXPLANATION:
<MASK>
Now we can see that number of ordered pairs (b,c) shall be equal to the factors of a×(a-1) i.e. fact(a)×fact(a-1). This is because fact(a) is multiplicative function and gcd(a,a-1)=1. Note here that if (b,c) satisfies equation so does (-b,-c) and similarly when (b,-c) satisfies equation so does (-b,c). Hence, actual answer will be 2×fact(a)×fact(a-1). But finding factors of a will give TLE. Hence we will use concept here.
<MASK>
If primality test comes out to be true that means a has left only one factor otherwise it has two factors. For large numbers we will use one of the probabilistic methods of primality test whose complexity is O(klogn) where k is number of iterations.
<MASK>
Those who don’t know how to find number of factors of large numbers can try problem given in Related Problems.
### EDITORIALIST’S SOLUTIONS:
Editorialist solution can be found here.
### RELATED PROBLEMS:
<MASK>
1 Like
Mention this as well -
Let fact(x) = no of factors of x.
And fact(x) is a multiplicative function.
Because gcd(a,a-1) is 1.
Therefore, fact(a*(a-1)) = fact(a)*fact(a-1).
2 Likes
|
<MASK>
Author: Vineet Shah
Editorialist: Vaibhav Jain
<MASK>
### PROBLEM:
Find the number of ordered pairs (b,c) such that it satisfies the following equation:
a+(b×c)=a×b+a×c
<MASK>
### EXPLANATION:
<MASK>
Now we can see that number of ordered pairs (b,c) shall be equal to the factors of a×(a-1) i.e. fact(a)×fact(a-1). This is because fact(a) is multiplicative function and gcd(a,a-1)=1. Note here that if (b,c) satisfies equation so does (-b,-c) and similarly when (b,-c) satisfies equation so does (-b,c). Hence, actual answer will be 2×fact(a)×fact(a-1). But finding factors of a will give TLE. Hence we will use concept here.
<MASK>
If primality test comes out to be true that means a has left only one factor otherwise it has two factors. For large numbers we will use one of the probabilistic methods of primality test whose complexity is O(klogn) where k is number of iterations.
<MASK>
Those who don’t know how to find number of factors of large numbers can try problem given in Related Problems.
### EDITORIALIST’S SOLUTIONS:
Editorialist solution can be found here.
### RELATED PROBLEMS:
<MASK>
1 Like
Mention this as well -
Let fact(x) = no of factors of x.
And fact(x) is a multiplicative function.
Because gcd(a,a-1) is 1.
Therefore, fact(a*(a-1)) = fact(a)*fact(a-1).
2 Likes
<UNMASK>
<MASK>
Author: Vineet Shah
Editorialist: Vaibhav Jain
<MASK>
### PROBLEM:
Find the number of ordered pairs (b,c) such that it satisfies the following equation:
a+(b×c)=a×b+a×c
### QUICK EXPLANATION:
We can rewrite equation as a(a-1)=(b-a)(c-a). Now the answer will be 2×fact(a)×fact(a-1) where fact(i) is the number of factors of i.
### EXPLANATION:
<MASK>
Now we can see that number of ordered pairs (b,c) shall be equal to the factors of a×(a-1) i.e. fact(a)×fact(a-1). This is because fact(a) is multiplicative function and gcd(a,a-1)=1. Note here that if (b,c) satisfies equation so does (-b,-c) and similarly when (b,-c) satisfies equation so does (-b,c). Hence, actual answer will be 2×fact(a)×fact(a-1). But finding factors of a will give TLE. Hence we will use concept here.
<MASK>
If primality test comes out to be true that means a has left only one factor otherwise it has two factors. For large numbers we will use one of the probabilistic methods of primality test whose complexity is O(klogn) where k is number of iterations.
Now only a=0 and a=1 will output -1. This is because 0 has infinite factors which means infinite pairs.
Those who don’t know how to find number of factors of large numbers can try problem given in Related Problems.
### EDITORIALIST’S SOLUTIONS:
Editorialist solution can be found here.
### RELATED PROBLEMS:
<MASK>
1 Like
Mention this as well -
Let fact(x) = no of factors of x.
And fact(x) is a multiplicative function.
Because gcd(a,a-1) is 1.
Therefore, fact(a*(a-1)) = fact(a)*fact(a-1).
2 Likes
|
<MASK>
Author: Vineet Shah
Editorialist: Vaibhav Jain
<MASK>
### PROBLEM:
Find the number of ordered pairs (b,c) such that it satisfies the following equation:
a+(b×c)=a×b+a×c
### QUICK EXPLANATION:
We can rewrite equation as a(a-1)=(b-a)(c-a). Now the answer will be 2×fact(a)×fact(a-1) where fact(i) is the number of factors of i.
### EXPLANATION:
<MASK>
Now we can see that number of ordered pairs (b,c) shall be equal to the factors of a×(a-1) i.e. fact(a)×fact(a-1). This is because fact(a) is multiplicative function and gcd(a,a-1)=1. Note here that if (b,c) satisfies equation so does (-b,-c) and similarly when (b,-c) satisfies equation so does (-b,c). Hence, actual answer will be 2×fact(a)×fact(a-1). But finding factors of a will give TLE. Hence we will use concept here.
<MASK>
If primality test comes out to be true that means a has left only one factor otherwise it has two factors. For large numbers we will use one of the probabilistic methods of primality test whose complexity is O(klogn) where k is number of iterations.
Now only a=0 and a=1 will output -1. This is because 0 has infinite factors which means infinite pairs.
Those who don’t know how to find number of factors of large numbers can try problem given in Related Problems.
### EDITORIALIST’S SOLUTIONS:
Editorialist solution can be found here.
### RELATED PROBLEMS:
<MASK>
1 Like
Mention this as well -
Let fact(x) = no of factors of x.
And fact(x) is a multiplicative function.
Because gcd(a,a-1) is 1.
Therefore, fact(a*(a-1)) = fact(a)*fact(a-1).
2 Likes
<UNMASK>
# SWAPSIGN(Code Melange) - EDITORIAL (UNOFFICIAL)
Author: Vineet Shah
Editorialist: Vaibhav Jain
MEDIUM
### PROBLEM:
Find the number of ordered pairs (b,c) such that it satisfies the following equation:
a+(b×c)=a×b+a×c
### QUICK EXPLANATION:
We can rewrite equation as a(a-1)=(b-a)(c-a). Now the answer will be 2×fact(a)×fact(a-1) where fact(i) is the number of factors of i.
### EXPLANATION:
Let us consider our equation:
a+(b×c)=a×b+a×c\tag*{}
Rewrite it as:
a+b×c-a×b-a×c=0\tag*{}
a+b×c-a×b-a×c+a^{2}=a^{2}\tag*{}
Taking common:
b×(c-a)+a×(a-c)=a^{2}-a\tag*{}
Rearranging:
(b-a)×(c-a)=a×(a-1)\tag*{}
Now we can see that number of ordered pairs (b,c) shall be equal to the factors of a×(a-1) i.e. fact(a)×fact(a-1). This is because fact(a) is multiplicative function and gcd(a,a-1)=1. Note here that if (b,c) satisfies equation so does (-b,-c) and similarly when (b,-c) satisfies equation so does (-b,c). Hence, actual answer will be 2×fact(a)×fact(a-1). But finding factors of a will give TLE. Hence we will use concept here.
We can precompute every prime number till 10^{6} using the old Sieve of Eratosthenes or whatever variant you prefer. Now we will divide a from every prime factor till 10^{6} to count the powers of distinct prime numbers to calculate number of factors. After doing this a must be either 1 or if not then a must have prime factor(s) greater than 10^{6}. Claim: In this case a can have either only one or two prime factors greater than 10^{6}. This is because if a will have three prime factors greater than 10^{6} it will exceed 10^{18}. Hence, to check whether a has only factor or two factors left can be done using primality test.
If primality test comes out to be true that means a has left only one factor otherwise it has two factors. For large numbers we will use one of the probabilistic methods of primality test whose complexity is O(klogn) where k is number of iterations.
Now only a=0 and a=1 will output -1. This is because 0 has infinite factors which means infinite pairs.
Those who don’t know how to find number of factors of large numbers can try problem given in Related Problems.
### EDITORIALIST’S SOLUTIONS:
Editorialist solution can be found here.
### RELATED PROBLEMS:
https://www.codechef.com/PRACTICE/problems/NUMFACT
1 Like
Mention this as well -
Let fact(x) = no of factors of x.
And fact(x) is a multiplicative function.
Because gcd(a,a-1) is 1.
Therefore, fact(a*(a-1)) = fact(a)*fact(a-1).
2 Likes
|
<MASK>
Differential equations make use of mathematical operations with derivatives. Solving differential equations is a very important, but also hard concept in calculus. There exist more methods of solving this kind of exercises.
Firstly, we will have a look on how first order differential equation can be solved. Below, you will find a series of methods of solving this kind of differential equations. The first one is the separation method of variables. Practically, as the name says itself, you have to separate the variables, obtaining an equality of two functions with different variables. Then you integrate the expression and you obtain an equality of two integrals. Then you will solve the integrals and find the solution.
<MASK>
<UNMASK>
# How to solve differential equations
Differential equations make use of mathematical operations with derivatives. Solving differential equations is a very important, but also hard concept in calculus. There exist more methods of solving this kind of exercises.
Firstly, we will have a look on how first order differential equation can be solved. Below, you will find a series of methods of solving this kind of differential equations. The first one is the separation method of variables. Practically, as the name says itself, you have to separate the variables, obtaining an equality of two functions with different variables. Then you integrate the expression and you obtain an equality of two integrals. Then you will solve the integrals and find the solution.
<MASK>
|
# How to solve differential equations
Differential equations make use of mathematical operations with derivatives. Solving differential equations is a very important, but also hard concept in calculus. There exist more methods of solving this kind of exercises.
Firstly, we will have a look on how first order differential equation can be solved. Below, you will find a series of methods of solving this kind of differential equations. The first one is the separation method of variables. Practically, as the name says itself, you have to separate the variables, obtaining an equality of two functions with different variables. Then you integrate the expression and you obtain an equality of two integrals. Then you will solve the integrals and find the solution.
<MASK>
<UNMASK>
# How to solve differential equations
Differential equations make use of mathematical operations with derivatives. Solving differential equations is a very important, but also hard concept in calculus. There exist more methods of solving this kind of exercises.
Firstly, we will have a look on how first order differential equation can be solved. Below, you will find a series of methods of solving this kind of differential equations. The first one is the separation method of variables. Practically, as the name says itself, you have to separate the variables, obtaining an equality of two functions with different variables. Then you integrate the expression and you obtain an equality of two integrals. Then you will solve the integrals and find the solution.
<MASK>
## Solving differential equations video lesson
Ian Roberts Engineer San Francisco, USA "If you're at school or you just deal with mathematics, you need to use Studygeek.org. This thing is really helpful." Lisa Jordan Math Teacher New-York, USA "I will recommend Studygeek to students, who have some chalenges in mathematics. This Site has bunch of great lessons and examples. " John Maloney Student, Designer Philadelphia, USA " I'm a geek, and I love this website. It really helped me during my math classes. Check it out) " Steve Karpesky Bookkeeper Vancuver, Canada "I use Studygeek.org a lot on a daily basis, helping my son with his geometry classes. Also, it has very cool math solver, which makes study process pretty fun"
|
# How to solve differential equations
Differential equations make use of mathematical operations with derivatives. Solving differential equations is a very important, but also hard concept in calculus. There exist more methods of solving this kind of exercises.
Firstly, we will have a look on how first order differential equation can be solved. Below, you will find a series of methods of solving this kind of differential equations. The first one is the separation method of variables. Practically, as the name says itself, you have to separate the variables, obtaining an equality of two functions with different variables. Then you integrate the expression and you obtain an equality of two integrals. Then you will solve the integrals and find the solution.
<MASK>
## Solving differential equations video lesson
Ian Roberts Engineer San Francisco, USA "If you're at school or you just deal with mathematics, you need to use Studygeek.org. This thing is really helpful." Lisa Jordan Math Teacher New-York, USA "I will recommend Studygeek to students, who have some chalenges in mathematics. This Site has bunch of great lessons and examples. " John Maloney Student, Designer Philadelphia, USA " I'm a geek, and I love this website. It really helped me during my math classes. Check it out) " Steve Karpesky Bookkeeper Vancuver, Canada "I use Studygeek.org a lot on a daily basis, helping my son with his geometry classes. Also, it has very cool math solver, which makes study process pretty fun"
<UNMASK>
# How to solve differential equations
Differential equations make use of mathematical operations with derivatives. Solving differential equations is a very important, but also hard concept in calculus. There exist more methods of solving this kind of exercises.
Firstly, we will have a look on how first order differential equation can be solved. Below, you will find a series of methods of solving this kind of differential equations. The first one is the separation method of variables. Practically, as the name says itself, you have to separate the variables, obtaining an equality of two functions with different variables. Then you integrate the expression and you obtain an equality of two integrals. Then you will solve the integrals and find the solution.
<MASK>
Another method to solve differential equation is the exact form method. You know your equation is in an exact form if it has the following form: M(x,y) dx + N(x,y) dy = 0, where M and N are the functions of x and y in such a way that:
A differential equation is called linear as long as the dependent variable and its derivatives occur in the first degree and are not multiplied together.
<MASK>
## Solving differential equations video lesson
Ian Roberts Engineer San Francisco, USA "If you're at school or you just deal with mathematics, you need to use Studygeek.org. This thing is really helpful." Lisa Jordan Math Teacher New-York, USA "I will recommend Studygeek to students, who have some chalenges in mathematics. This Site has bunch of great lessons and examples. " John Maloney Student, Designer Philadelphia, USA " I'm a geek, and I love this website. It really helped me during my math classes. Check it out) " Steve Karpesky Bookkeeper Vancuver, Canada "I use Studygeek.org a lot on a daily basis, helping my son with his geometry classes. Also, it has very cool math solver, which makes study process pretty fun"
|
# How to solve differential equations
Differential equations make use of mathematical operations with derivatives. Solving differential equations is a very important, but also hard concept in calculus. There exist more methods of solving this kind of exercises.
Firstly, we will have a look on how first order differential equation can be solved. Below, you will find a series of methods of solving this kind of differential equations. The first one is the separation method of variables. Practically, as the name says itself, you have to separate the variables, obtaining an equality of two functions with different variables. Then you integrate the expression and you obtain an equality of two integrals. Then you will solve the integrals and find the solution.
<MASK>
Another method to solve differential equation is the exact form method. You know your equation is in an exact form if it has the following form: M(x,y) dx + N(x,y) dy = 0, where M and N are the functions of x and y in such a way that:
A differential equation is called linear as long as the dependent variable and its derivatives occur in the first degree and are not multiplied together.
<MASK>
## Solving differential equations video lesson
Ian Roberts Engineer San Francisco, USA "If you're at school or you just deal with mathematics, you need to use Studygeek.org. This thing is really helpful." Lisa Jordan Math Teacher New-York, USA "I will recommend Studygeek to students, who have some chalenges in mathematics. This Site has bunch of great lessons and examples. " John Maloney Student, Designer Philadelphia, USA " I'm a geek, and I love this website. It really helped me during my math classes. Check it out) " Steve Karpesky Bookkeeper Vancuver, Canada "I use Studygeek.org a lot on a daily basis, helping my son with his geometry classes. Also, it has very cool math solver, which makes study process pretty fun"
<UNMASK>
# How to solve differential equations
Differential equations make use of mathematical operations with derivatives. Solving differential equations is a very important, but also hard concept in calculus. There exist more methods of solving this kind of exercises.
Firstly, we will have a look on how first order differential equation can be solved. Below, you will find a series of methods of solving this kind of differential equations. The first one is the separation method of variables. Practically, as the name says itself, you have to separate the variables, obtaining an equality of two functions with different variables. Then you integrate the expression and you obtain an equality of two integrals. Then you will solve the integrals and find the solution.
For the homogeneous differential equations, we use the substitution method and we reduce the equation to the variable separable. Having an exercise in which you have to solve the differential equation, you firstly have to figure out what kind of differential equation is the equation, so you know what method it's better to use.
Another method to solve differential equation is the exact form method. You know your equation is in an exact form if it has the following form: M(x,y) dx + N(x,y) dy = 0, where M and N are the functions of x and y in such a way that:
A differential equation is called linear as long as the dependent variable and its derivatives occur in the first degree and are not multiplied together.
f to fn are the functions of x.
In order to figure out how to solve differential equation, you firstly have to determine the order of the differential equation. For example, for the second order differential equation there is a more special method of finding the solution: divide the second order differential equation in 2 parts: Q(x)=0 and Q(x) is a function of x. For both members calculate the auxiliary equation and find the complementary function. Next, if Q(x) is a part of the equation, find the particular integral of the equation. In the end, sum up the complementary function with the particular integral.
## Solving differential equations video lesson
Ian Roberts Engineer San Francisco, USA "If you're at school or you just deal with mathematics, you need to use Studygeek.org. This thing is really helpful." Lisa Jordan Math Teacher New-York, USA "I will recommend Studygeek to students, who have some chalenges in mathematics. This Site has bunch of great lessons and examples. " John Maloney Student, Designer Philadelphia, USA " I'm a geek, and I love this website. It really helped me during my math classes. Check it out) " Steve Karpesky Bookkeeper Vancuver, Canada "I use Studygeek.org a lot on a daily basis, helping my son with his geometry classes. Also, it has very cool math solver, which makes study process pretty fun"
|
<MASK>
Below is a radius calculator, which will compute a circle's diameter, circumference, and area if you know the radius.
<MASK>
r=c/2\pi
Area, on the other hand, is all the space contained inside the circle. It's also straightforward to find the area if you know the radius:
a = \pi r^2
You can enter the radius and then compute diameter and circumference in mils, inches, feet, yards, miles, millimeters, centimeters, meters and kilometers.
<MASK>
To run the computations, hit the 'Calculate Circle Dimensions' button when you have entered the known radius.
<MASK>
<UNMASK>
# Radius Calculator: Compute Dimensions of a Circle
Written by:
PK
Below is a radius calculator, which will compute a circle's diameter, circumference, and area if you know the radius.
<MASK>
For a circle, three lengths most commonly are applied:
<MASK>
If you know the radius, it is straightforward to compute the other two.
<MASK>
If you know the circumference it is a bit harder, but not too bad:
r=c/2\pi
Area, on the other hand, is all the space contained inside the circle. It's also straightforward to find the area if you know the radius:
a = \pi r^2
You can enter the radius and then compute diameter and circumference in mils, inches, feet, yards, miles, millimeters, centimeters, meters and kilometers.
<MASK>
To run the computations, hit the 'Calculate Circle Dimensions' button when you have entered the known radius.
<MASK>
|
# Radius Calculator: Compute Dimensions of a Circle
Written by:
PK
Below is a radius calculator, which will compute a circle's diameter, circumference, and area if you know the radius.
<MASK>
For a circle, three lengths most commonly are applied:
<MASK>
If you know the radius, it is straightforward to compute the other two.
<MASK>
If you know the circumference it is a bit harder, but not too bad:
r=c/2\pi
Area, on the other hand, is all the space contained inside the circle. It's also straightforward to find the area if you know the radius:
a = \pi r^2
You can enter the radius and then compute diameter and circumference in mils, inches, feet, yards, miles, millimeters, centimeters, meters and kilometers.
<MASK>
To run the computations, hit the 'Calculate Circle Dimensions' button when you have entered the known radius.
<MASK>
<UNMASK>
# Radius Calculator: Compute Dimensions of a Circle
Written by:
PK
Below is a radius calculator, which will compute a circle's diameter, circumference, and area if you know the radius.
Do you know a different dimension? Instead try one of the other circle dimension calculators:
<MASK>
For a circle, three lengths most commonly are applied:
• The radius – defined above
• The diameter – the distance from edge to edge of a circle passing through its origin or center. Twice the length of a circle's radius
• The circumference – the length of the outside boundaries of the circle
If you know the radius, it is straightforward to compute the other two.
To find the radius from the diameter, you only have to divide by two:
<MASK>
If you know the circumference it is a bit harder, but not too bad:
r=c/2\pi
Area, on the other hand, is all the space contained inside the circle. It's also straightforward to find the area if you know the radius:
a = \pi r^2
You can enter the radius and then compute diameter and circumference in mils, inches, feet, yards, miles, millimeters, centimeters, meters and kilometers.
Area has different units, but you can use: square mils, square inches, square feet, square yards, square miles, acres, hectares, square millimeters, square centimeters, square meters, and square kilometers.
To run the computations, hit the 'Calculate Circle Dimensions' button when you have entered the known radius.
<MASK>
DQYDJ may be compensated by our partners if you make purchases through links. See our disclosures page. As an Amazon Associate we earn from qualifying purchases.
|
# Radius Calculator: Compute Dimensions of a Circle
Written by:
PK
Below is a radius calculator, which will compute a circle's diameter, circumference, and area if you know the radius.
Do you know a different dimension? Instead try one of the other circle dimension calculators:
<MASK>
For a circle, three lengths most commonly are applied:
• The radius – defined above
• The diameter – the distance from edge to edge of a circle passing through its origin or center. Twice the length of a circle's radius
• The circumference – the length of the outside boundaries of the circle
If you know the radius, it is straightforward to compute the other two.
To find the radius from the diameter, you only have to divide by two:
<MASK>
If you know the circumference it is a bit harder, but not too bad:
r=c/2\pi
Area, on the other hand, is all the space contained inside the circle. It's also straightforward to find the area if you know the radius:
a = \pi r^2
You can enter the radius and then compute diameter and circumference in mils, inches, feet, yards, miles, millimeters, centimeters, meters and kilometers.
Area has different units, but you can use: square mils, square inches, square feet, square yards, square miles, acres, hectares, square millimeters, square centimeters, square meters, and square kilometers.
To run the computations, hit the 'Calculate Circle Dimensions' button when you have entered the known radius.
<MASK>
DQYDJ may be compensated by our partners if you make purchases through links. See our disclosures page. As an Amazon Associate we earn from qualifying purchases.
<UNMASK>
# Radius Calculator: Compute Dimensions of a Circle
Written by:
PK
Below is a radius calculator, which will compute a circle's diameter, circumference, and area if you know the radius.
Do you know a different dimension? Instead try one of the other circle dimension calculators:
## What is a Circle's Radius?
<MASK>
## Dimensions of a Circle
For a circle, three lengths most commonly are applied:
• The radius – defined above
• The diameter – the distance from edge to edge of a circle passing through its origin or center. Twice the length of a circle's radius
• The circumference – the length of the outside boundaries of the circle
If you know the radius, it is straightforward to compute the other two.
To find the radius from the diameter, you only have to divide by two:
<MASK>
If you know the circumference it is a bit harder, but not too bad:
r=c/2\pi
Area, on the other hand, is all the space contained inside the circle. It's also straightforward to find the area if you know the radius:
a = \pi r^2
You can enter the radius and then compute diameter and circumference in mils, inches, feet, yards, miles, millimeters, centimeters, meters and kilometers.
Area has different units, but you can use: square mils, square inches, square feet, square yards, square miles, acres, hectares, square millimeters, square centimeters, square meters, and square kilometers.
To run the computations, hit the 'Calculate Circle Dimensions' button when you have entered the known radius.
<MASK>
PK started DQYDJ in 2009 to research and discuss finance and investing and help answer financial questions. He's expanded DQYDJ to build visualizations, calculators, and interactive tools.
<MASK>
DQYDJ may be compensated by our partners if you make purchases through links. See our disclosures page. As an Amazon Associate we earn from qualifying purchases.
|
# Radius Calculator: Compute Dimensions of a Circle
Written by:
PK
Below is a radius calculator, which will compute a circle's diameter, circumference, and area if you know the radius.
Do you know a different dimension? Instead try one of the other circle dimension calculators:
## What is a Circle's Radius?
<MASK>
## Dimensions of a Circle
For a circle, three lengths most commonly are applied:
• The radius – defined above
• The diameter – the distance from edge to edge of a circle passing through its origin or center. Twice the length of a circle's radius
• The circumference – the length of the outside boundaries of the circle
If you know the radius, it is straightforward to compute the other two.
To find the radius from the diameter, you only have to divide by two:
<MASK>
If you know the circumference it is a bit harder, but not too bad:
r=c/2\pi
Area, on the other hand, is all the space contained inside the circle. It's also straightforward to find the area if you know the radius:
a = \pi r^2
You can enter the radius and then compute diameter and circumference in mils, inches, feet, yards, miles, millimeters, centimeters, meters and kilometers.
Area has different units, but you can use: square mils, square inches, square feet, square yards, square miles, acres, hectares, square millimeters, square centimeters, square meters, and square kilometers.
To run the computations, hit the 'Calculate Circle Dimensions' button when you have entered the known radius.
<MASK>
PK started DQYDJ in 2009 to research and discuss finance and investing and help answer financial questions. He's expanded DQYDJ to build visualizations, calculators, and interactive tools.
<MASK>
DQYDJ may be compensated by our partners if you make purchases through links. See our disclosures page. As an Amazon Associate we earn from qualifying purchases.
<UNMASK>
# Radius Calculator: Compute Dimensions of a Circle
Written by:
PK
Below is a radius calculator, which will compute a circle's diameter, circumference, and area if you know the radius.
Do you know a different dimension? Instead try one of the other circle dimension calculators:
## What is a Circle's Radius?
The radius of a circle is the distance from a circle's origin or center to its edge.
Conveniently, it is half as long as the diameter of a circle. A diameter is just two radiuses drawn in opposing directions from the circle's origin.
## Dimensions of a Circle
For a circle, three lengths most commonly are applied:
• The radius – defined above
• The diameter – the distance from edge to edge of a circle passing through its origin or center. Twice the length of a circle's radius
• The circumference – the length of the outside boundaries of the circle
If you know the radius, it is straightforward to compute the other two.
To find the radius from the diameter, you only have to divide by two:
r=d/2
If you know the circumference it is a bit harder, but not too bad:
r=c/2\pi
Area, on the other hand, is all the space contained inside the circle. It's also straightforward to find the area if you know the radius:
a = \pi r^2
You can enter the radius and then compute diameter and circumference in mils, inches, feet, yards, miles, millimeters, centimeters, meters and kilometers.
Area has different units, but you can use: square mils, square inches, square feet, square yards, square miles, acres, hectares, square millimeters, square centimeters, square meters, and square kilometers.
To run the computations, hit the 'Calculate Circle Dimensions' button when you have entered the known radius.
### PK
PK started DQYDJ in 2009 to research and discuss finance and investing and help answer financial questions. He's expanded DQYDJ to build visualizations, calculators, and interactive tools.
PK lives in New Hampshire with his wife, kids, and dog.
### Don't Quit Your Day Job...
DQYDJ may be compensated by our partners if you make purchases through links. See our disclosures page. As an Amazon Associate we earn from qualifying purchases.
|
<MASK>
<UNMASK>
<MASK>
## Questions Asked in IBPS SO Prelims Exam 31 December 2017
The final exam of this year , IBPS SO Prelims Exam 2017 has begun and we know you have been eagerly waiting for the review and analysis of the IBPS SO Prelims Exam 2017. Considering the level of difficulty of the exam, this exam has got a major attraction, since this Specialist Officer Exam is one of the prominent Exam of the year and since this is the last exam of the year, this has caught the major attraction among the other exams.
<MASK>
Q 1. Number Series Questions Asked in IBPS SO Pre Exam 31 Dec 2017
1. 11, 13, 111, 257, 427, ? Answers - 609
2. 38, 51, 25, 64, 12, ? Answers - 77
3. 4, 14, 31, 57, 94, ? Answers - 50
4. 4, 2.5, 3.5, 9, ?, 328 Answers - 40
5. 6, 5, 9, 26, ?, 514 Answers - 103
Q 2. A B C D are four colleges there are total 2000 students in and in college d there are 380 female students and and in college a there are 260 female students and in college a female students are 50 more then male students and in college d male are 20% less then female and the ratio of students in college b and c are 7:13 and in college b and c the no of male students are 35 and 15 less then the avg no of male students ?
Q 3. Root (2025-x)/25 =16
<MASK>
1. Reading Comprehension – 2 Sets, 10 Qs (each consisting of 5 Qs)
– 1 Passage was based on “Scope on Economy”
– Other passage was based on “Steps that should be taken to stop the corruption”
– Meaning of “Narrowing” was asked. Also antonym of “Dearer” was asked in exam.
2. Error Spotting 10 Qs – Asked in a different way.
A sentence was divided into 5 parts. It was given that the last part of the sentence is error free. You have to find error free part of the sentence among the 4 parts.
3. Fill in the blanks – 5 Qs – Vocab Based.
2 Sentences were given in which you had to choose a word which correctly fits in both the sentences.
Eg. i) He is _______ at playing the piano.
ii) Exercise is _______ for health.
<MASK>
2. FIFA U-19 was held at?
3. 1 Qs based on Basel committee?
5. Repo rate?
update soon
## Reasoning Questions Asked in IBPS SO Pre Exam 31 Dec 2017
<MASK>
## IBPS SO Prelims Exam Analysis 31 December 2017 (Shift-2)
We anticipate your Best performance in the exam. Feel free to drop any of your queries/ suggestions in the comments section below. You can also share the questions asked in the exam in the same place. Stay connected for more information regarding IBPS PO Pre Exam Analysis & Questions 22 September 2017. Follow us on www.ibtsindia.com or www.ibtsindia.com/ibtsinstitute.com
|
<MASK>
## Questions Asked in IBPS SO Prelims Exam 31 December 2017
The final exam of this year , IBPS SO Prelims Exam 2017 has begun and we know you have been eagerly waiting for the review and analysis of the IBPS SO Prelims Exam 2017. Considering the level of difficulty of the exam, this exam has got a major attraction, since this Specialist Officer Exam is one of the prominent Exam of the year and since this is the last exam of the year, this has caught the major attraction among the other exams.
<MASK>
Q 1. Number Series Questions Asked in IBPS SO Pre Exam 31 Dec 2017
1. 11, 13, 111, 257, 427, ? Answers - 609
2. 38, 51, 25, 64, 12, ? Answers - 77
3. 4, 14, 31, 57, 94, ? Answers - 50
4. 4, 2.5, 3.5, 9, ?, 328 Answers - 40
5. 6, 5, 9, 26, ?, 514 Answers - 103
Q 2. A B C D are four colleges there are total 2000 students in and in college d there are 380 female students and and in college a there are 260 female students and in college a female students are 50 more then male students and in college d male are 20% less then female and the ratio of students in college b and c are 7:13 and in college b and c the no of male students are 35 and 15 less then the avg no of male students ?
Q 3. Root (2025-x)/25 =16
<MASK>
1. Reading Comprehension – 2 Sets, 10 Qs (each consisting of 5 Qs)
– 1 Passage was based on “Scope on Economy”
– Other passage was based on “Steps that should be taken to stop the corruption”
– Meaning of “Narrowing” was asked. Also antonym of “Dearer” was asked in exam.
2. Error Spotting 10 Qs – Asked in a different way.
A sentence was divided into 5 parts. It was given that the last part of the sentence is error free. You have to find error free part of the sentence among the 4 parts.
3. Fill in the blanks – 5 Qs – Vocab Based.
2 Sentences were given in which you had to choose a word which correctly fits in both the sentences.
Eg. i) He is _______ at playing the piano.
ii) Exercise is _______ for health.
<MASK>
2. FIFA U-19 was held at?
3. 1 Qs based on Basel committee?
5. Repo rate?
update soon
## Reasoning Questions Asked in IBPS SO Pre Exam 31 Dec 2017
<MASK>
## IBPS SO Prelims Exam Analysis 31 December 2017 (Shift-2)
We anticipate your Best performance in the exam. Feel free to drop any of your queries/ suggestions in the comments section below. You can also share the questions asked in the exam in the same place. Stay connected for more information regarding IBPS PO Pre Exam Analysis & Questions 22 September 2017. Follow us on www.ibtsindia.com or www.ibtsindia.com/ibtsinstitute.com
<UNMASK>
<MASK>
## Questions Asked in IBPS SO Prelims Exam 31 December 2017
The final exam of this year , IBPS SO Prelims Exam 2017 has begun and we know you have been eagerly waiting for the review and analysis of the IBPS SO Prelims Exam 2017. Considering the level of difficulty of the exam, this exam has got a major attraction, since this Specialist Officer Exam is one of the prominent Exam of the year and since this is the last exam of the year, this has caught the major attraction among the other exams.
<MASK>
Q 1. Number Series Questions Asked in IBPS SO Pre Exam 31 Dec 2017
1. 11, 13, 111, 257, 427, ? Answers - 609
2. 38, 51, 25, 64, 12, ? Answers - 77
3. 4, 14, 31, 57, 94, ? Answers - 50
4. 4, 2.5, 3.5, 9, ?, 328 Answers - 40
5. 6, 5, 9, 26, ?, 514 Answers - 103
Q 2. A B C D are four colleges there are total 2000 students in and in college d there are 380 female students and and in college a there are 260 female students and in college a female students are 50 more then male students and in college d male are 20% less then female and the ratio of students in college b and c are 7:13 and in college b and c the no of male students are 35 and 15 less then the avg no of male students ?
Q 3. Root (2025-x)/25 =16
<MASK>
1. Reading Comprehension – 2 Sets, 10 Qs (each consisting of 5 Qs)
– 1 Passage was based on “Scope on Economy”
– Other passage was based on “Steps that should be taken to stop the corruption”
– Meaning of “Narrowing” was asked. Also antonym of “Dearer” was asked in exam.
2. Error Spotting 10 Qs – Asked in a different way.
A sentence was divided into 5 parts. It was given that the last part of the sentence is error free. You have to find error free part of the sentence among the 4 parts.
3. Fill in the blanks – 5 Qs – Vocab Based.
2 Sentences were given in which you had to choose a word which correctly fits in both the sentences.
Eg. i) He is _______ at playing the piano.
ii) Exercise is _______ for health.
<MASK>
2. FIFA U-19 was held at?
3. 1 Qs based on Basel committee?
5. Repo rate?
update soon
## Reasoning Questions Asked in IBPS SO Pre Exam 31 Dec 2017
<MASK>
## English Questions Asked in IBPS SO Pre Exam 31 Dec 2017
update soon
## IBPS SO Prelims Exam Analysis 31 December 2017 (Shift-2)
We anticipate your Best performance in the exam. Feel free to drop any of your queries/ suggestions in the comments section below. You can also share the questions asked in the exam in the same place. Stay connected for more information regarding IBPS PO Pre Exam Analysis & Questions 22 September 2017. Follow us on www.ibtsindia.com or www.ibtsindia.com/ibtsinstitute.com
|
<MASK>
## Questions Asked in IBPS SO Prelims Exam 31 December 2017
The final exam of this year , IBPS SO Prelims Exam 2017 has begun and we know you have been eagerly waiting for the review and analysis of the IBPS SO Prelims Exam 2017. Considering the level of difficulty of the exam, this exam has got a major attraction, since this Specialist Officer Exam is one of the prominent Exam of the year and since this is the last exam of the year, this has caught the major attraction among the other exams.
<MASK>
Q 1. Number Series Questions Asked in IBPS SO Pre Exam 31 Dec 2017
1. 11, 13, 111, 257, 427, ? Answers - 609
2. 38, 51, 25, 64, 12, ? Answers - 77
3. 4, 14, 31, 57, 94, ? Answers - 50
4. 4, 2.5, 3.5, 9, ?, 328 Answers - 40
5. 6, 5, 9, 26, ?, 514 Answers - 103
Q 2. A B C D are four colleges there are total 2000 students in and in college d there are 380 female students and and in college a there are 260 female students and in college a female students are 50 more then male students and in college d male are 20% less then female and the ratio of students in college b and c are 7:13 and in college b and c the no of male students are 35 and 15 less then the avg no of male students ?
Q 3. Root (2025-x)/25 =16
<MASK>
1. Reading Comprehension – 2 Sets, 10 Qs (each consisting of 5 Qs)
– 1 Passage was based on “Scope on Economy”
– Other passage was based on “Steps that should be taken to stop the corruption”
– Meaning of “Narrowing” was asked. Also antonym of “Dearer” was asked in exam.
2. Error Spotting 10 Qs – Asked in a different way.
A sentence was divided into 5 parts. It was given that the last part of the sentence is error free. You have to find error free part of the sentence among the 4 parts.
3. Fill in the blanks – 5 Qs – Vocab Based.
2 Sentences were given in which you had to choose a word which correctly fits in both the sentences.
Eg. i) He is _______ at playing the piano.
ii) Exercise is _______ for health.
<MASK>
2. FIFA U-19 was held at?
3. 1 Qs based on Basel committee?
5. Repo rate?
update soon
## Reasoning Questions Asked in IBPS SO Pre Exam 31 Dec 2017
<MASK>
## English Questions Asked in IBPS SO Pre Exam 31 Dec 2017
update soon
## IBPS SO Prelims Exam Analysis 31 December 2017 (Shift-2)
We anticipate your Best performance in the exam. Feel free to drop any of your queries/ suggestions in the comments section below. You can also share the questions asked in the exam in the same place. Stay connected for more information regarding IBPS PO Pre Exam Analysis & Questions 22 September 2017. Follow us on www.ibtsindia.com or www.ibtsindia.com/ibtsinstitute.com
<UNMASK>
<MASK>
## Questions Asked in IBPS SO Prelims Exam 31 December 2017
The final exam of this year , IBPS SO Prelims Exam 2017 has begun and we know you have been eagerly waiting for the review and analysis of the IBPS SO Prelims Exam 2017. Considering the level of difficulty of the exam, this exam has got a major attraction, since this Specialist Officer Exam is one of the prominent Exam of the year and since this is the last exam of the year, this has caught the major attraction among the other exams.
Here is the detailed Questions Asked in IBPS SO Prelims Exam 31 December 2017. Read today’s Exam Analysis along with this article to know what kind of questions are asked in today’s exam. This will be very beneficial for you if you haven’t appeared for your exam yet. Even if you did appear for the exam already, you can analyze your performance by having a look at the analysis published.
<MASK>
Q 1. Number Series Questions Asked in IBPS SO Pre Exam 31 Dec 2017
1. 11, 13, 111, 257, 427, ? Answers - 609
2. 38, 51, 25, 64, 12, ? Answers - 77
3. 4, 14, 31, 57, 94, ? Answers - 50
4. 4, 2.5, 3.5, 9, ?, 328 Answers - 40
5. 6, 5, 9, 26, ?, 514 Answers - 103
Q 2. A B C D are four colleges there are total 2000 students in and in college d there are 380 female students and and in college a there are 260 female students and in college a female students are 50 more then male students and in college d male are 20% less then female and the ratio of students in college b and c are 7:13 and in college b and c the no of male students are 35 and 15 less then the avg no of male students ?
Q 3. Root (2025-x)/25 =16
## Reasoning Questions Asked in IBPS SO Pre Exam 31 Dec 2017
Directions (1 – 5): Answer the questions on the basis of the information given below.
Eight friends A, B, C, D, E, F, G and H are seating around a circular table. Some of them are facing inside & others are facing outside. Opposite direction means if one is facing inside the centre, second one face outside the centre, and vice versa.
Three people are seating between F and D and both face opposite direction to each other. E is seating second to right of both D and F, and face opposite direction as F faces. C is seating third to right of E, who is not opposite to B. G is neighbour of both E and D, and seating second to right of B, who is not a neighbour of E. G is third to left of A, who is third to right of both H and G. B face towards the centre and is seating second to left of C.
1. If all the friends are seating according to alphabetical order in anti clock wise direction, starting from A, how many friends remain at same position (excluding A)?
A) One
B) Two
C) Three
D) Four
E) None
2. If D interchanged his position with H and, F interchanged his position with G, who sits immediately right of H in new arrangement?
A) F
B) G
C) E
D) A
E) None
3. Who is seating third to left of B?
A) H
B) C
C) A
D) F
E) None
4. How many persons are facing away from the centre?
A) Three
B) Four
C) Five
D) Two
E) None
5. Who among the following pairs are facing same direction and seating opposite to each other ?
A) H, B
B) C, G
C) F, D
D) E, A
E) None
Directions (6 – 10): Answer the questions on the basis of the information given below.
Ten friends are sitting on twelve seats in two parallel rows containing five people each, in such a way that there is an equal distance between adjacent persons. In Row 1: A, B, C, D and E are seated and all of them are facing south, and in Row 2: P, Q, R, S and T are sitting and all of them are facing north. One seat is vacant in each row. Therefore, in the given seating arrangement each member seated in a row faces another member of the other row.
All of them like different colors – Red, Green, Black, Yellow, White, Blue, Brown, Purple, Pink and Grey, but not necessarily in the same order.
There are two seats between Q and the vacant seat. Q does not like White, Red and Purple. E is not an immediate neighbor of C. B likes Grey. Vacant seat of row 1 is not opposite to S and is also not at any of the extreme ends of Row-1.The one who likes Black sits opposite to the one, who sits third to the right of the seat, which is opposite to S. C is not an immediate neighbor of D. T, who likes neither White nor Blue, does not face vacant seat. D faces R. The vacant seats are not opposite to each other. Two seats are there between C and B, who sits third right of the seat, on which the person who likes Brown is sitting. S sits third to the right of seat on which R sits and likes Yellow. The one who likes Pink faces the one who likes Yellow. The persons who like Red and Purple are adjacent to each other. The vacant seat in row 1 is not adjacent to D.Q sits at one of the extreme ends. E neither likes Pink nor faces the seat which is adjacent to the one who likes Blue. The one who likes White is not to the immediate right of the one who likes Yellow. The person who likes Green doesn’t face the person who likes Purple.
6. How many persons are sitting between T and the one who likes yellow color?
A) None
B) One
C) Two
D) Three
E) None of these
7. Which of the following faces the vacant seat of Row – 2?
A) The one who like white color
B) A
C) D
D) The one who likes grey color
E) Cannot be determined
8. Who is sitting at the immediate left of person who likes purple color?
A) E
B) D
C) The one who likes black color
D) The one who likes green color
E) The one who likes grey color
9. Who amongst the following sits at the extreme end of the row?
A) R, Q
B) E, S
C) T, C
D) C, D
E) None of these
10. If Q is made to sit on vacant seat of his row, then how many persons are there between the persons who sit opposite to Q now and who sat opposite to Q previously?
A) Two
B) Three
C) Four
D) None
E) One
Directions (Q. 11–15): Study the following arrangement of series carefully and answer the questions given below:
C # E N 4 \$ F 3 I L 8 @ G © P O V 5 2 A X % J 9 * W K 6 Z 7&2 S
11. How many such symbols are there in the above arrangement each of which is either immediately preceded by a letter or immediately followed by a letter but not both?
A) None
B) Three
C) One
D) More than three
E) Two
12. If all the symbols in the above arrangement are dropped which of the following will be twelfth from the left end?
A) 8
B) 2
C) A
D) Other than given options
E) O
13. How many such numbers are there in the above arrangements each of which is immediately followed by a consonant but not immediately preceded by a letter?
A) None
B) Two
C) One
D) Three
E) More than three
14. Four of the following five are alike in a certain way based on their positions in the above arrangement and so form a group. Which is the one that does not belong to this group?
A) 6 &*
B) 5 X O
C) F L 4
D) G O 8
E) 9 K %
15. Which of the following is the seventh to the right of the eighteenth from the right end of the above arrangement?
A) I
B) Other than given options
C) 8
D) J
E) *
<MASK>
## English Questions Asked in IBPS SO Pre Exam 31 Dec 2017
1. Reading Comprehension – 2 Sets, 10 Qs (each consisting of 5 Qs)
– 1 Passage was based on “Scope on Economy”
– Other passage was based on “Steps that should be taken to stop the corruption”
– Meaning of “Narrowing” was asked. Also antonym of “Dearer” was asked in exam.
2. Error Spotting 10 Qs – Asked in a different way.
A sentence was divided into 5 parts. It was given that the last part of the sentence is error free. You have to find error free part of the sentence among the 4 parts.
3. Fill in the blanks – 5 Qs – Vocab Based.
2 Sentences were given in which you had to choose a word which correctly fits in both the sentences.
Eg. i) He is _______ at playing the piano.
ii) Exercise is _______ for health.
## General Awareness Asked in IBPS SO Pre Exam 31 Dec 2017
2. FIFA U-19 was held at?
3. 1 Qs based on Basel committee?
5. Repo rate?
update soon
## Reasoning Questions Asked in IBPS SO Pre Exam 31 Dec 2017
<MASK>
## English Questions Asked in IBPS SO Pre Exam 31 Dec 2017
update soon
## IBPS SO Prelims Exam Analysis 31 December 2017 (Shift-2)
We anticipate your Best performance in the exam. Feel free to drop any of your queries/ suggestions in the comments section below. You can also share the questions asked in the exam in the same place. Stay connected for more information regarding IBPS PO Pre Exam Analysis & Questions 22 September 2017. Follow us on www.ibtsindia.com or www.ibtsindia.com/ibtsinstitute.com
|
<MASK>
## Questions Asked in IBPS SO Prelims Exam 31 December 2017
The final exam of this year , IBPS SO Prelims Exam 2017 has begun and we know you have been eagerly waiting for the review and analysis of the IBPS SO Prelims Exam 2017. Considering the level of difficulty of the exam, this exam has got a major attraction, since this Specialist Officer Exam is one of the prominent Exam of the year and since this is the last exam of the year, this has caught the major attraction among the other exams.
Here is the detailed Questions Asked in IBPS SO Prelims Exam 31 December 2017. Read today’s Exam Analysis along with this article to know what kind of questions are asked in today’s exam. This will be very beneficial for you if you haven’t appeared for your exam yet. Even if you did appear for the exam already, you can analyze your performance by having a look at the analysis published.
<MASK>
Q 1. Number Series Questions Asked in IBPS SO Pre Exam 31 Dec 2017
1. 11, 13, 111, 257, 427, ? Answers - 609
2. 38, 51, 25, 64, 12, ? Answers - 77
3. 4, 14, 31, 57, 94, ? Answers - 50
4. 4, 2.5, 3.5, 9, ?, 328 Answers - 40
5. 6, 5, 9, 26, ?, 514 Answers - 103
Q 2. A B C D are four colleges there are total 2000 students in and in college d there are 380 female students and and in college a there are 260 female students and in college a female students are 50 more then male students and in college d male are 20% less then female and the ratio of students in college b and c are 7:13 and in college b and c the no of male students are 35 and 15 less then the avg no of male students ?
Q 3. Root (2025-x)/25 =16
## Reasoning Questions Asked in IBPS SO Pre Exam 31 Dec 2017
Directions (1 – 5): Answer the questions on the basis of the information given below.
Eight friends A, B, C, D, E, F, G and H are seating around a circular table. Some of them are facing inside & others are facing outside. Opposite direction means if one is facing inside the centre, second one face outside the centre, and vice versa.
Three people are seating between F and D and both face opposite direction to each other. E is seating second to right of both D and F, and face opposite direction as F faces. C is seating third to right of E, who is not opposite to B. G is neighbour of both E and D, and seating second to right of B, who is not a neighbour of E. G is third to left of A, who is third to right of both H and G. B face towards the centre and is seating second to left of C.
1. If all the friends are seating according to alphabetical order in anti clock wise direction, starting from A, how many friends remain at same position (excluding A)?
A) One
B) Two
C) Three
D) Four
E) None
2. If D interchanged his position with H and, F interchanged his position with G, who sits immediately right of H in new arrangement?
A) F
B) G
C) E
D) A
E) None
3. Who is seating third to left of B?
A) H
B) C
C) A
D) F
E) None
4. How many persons are facing away from the centre?
A) Three
B) Four
C) Five
D) Two
E) None
5. Who among the following pairs are facing same direction and seating opposite to each other ?
A) H, B
B) C, G
C) F, D
D) E, A
E) None
Directions (6 – 10): Answer the questions on the basis of the information given below.
Ten friends are sitting on twelve seats in two parallel rows containing five people each, in such a way that there is an equal distance between adjacent persons. In Row 1: A, B, C, D and E are seated and all of them are facing south, and in Row 2: P, Q, R, S and T are sitting and all of them are facing north. One seat is vacant in each row. Therefore, in the given seating arrangement each member seated in a row faces another member of the other row.
All of them like different colors – Red, Green, Black, Yellow, White, Blue, Brown, Purple, Pink and Grey, but not necessarily in the same order.
There are two seats between Q and the vacant seat. Q does not like White, Red and Purple. E is not an immediate neighbor of C. B likes Grey. Vacant seat of row 1 is not opposite to S and is also not at any of the extreme ends of Row-1.The one who likes Black sits opposite to the one, who sits third to the right of the seat, which is opposite to S. C is not an immediate neighbor of D. T, who likes neither White nor Blue, does not face vacant seat. D faces R. The vacant seats are not opposite to each other. Two seats are there between C and B, who sits third right of the seat, on which the person who likes Brown is sitting. S sits third to the right of seat on which R sits and likes Yellow. The one who likes Pink faces the one who likes Yellow. The persons who like Red and Purple are adjacent to each other. The vacant seat in row 1 is not adjacent to D.Q sits at one of the extreme ends. E neither likes Pink nor faces the seat which is adjacent to the one who likes Blue. The one who likes White is not to the immediate right of the one who likes Yellow. The person who likes Green doesn’t face the person who likes Purple.
6. How many persons are sitting between T and the one who likes yellow color?
A) None
B) One
C) Two
D) Three
E) None of these
7. Which of the following faces the vacant seat of Row – 2?
A) The one who like white color
B) A
C) D
D) The one who likes grey color
E) Cannot be determined
8. Who is sitting at the immediate left of person who likes purple color?
A) E
B) D
C) The one who likes black color
D) The one who likes green color
E) The one who likes grey color
9. Who amongst the following sits at the extreme end of the row?
A) R, Q
B) E, S
C) T, C
D) C, D
E) None of these
10. If Q is made to sit on vacant seat of his row, then how many persons are there between the persons who sit opposite to Q now and who sat opposite to Q previously?
A) Two
B) Three
C) Four
D) None
E) One
Directions (Q. 11–15): Study the following arrangement of series carefully and answer the questions given below:
C # E N 4 \$ F 3 I L 8 @ G © P O V 5 2 A X % J 9 * W K 6 Z 7&2 S
11. How many such symbols are there in the above arrangement each of which is either immediately preceded by a letter or immediately followed by a letter but not both?
A) None
B) Three
C) One
D) More than three
E) Two
12. If all the symbols in the above arrangement are dropped which of the following will be twelfth from the left end?
A) 8
B) 2
C) A
D) Other than given options
E) O
13. How many such numbers are there in the above arrangements each of which is immediately followed by a consonant but not immediately preceded by a letter?
A) None
B) Two
C) One
D) Three
E) More than three
14. Four of the following five are alike in a certain way based on their positions in the above arrangement and so form a group. Which is the one that does not belong to this group?
A) 6 &*
B) 5 X O
C) F L 4
D) G O 8
E) 9 K %
15. Which of the following is the seventh to the right of the eighteenth from the right end of the above arrangement?
A) I
B) Other than given options
C) 8
D) J
E) *
<MASK>
## English Questions Asked in IBPS SO Pre Exam 31 Dec 2017
1. Reading Comprehension – 2 Sets, 10 Qs (each consisting of 5 Qs)
– 1 Passage was based on “Scope on Economy”
– Other passage was based on “Steps that should be taken to stop the corruption”
– Meaning of “Narrowing” was asked. Also antonym of “Dearer” was asked in exam.
2. Error Spotting 10 Qs – Asked in a different way.
A sentence was divided into 5 parts. It was given that the last part of the sentence is error free. You have to find error free part of the sentence among the 4 parts.
3. Fill in the blanks – 5 Qs – Vocab Based.
2 Sentences were given in which you had to choose a word which correctly fits in both the sentences.
Eg. i) He is _______ at playing the piano.
ii) Exercise is _______ for health.
## General Awareness Asked in IBPS SO Pre Exam 31 Dec 2017
2. FIFA U-19 was held at?
3. 1 Qs based on Basel committee?
5. Repo rate?
update soon
## Reasoning Questions Asked in IBPS SO Pre Exam 31 Dec 2017
<MASK>
## English Questions Asked in IBPS SO Pre Exam 31 Dec 2017
update soon
## IBPS SO Prelims Exam Analysis 31 December 2017 (Shift-2)
We anticipate your Best performance in the exam. Feel free to drop any of your queries/ suggestions in the comments section below. You can also share the questions asked in the exam in the same place. Stay connected for more information regarding IBPS PO Pre Exam Analysis & Questions 22 September 2017. Follow us on www.ibtsindia.com or www.ibtsindia.com/ibtsinstitute.com
<UNMASK>
# Questions Asked in IBPS SO Prelims Exam 31 December 2017
## Questions Asked in IBPS SO Prelims Exam 31 December 2017
The final exam of this year , IBPS SO Prelims Exam 2017 has begun and we know you have been eagerly waiting for the review and analysis of the IBPS SO Prelims Exam 2017. Considering the level of difficulty of the exam, this exam has got a major attraction, since this Specialist Officer Exam is one of the prominent Exam of the year and since this is the last exam of the year, this has caught the major attraction among the other exams.
Here is the detailed Questions Asked in IBPS SO Prelims Exam 31 December 2017. Read today’s Exam Analysis along with this article to know what kind of questions are asked in today’s exam. This will be very beneficial for you if you haven’t appeared for your exam yet. Even if you did appear for the exam already, you can analyze your performance by having a look at the analysis published.
## Quant Questions Asked in IBPS SO Pre Exam 31 Dec 2017
Q 1. Number Series Questions Asked in IBPS SO Pre Exam 31 Dec 2017
1. 11, 13, 111, 257, 427, ? Answers - 609
2. 38, 51, 25, 64, 12, ? Answers - 77
3. 4, 14, 31, 57, 94, ? Answers - 50
4. 4, 2.5, 3.5, 9, ?, 328 Answers - 40
5. 6, 5, 9, 26, ?, 514 Answers - 103
Q 2. A B C D are four colleges there are total 2000 students in and in college d there are 380 female students and and in college a there are 260 female students and in college a female students are 50 more then male students and in college d male are 20% less then female and the ratio of students in college b and c are 7:13 and in college b and c the no of male students are 35 and 15 less then the avg no of male students ?
Q 3. Root (2025-x)/25 =16
## Reasoning Questions Asked in IBPS SO Pre Exam 31 Dec 2017
Directions (1 – 5): Answer the questions on the basis of the information given below.
Eight friends A, B, C, D, E, F, G and H are seating around a circular table. Some of them are facing inside & others are facing outside. Opposite direction means if one is facing inside the centre, second one face outside the centre, and vice versa.
Three people are seating between F and D and both face opposite direction to each other. E is seating second to right of both D and F, and face opposite direction as F faces. C is seating third to right of E, who is not opposite to B. G is neighbour of both E and D, and seating second to right of B, who is not a neighbour of E. G is third to left of A, who is third to right of both H and G. B face towards the centre and is seating second to left of C.
1. If all the friends are seating according to alphabetical order in anti clock wise direction, starting from A, how many friends remain at same position (excluding A)?
A) One
B) Two
C) Three
D) Four
E) None
2. If D interchanged his position with H and, F interchanged his position with G, who sits immediately right of H in new arrangement?
A) F
B) G
C) E
D) A
E) None
3. Who is seating third to left of B?
A) H
B) C
C) A
D) F
E) None
4. How many persons are facing away from the centre?
A) Three
B) Four
C) Five
D) Two
E) None
5. Who among the following pairs are facing same direction and seating opposite to each other ?
A) H, B
B) C, G
C) F, D
D) E, A
E) None
Directions (6 – 10): Answer the questions on the basis of the information given below.
Ten friends are sitting on twelve seats in two parallel rows containing five people each, in such a way that there is an equal distance between adjacent persons. In Row 1: A, B, C, D and E are seated and all of them are facing south, and in Row 2: P, Q, R, S and T are sitting and all of them are facing north. One seat is vacant in each row. Therefore, in the given seating arrangement each member seated in a row faces another member of the other row.
All of them like different colors – Red, Green, Black, Yellow, White, Blue, Brown, Purple, Pink and Grey, but not necessarily in the same order.
There are two seats between Q and the vacant seat. Q does not like White, Red and Purple. E is not an immediate neighbor of C. B likes Grey. Vacant seat of row 1 is not opposite to S and is also not at any of the extreme ends of Row-1.The one who likes Black sits opposite to the one, who sits third to the right of the seat, which is opposite to S. C is not an immediate neighbor of D. T, who likes neither White nor Blue, does not face vacant seat. D faces R. The vacant seats are not opposite to each other. Two seats are there between C and B, who sits third right of the seat, on which the person who likes Brown is sitting. S sits third to the right of seat on which R sits and likes Yellow. The one who likes Pink faces the one who likes Yellow. The persons who like Red and Purple are adjacent to each other. The vacant seat in row 1 is not adjacent to D.Q sits at one of the extreme ends. E neither likes Pink nor faces the seat which is adjacent to the one who likes Blue. The one who likes White is not to the immediate right of the one who likes Yellow. The person who likes Green doesn’t face the person who likes Purple.
6. How many persons are sitting between T and the one who likes yellow color?
A) None
B) One
C) Two
D) Three
E) None of these
7. Which of the following faces the vacant seat of Row – 2?
A) The one who like white color
B) A
C) D
D) The one who likes grey color
E) Cannot be determined
8. Who is sitting at the immediate left of person who likes purple color?
A) E
B) D
C) The one who likes black color
D) The one who likes green color
E) The one who likes grey color
9. Who amongst the following sits at the extreme end of the row?
A) R, Q
B) E, S
C) T, C
D) C, D
E) None of these
10. If Q is made to sit on vacant seat of his row, then how many persons are there between the persons who sit opposite to Q now and who sat opposite to Q previously?
A) Two
B) Three
C) Four
D) None
E) One
Directions (Q. 11–15): Study the following arrangement of series carefully and answer the questions given below:
C # E N 4 \$ F 3 I L 8 @ G © P O V 5 2 A X % J 9 * W K 6 Z 7&2 S
11. How many such symbols are there in the above arrangement each of which is either immediately preceded by a letter or immediately followed by a letter but not both?
A) None
B) Three
C) One
D) More than three
E) Two
12. If all the symbols in the above arrangement are dropped which of the following will be twelfth from the left end?
A) 8
B) 2
C) A
D) Other than given options
E) O
13. How many such numbers are there in the above arrangements each of which is immediately followed by a consonant but not immediately preceded by a letter?
A) None
B) Two
C) One
D) Three
E) More than three
14. Four of the following five are alike in a certain way based on their positions in the above arrangement and so form a group. Which is the one that does not belong to this group?
A) 6 &*
B) 5 X O
C) F L 4
D) G O 8
E) 9 K %
15. Which of the following is the seventh to the right of the eighteenth from the right end of the above arrangement?
A) I
B) Other than given options
C) 8
D) J
E) *
1. B
2. A
3. D
4. B
5. A
6. C
7. D
8. E
9. C
10. E
11. B
12. E
13. C
14. A
15. D
## English Questions Asked in IBPS SO Pre Exam 31 Dec 2017
1. Reading Comprehension – 2 Sets, 10 Qs (each consisting of 5 Qs)
– 1 Passage was based on “Scope on Economy”
– Other passage was based on “Steps that should be taken to stop the corruption”
– Meaning of “Narrowing” was asked. Also antonym of “Dearer” was asked in exam.
2. Error Spotting 10 Qs – Asked in a different way.
A sentence was divided into 5 parts. It was given that the last part of the sentence is error free. You have to find error free part of the sentence among the 4 parts.
3. Fill in the blanks – 5 Qs – Vocab Based.
2 Sentences were given in which you had to choose a word which correctly fits in both the sentences.
Eg. i) He is _______ at playing the piano.
ii) Exercise is _______ for health.
## General Awareness Asked in IBPS SO Pre Exam 31 Dec 2017
2. FIFA U-19 was held at?
3. 1 Qs based on Basel committee?
5. Repo rate?
update soon
## Reasoning Questions Asked in IBPS SO Pre Exam 31 Dec 2017
update soon
## English Questions Asked in IBPS SO Pre Exam 31 Dec 2017
update soon
## IBPS SO Prelims Exam Analysis 31 December 2017 (Shift-2)
We anticipate your Best performance in the exam. Feel free to drop any of your queries/ suggestions in the comments section below. You can also share the questions asked in the exam in the same place. Stay connected for more information regarding IBPS PO Pre Exam Analysis & Questions 22 September 2017. Follow us on www.ibtsindia.com or www.ibtsindia.com/ibtsinstitute.com
|
# 711 US dry barrels in dry quarts
<MASK>
It can also be expressed as: 711 US dry barrels is equal to $\frac{1}{\mathrm{1.339507468482e-05}}$ dry quarts.
<MASK>
[1] The precision is 15 significant digits (fourteen digits to the right of the decimal point).
<MASK>
<UNMASK>
# 711 US dry barrels in dry quarts
<MASK>
It can also be expressed as: 711 US dry barrels is equal to $\frac{1}{\mathrm{1.339507468482e-05}}$ dry quarts.
<MASK>
An approximate numerical result would be: seven hundred and eleven US dry barrels is about seventy-four thousand, six hundred and fifty-four point three zero dry quarts, or alternatively, a dry quart is about zero times seven hundred and eleven US dry barrels.
<MASK>
[1] The precision is 15 significant digits (fourteen digits to the right of the decimal point).
<MASK>
|
# 711 US dry barrels in dry quarts
<MASK>
It can also be expressed as: 711 US dry barrels is equal to $\frac{1}{\mathrm{1.339507468482e-05}}$ dry quarts.
<MASK>
An approximate numerical result would be: seven hundred and eleven US dry barrels is about seventy-four thousand, six hundred and fifty-four point three zero dry quarts, or alternatively, a dry quart is about zero times seven hundred and eleven US dry barrels.
<MASK>
[1] The precision is 15 significant digits (fourteen digits to the right of the decimal point).
<MASK>
<UNMASK>
# 711 US dry barrels in dry quarts
<MASK>
## Conversion formula How to convert 711 US dry barrels to dry quarts?
<MASK>
We can set up a proportion to solve for the number of dry quarts.
$1 drybarrel 711 drybarrel ≈ 104.999023446583 dryquart x dryquart$
<MASK>
$x\mathrm{dryquart}\approx \frac{711\mathrm{drybarrel}}{1\mathrm{drybarrel}}*104.999023446583\mathrm{dryquart}\to x\mathrm{dryquart}\approx 74654.30567052051\mathrm{dryquart}$
<MASK>
It can also be expressed as: 711 US dry barrels is equal to $\frac{1}{\mathrm{1.339507468482e-05}}$ dry quarts.
<MASK>
An approximate numerical result would be: seven hundred and eleven US dry barrels is about seventy-four thousand, six hundred and fifty-four point three zero dry quarts, or alternatively, a dry quart is about zero times seven hundred and eleven US dry barrels.
<MASK>
[1] The precision is 15 significant digits (fourteen digits to the right of the decimal point).
Results may contain small errors due to the use of floating point arithmetic.
<MASK>
|
# 711 US dry barrels in dry quarts
<MASK>
## Conversion formula How to convert 711 US dry barrels to dry quarts?
<MASK>
We can set up a proportion to solve for the number of dry quarts.
$1 drybarrel 711 drybarrel ≈ 104.999023446583 dryquart x dryquart$
<MASK>
$x\mathrm{dryquart}\approx \frac{711\mathrm{drybarrel}}{1\mathrm{drybarrel}}*104.999023446583\mathrm{dryquart}\to x\mathrm{dryquart}\approx 74654.30567052051\mathrm{dryquart}$
<MASK>
It can also be expressed as: 711 US dry barrels is equal to $\frac{1}{\mathrm{1.339507468482e-05}}$ dry quarts.
<MASK>
An approximate numerical result would be: seven hundred and eleven US dry barrels is about seventy-four thousand, six hundred and fifty-four point three zero dry quarts, or alternatively, a dry quart is about zero times seven hundred and eleven US dry barrels.
<MASK>
[1] The precision is 15 significant digits (fourteen digits to the right of the decimal point).
Results may contain small errors due to the use of floating point arithmetic.
<MASK>
<UNMASK>
# 711 US dry barrels in dry quarts
## Conversion
<MASK>
## Conversion formula How to convert 711 US dry barrels to dry quarts?
<MASK>
We can set up a proportion to solve for the number of dry quarts.
$1 drybarrel 711 drybarrel ≈ 104.999023446583 dryquart x dryquart$
Now, we cross multiply to solve for our unknown $x$:
$x\mathrm{dryquart}\approx \frac{711\mathrm{drybarrel}}{1\mathrm{drybarrel}}*104.999023446583\mathrm{dryquart}\to x\mathrm{dryquart}\approx 74654.30567052051\mathrm{dryquart}$
Conclusion: $711 drybarrel ≈ 74654.30567052051 dryquart$
<MASK>
The inverse of the conversion factor is that 1 dry quart is equal to 1.339507468482e-05 times 711 US dry barrels.
It can also be expressed as: 711 US dry barrels is equal to $\frac{1}{\mathrm{1.339507468482e-05}}$ dry quarts.
## Approximation
An approximate numerical result would be: seven hundred and eleven US dry barrels is about seventy-four thousand, six hundred and fifty-four point three zero dry quarts, or alternatively, a dry quart is about zero times seven hundred and eleven US dry barrels.
## Footnotes
[1] The precision is 15 significant digits (fourteen digits to the right of the decimal point).
Results may contain small errors due to the use of floating point arithmetic.
<MASK>
|
# 711 US dry barrels in dry quarts
## Conversion
<MASK>
## Conversion formula How to convert 711 US dry barrels to dry quarts?
<MASK>
We can set up a proportion to solve for the number of dry quarts.
$1 drybarrel 711 drybarrel ≈ 104.999023446583 dryquart x dryquart$
Now, we cross multiply to solve for our unknown $x$:
$x\mathrm{dryquart}\approx \frac{711\mathrm{drybarrel}}{1\mathrm{drybarrel}}*104.999023446583\mathrm{dryquart}\to x\mathrm{dryquart}\approx 74654.30567052051\mathrm{dryquart}$
Conclusion: $711 drybarrel ≈ 74654.30567052051 dryquart$
<MASK>
The inverse of the conversion factor is that 1 dry quart is equal to 1.339507468482e-05 times 711 US dry barrels.
It can also be expressed as: 711 US dry barrels is equal to $\frac{1}{\mathrm{1.339507468482e-05}}$ dry quarts.
## Approximation
An approximate numerical result would be: seven hundred and eleven US dry barrels is about seventy-four thousand, six hundred and fifty-four point three zero dry quarts, or alternatively, a dry quart is about zero times seven hundred and eleven US dry barrels.
## Footnotes
[1] The precision is 15 significant digits (fourteen digits to the right of the decimal point).
Results may contain small errors due to the use of floating point arithmetic.
<MASK>
<UNMASK>
# 711 US dry barrels in dry quarts
## Conversion
711 US dry barrels is equivalent to 74654.3056705202 dry quarts.[1]
## Conversion formula How to convert 711 US dry barrels to dry quarts?
We know (by definition) that: $1\mathrm{drybarrel}\approx 104.999023446583\mathrm{dryquart}$
We can set up a proportion to solve for the number of dry quarts.
$1 drybarrel 711 drybarrel ≈ 104.999023446583 dryquart x dryquart$
Now, we cross multiply to solve for our unknown $x$:
$x\mathrm{dryquart}\approx \frac{711\mathrm{drybarrel}}{1\mathrm{drybarrel}}*104.999023446583\mathrm{dryquart}\to x\mathrm{dryquart}\approx 74654.30567052051\mathrm{dryquart}$
Conclusion: $711 drybarrel ≈ 74654.30567052051 dryquart$
## Conversion in the opposite direction
The inverse of the conversion factor is that 1 dry quart is equal to 1.339507468482e-05 times 711 US dry barrels.
It can also be expressed as: 711 US dry barrels is equal to $\frac{1}{\mathrm{1.339507468482e-05}}$ dry quarts.
## Approximation
An approximate numerical result would be: seven hundred and eleven US dry barrels is about seventy-four thousand, six hundred and fifty-four point three zero dry quarts, or alternatively, a dry quart is about zero times seven hundred and eleven US dry barrels.
## Footnotes
[1] The precision is 15 significant digits (fourteen digits to the right of the decimal point).
Results may contain small errors due to the use of floating point arithmetic.
Was it helpful? Share it!
|
Introduction Astronomy Tools Concepts 1. Electromagnetic Spectrum 2. Atmosphere Limitations 3. Space Observations Equipment 1. Telescopes 2. Radio 3. Space Tools 4. Photography 5. Spectroscopy 6. Computers 7. Advanced Methods 8. Radio Astronomy Basic Mathematics Algebra Statistics Geometry Scientific Notation Log Scales Calculus Physics Concepts - Basic Units of Measure - Mass & Density - Temperature - Velocity & Acceleration - Force, Pressure & Energy - Atoms - Quantum Physics - Nature of Light Formulas - Brightness - Cepheid Rulers - Distance - Doppler Shift - Frequency & Wavelength - Hubble's Law - Inverse Square Law - Kinetic Energy - Luminosity - Magnitudes - Convert Mass to Energy - Kepler & Newton - Orbits - Parallax - Planck's Law - Relativistic Redshift - Relativity - Schwarzschild Radius - Synodic & Sidereal Periods - Sidereal Time - Small Angle Formula - Stellar Properties - Stephan-Boltzmann Law - Telescope Related - Temperature - Tidal Forces - Wien's Law Constants Computer Models Additional Resources 1. Advanced Topics 2. Guest Contributions
Physics - Concepts - temperature
<MASK>
In an effort to standardize the temperature scale, Kelvin's are preferably used as 0 (zero) on the Kelvin scale is absolute zero. As such:
<MASK>
• 0 degree Celsius = Freezing point of water
<MASK>
The Fahrenheit scale is used prominently in the United States, and mostly for weather and cooking measurements. This scale is not at all used in science.
<MASK>
<UNMASK>
Introduction Astronomy Tools Concepts 1. Electromagnetic Spectrum 2. Atmosphere Limitations 3. Space Observations Equipment 1. Telescopes 2. Radio 3. Space Tools 4. Photography 5. Spectroscopy 6. Computers 7. Advanced Methods 8. Radio Astronomy Basic Mathematics Algebra Statistics Geometry Scientific Notation Log Scales Calculus Physics Concepts - Basic Units of Measure - Mass & Density - Temperature - Velocity & Acceleration - Force, Pressure & Energy - Atoms - Quantum Physics - Nature of Light Formulas - Brightness - Cepheid Rulers - Distance - Doppler Shift - Frequency & Wavelength - Hubble's Law - Inverse Square Law - Kinetic Energy - Luminosity - Magnitudes - Convert Mass to Energy - Kepler & Newton - Orbits - Parallax - Planck's Law - Relativistic Redshift - Relativity - Schwarzschild Radius - Synodic & Sidereal Periods - Sidereal Time - Small Angle Formula - Stellar Properties - Stephan-Boltzmann Law - Telescope Related - Temperature - Tidal Forces - Wien's Law Constants Computer Models Additional Resources 1. Advanced Topics 2. Guest Contributions
Physics - Concepts - temperature
<MASK>
In an effort to standardize the temperature scale, Kelvin's are preferably used as 0 (zero) on the Kelvin scale is absolute zero. As such:
<MASK>
• 0 degree Celsius = Freezing point of water
<MASK>
The Fahrenheit scale is used prominently in the United States, and mostly for weather and cooking measurements. This scale is not at all used in science.
<MASK>
In science, DO convert Fahrenheit to Celsius and avoid using Fahrenheit altogether.
|
Introduction Astronomy Tools Concepts 1. Electromagnetic Spectrum 2. Atmosphere Limitations 3. Space Observations Equipment 1. Telescopes 2. Radio 3. Space Tools 4. Photography 5. Spectroscopy 6. Computers 7. Advanced Methods 8. Radio Astronomy Basic Mathematics Algebra Statistics Geometry Scientific Notation Log Scales Calculus Physics Concepts - Basic Units of Measure - Mass & Density - Temperature - Velocity & Acceleration - Force, Pressure & Energy - Atoms - Quantum Physics - Nature of Light Formulas - Brightness - Cepheid Rulers - Distance - Doppler Shift - Frequency & Wavelength - Hubble's Law - Inverse Square Law - Kinetic Energy - Luminosity - Magnitudes - Convert Mass to Energy - Kepler & Newton - Orbits - Parallax - Planck's Law - Relativistic Redshift - Relativity - Schwarzschild Radius - Synodic & Sidereal Periods - Sidereal Time - Small Angle Formula - Stellar Properties - Stephan-Boltzmann Law - Telescope Related - Temperature - Tidal Forces - Wien's Law Constants Computer Models Additional Resources 1. Advanced Topics 2. Guest Contributions
Physics - Concepts - temperature
<MASK>
In an effort to standardize the temperature scale, Kelvin's are preferably used as 0 (zero) on the Kelvin scale is absolute zero. As such:
<MASK>
• 0 degree Celsius = Freezing point of water
<MASK>
The Fahrenheit scale is used prominently in the United States, and mostly for weather and cooking measurements. This scale is not at all used in science.
<MASK>
In science, DO convert Fahrenheit to Celsius and avoid using Fahrenheit altogether.
<UNMASK>
Introduction Astronomy Tools Concepts 1. Electromagnetic Spectrum 2. Atmosphere Limitations 3. Space Observations Equipment 1. Telescopes 2. Radio 3. Space Tools 4. Photography 5. Spectroscopy 6. Computers 7. Advanced Methods 8. Radio Astronomy Basic Mathematics Algebra Statistics Geometry Scientific Notation Log Scales Calculus Physics Concepts - Basic Units of Measure - Mass & Density - Temperature - Velocity & Acceleration - Force, Pressure & Energy - Atoms - Quantum Physics - Nature of Light Formulas - Brightness - Cepheid Rulers - Distance - Doppler Shift - Frequency & Wavelength - Hubble's Law - Inverse Square Law - Kinetic Energy - Luminosity - Magnitudes - Convert Mass to Energy - Kepler & Newton - Orbits - Parallax - Planck's Law - Relativistic Redshift - Relativity - Schwarzschild Radius - Synodic & Sidereal Periods - Sidereal Time - Small Angle Formula - Stellar Properties - Stephan-Boltzmann Law - Telescope Related - Temperature - Tidal Forces - Wien's Law Constants Computer Models Additional Resources 1. Advanced Topics 2. Guest Contributions
Physics - Concepts - temperature
We often associate temperature with how hot or cold something is. If we see a piece of metal glowing red, that would be hot to the touch. A ground covered with snow would be cold to touch. This is only part of the story.
<MASK>
The image above shows an example of this. If this image represents a gas, the particles that constitute the gas would move depending on temperature. In a hot gas, the molecules (represented by the dots) are free to move - the faster they move, the hotter the gas. Alternately, a cold gas would equal slow or very minimal movement of molecules.
<MASK>
In an effort to standardize the temperature scale, Kelvin's are preferably used as 0 (zero) on the Kelvin scale is absolute zero. As such:
Celsius has by no means been abandoned. Much of our standards are based on the Celsius scale:
• 0 degree Celsius = Freezing point of water
• 100 degrees Celsius = Boiling point of water
The Fahrenheit scale is used prominently in the United States, and mostly for weather and cooking measurements. This scale is not at all used in science.
• 32 degrees Fahrenheit = Freezing point of water
<MASK>
The temperatures above do not follow a nice neat pattern like the Celsius scale, but of special note:
<MASK>
The symbol for degrees or the word degree IS to be used for Celsius or Fahrenheit temperature scales.
In science, DO convert Fahrenheit to Celsius and avoid using Fahrenheit altogether.
|
Introduction Astronomy Tools Concepts 1. Electromagnetic Spectrum 2. Atmosphere Limitations 3. Space Observations Equipment 1. Telescopes 2. Radio 3. Space Tools 4. Photography 5. Spectroscopy 6. Computers 7. Advanced Methods 8. Radio Astronomy Basic Mathematics Algebra Statistics Geometry Scientific Notation Log Scales Calculus Physics Concepts - Basic Units of Measure - Mass & Density - Temperature - Velocity & Acceleration - Force, Pressure & Energy - Atoms - Quantum Physics - Nature of Light Formulas - Brightness - Cepheid Rulers - Distance - Doppler Shift - Frequency & Wavelength - Hubble's Law - Inverse Square Law - Kinetic Energy - Luminosity - Magnitudes - Convert Mass to Energy - Kepler & Newton - Orbits - Parallax - Planck's Law - Relativistic Redshift - Relativity - Schwarzschild Radius - Synodic & Sidereal Periods - Sidereal Time - Small Angle Formula - Stellar Properties - Stephan-Boltzmann Law - Telescope Related - Temperature - Tidal Forces - Wien's Law Constants Computer Models Additional Resources 1. Advanced Topics 2. Guest Contributions
Physics - Concepts - temperature
We often associate temperature with how hot or cold something is. If we see a piece of metal glowing red, that would be hot to the touch. A ground covered with snow would be cold to touch. This is only part of the story.
<MASK>
The image above shows an example of this. If this image represents a gas, the particles that constitute the gas would move depending on temperature. In a hot gas, the molecules (represented by the dots) are free to move - the faster they move, the hotter the gas. Alternately, a cold gas would equal slow or very minimal movement of molecules.
<MASK>
In an effort to standardize the temperature scale, Kelvin's are preferably used as 0 (zero) on the Kelvin scale is absolute zero. As such:
Celsius has by no means been abandoned. Much of our standards are based on the Celsius scale:
• 0 degree Celsius = Freezing point of water
• 100 degrees Celsius = Boiling point of water
The Fahrenheit scale is used prominently in the United States, and mostly for weather and cooking measurements. This scale is not at all used in science.
• 32 degrees Fahrenheit = Freezing point of water
<MASK>
The temperatures above do not follow a nice neat pattern like the Celsius scale, but of special note:
<MASK>
The symbol for degrees or the word degree IS to be used for Celsius or Fahrenheit temperature scales.
In science, DO convert Fahrenheit to Celsius and avoid using Fahrenheit altogether.
<UNMASK>
Introduction Astronomy Tools Concepts 1. Electromagnetic Spectrum 2. Atmosphere Limitations 3. Space Observations Equipment 1. Telescopes 2. Radio 3. Space Tools 4. Photography 5. Spectroscopy 6. Computers 7. Advanced Methods 8. Radio Astronomy Basic Mathematics Algebra Statistics Geometry Scientific Notation Log Scales Calculus Physics Concepts - Basic Units of Measure - Mass & Density - Temperature - Velocity & Acceleration - Force, Pressure & Energy - Atoms - Quantum Physics - Nature of Light Formulas - Brightness - Cepheid Rulers - Distance - Doppler Shift - Frequency & Wavelength - Hubble's Law - Inverse Square Law - Kinetic Energy - Luminosity - Magnitudes - Convert Mass to Energy - Kepler & Newton - Orbits - Parallax - Planck's Law - Relativistic Redshift - Relativity - Schwarzschild Radius - Synodic & Sidereal Periods - Sidereal Time - Small Angle Formula - Stellar Properties - Stephan-Boltzmann Law - Telescope Related - Temperature - Tidal Forces - Wien's Law Constants Computer Models Additional Resources 1. Advanced Topics 2. Guest Contributions
Physics - Concepts - temperature
We often associate temperature with how hot or cold something is. If we see a piece of metal glowing red, that would be hot to the touch. A ground covered with snow would be cold to touch. This is only part of the story.
In physics, temperature is the measure of how internal particles move.
The image above shows an example of this. If this image represents a gas, the particles that constitute the gas would move depending on temperature. In a hot gas, the molecules (represented by the dots) are free to move - the faster they move, the hotter the gas. Alternately, a cold gas would equal slow or very minimal movement of molecules.
<MASK>
In an effort to standardize the temperature scale, Kelvin's are preferably used as 0 (zero) on the Kelvin scale is absolute zero. As such:
Celsius has by no means been abandoned. Much of our standards are based on the Celsius scale:
• 0 degree Celsius = Freezing point of water
• 100 degrees Celsius = Boiling point of water
The Fahrenheit scale is used prominently in the United States, and mostly for weather and cooking measurements. This scale is not at all used in science.
• 32 degrees Fahrenheit = Freezing point of water
• 212 degrees Fahrenheit = Boiling point of water
The temperatures above do not follow a nice neat pattern like the Celsius scale, but of special note:
<MASK>
Absolute zero = 0 K
The symbol for degrees or the word degree IS to be used for Celsius or Fahrenheit temperature scales.
In science, DO convert Fahrenheit to Celsius and avoid using Fahrenheit altogether.
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Introduction Astronomy Tools Concepts 1. Electromagnetic Spectrum 2. Atmosphere Limitations 3. Space Observations Equipment 1. Telescopes 2. Radio 3. Space Tools 4. Photography 5. Spectroscopy 6. Computers 7. Advanced Methods 8. Radio Astronomy Basic Mathematics Algebra Statistics Geometry Scientific Notation Log Scales Calculus Physics Concepts - Basic Units of Measure - Mass & Density - Temperature - Velocity & Acceleration - Force, Pressure & Energy - Atoms - Quantum Physics - Nature of Light Formulas - Brightness - Cepheid Rulers - Distance - Doppler Shift - Frequency & Wavelength - Hubble's Law - Inverse Square Law - Kinetic Energy - Luminosity - Magnitudes - Convert Mass to Energy - Kepler & Newton - Orbits - Parallax - Planck's Law - Relativistic Redshift - Relativity - Schwarzschild Radius - Synodic & Sidereal Periods - Sidereal Time - Small Angle Formula - Stellar Properties - Stephan-Boltzmann Law - Telescope Related - Temperature - Tidal Forces - Wien's Law Constants Computer Models Additional Resources 1. Advanced Topics 2. Guest Contributions
Physics - Concepts - temperature
We often associate temperature with how hot or cold something is. If we see a piece of metal glowing red, that would be hot to the touch. A ground covered with snow would be cold to touch. This is only part of the story.
In physics, temperature is the measure of how internal particles move.
The image above shows an example of this. If this image represents a gas, the particles that constitute the gas would move depending on temperature. In a hot gas, the molecules (represented by the dots) are free to move - the faster they move, the hotter the gas. Alternately, a cold gas would equal slow or very minimal movement of molecules.
<MASK>
In an effort to standardize the temperature scale, Kelvin's are preferably used as 0 (zero) on the Kelvin scale is absolute zero. As such:
Celsius has by no means been abandoned. Much of our standards are based on the Celsius scale:
• 0 degree Celsius = Freezing point of water
• 100 degrees Celsius = Boiling point of water
The Fahrenheit scale is used prominently in the United States, and mostly for weather and cooking measurements. This scale is not at all used in science.
• 32 degrees Fahrenheit = Freezing point of water
• 212 degrees Fahrenheit = Boiling point of water
The temperatures above do not follow a nice neat pattern like the Celsius scale, but of special note:
<MASK>
Absolute zero = 0 K
The symbol for degrees or the word degree IS to be used for Celsius or Fahrenheit temperature scales.
In science, DO convert Fahrenheit to Celsius and avoid using Fahrenheit altogether.
<UNMASK>
Introduction Astronomy Tools Concepts 1. Electromagnetic Spectrum 2. Atmosphere Limitations 3. Space Observations Equipment 1. Telescopes 2. Radio 3. Space Tools 4. Photography 5. Spectroscopy 6. Computers 7. Advanced Methods 8. Radio Astronomy Basic Mathematics Algebra Statistics Geometry Scientific Notation Log Scales Calculus Physics Concepts - Basic Units of Measure - Mass & Density - Temperature - Velocity & Acceleration - Force, Pressure & Energy - Atoms - Quantum Physics - Nature of Light Formulas - Brightness - Cepheid Rulers - Distance - Doppler Shift - Frequency & Wavelength - Hubble's Law - Inverse Square Law - Kinetic Energy - Luminosity - Magnitudes - Convert Mass to Energy - Kepler & Newton - Orbits - Parallax - Planck's Law - Relativistic Redshift - Relativity - Schwarzschild Radius - Synodic & Sidereal Periods - Sidereal Time - Small Angle Formula - Stellar Properties - Stephan-Boltzmann Law - Telescope Related - Temperature - Tidal Forces - Wien's Law Constants Computer Models Additional Resources 1. Advanced Topics 2. Guest Contributions
Physics - Concepts - temperature
We often associate temperature with how hot or cold something is. If we see a piece of metal glowing red, that would be hot to the touch. A ground covered with snow would be cold to touch. This is only part of the story.
In physics, temperature is the measure of how internal particles move.
The image above shows an example of this. If this image represents a gas, the particles that constitute the gas would move depending on temperature. In a hot gas, the molecules (represented by the dots) are free to move - the faster they move, the hotter the gas. Alternately, a cold gas would equal slow or very minimal movement of molecules.
Absolute zero is defined as molecules in a medium that are not moving at all. In the Celsius temperature scale, absolute zero is -273.2 degrees C.
In an effort to standardize the temperature scale, Kelvin's are preferably used as 0 (zero) on the Kelvin scale is absolute zero. As such:
Celsius has by no means been abandoned. Much of our standards are based on the Celsius scale:
• 0 degree Celsius = Freezing point of water
• 100 degrees Celsius = Boiling point of water
The Fahrenheit scale is used prominently in the United States, and mostly for weather and cooking measurements. This scale is not at all used in science.
• 32 degrees Fahrenheit = Freezing point of water
• 212 degrees Fahrenheit = Boiling point of water
The temperatures above do not follow a nice neat pattern like the Celsius scale, but of special note:
-40 degrees Fahrenheit = -40 degrees Celsius
To convert Fahrenheit to Celsius:
Other important notes:
The symbol for degrees (o) or the word degree is NOT used on the Kelvin scale:
Absolute zero = 0 K
The symbol for degrees or the word degree IS to be used for Celsius or Fahrenheit temperature scales.
In science, DO convert Fahrenheit to Celsius and avoid using Fahrenheit altogether.
|
<MASK>
and time taken be man to swim z distance in normal direction t= z/2............1
<MASK>
ds =(2t)1/2 dt
<MASK>
<UNMASK>
<MASK>
velocity of man =2 m/sec
<MASK>
and time taken be man to swim z distance in normal direction t= z/2............1
<MASK>
ds =(2t)1/2 dt
<MASK>
Please feel free to post as many doubts on our discussion forum as you can.
If you find any question Difficult to understand - post it here and we will get you
the answer and detailed solution very quickly.
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<MASK>
velocity of man =2 m/sec
<MASK>
and time taken be man to swim z distance in normal direction t= z/2............1
<MASK>
ds =(2t)1/2 dt
<MASK>
Please feel free to post as many doubts on our discussion forum as you can.
If you find any question Difficult to understand - post it here and we will get you
the answer and detailed solution very quickly.
<MASK>
<UNMASK>
<MASK>
Dear vardaan
<MASK>
velocity of man =2 m/sec
<MASK>
and time taken be man to swim z distance in normal direction t= z/2............1
<MASK>
ds =(2t)1/2 dt
S =√2 t3/2 /3/2 limit o to 10
<MASK>
Please feel free to post as many doubts on our discussion forum as you can.
If you find any question Difficult to understand - post it here and we will get you
the answer and detailed solution very quickly.
We are all IITians and here to help you in your IIT JEE preparation.
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Regards,
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<MASK>
Dear vardaan
<MASK>
velocity of man =2 m/sec
<MASK>
and time taken be man to swim z distance in normal direction t= z/2............1
<MASK>
ds =(2t)1/2 dt
S =√2 t3/2 /3/2 limit o to 10
<MASK>
Please feel free to post as many doubts on our discussion forum as you can.
If you find any question Difficult to understand - post it here and we will get you
the answer and detailed solution very quickly.
We are all IITians and here to help you in your IIT JEE preparation.
<MASK>
Regards,
<UNMASK>
<MASK>
Dear vardaan
<MASK>
velocity of man =2 m/sec
<MASK>
and time taken be man to swim z distance in normal direction t= z/2............1
<MASK>
ds =(2t)1/2 dt
S =√2 t3/2 /3/2 limit o to 10
<MASK>
Please feel free to post as many doubts on our discussion forum as you can.
If you find any question Difficult to understand - post it here and we will get you
the answer and detailed solution very quickly.
We are all IITians and here to help you in your IIT JEE preparation.
All the best.
Regards,
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<MASK>
Dear vardaan
<MASK>
velocity of man =2 m/sec
<MASK>
and time taken be man to swim z distance in normal direction t= z/2............1
<MASK>
ds =(2t)1/2 dt
S =√2 t3/2 /3/2 limit o to 10
<MASK>
Please feel free to post as many doubts on our discussion forum as you can.
If you find any question Difficult to understand - post it here and we will get you
the answer and detailed solution very quickly.
We are all IITians and here to help you in your IIT JEE preparation.
All the best.
Regards,
<UNMASK>
# THE FLOW SPEED OF WATER VARIES WITH THE DISTANCE FROM ITS STRAIGHT BANK AS V=(1m1/2s).z1/ 2 where z is the perpendicular distance in m from the bank A.SWIMMER ENTERS THE RIVER AT A POINT A ON THE BANK AND SWIMS AT 2m/s, IN A DIRN NORMAL TO RIVER FLOW. WHAT IS HIS DRIFT DOWNSTREAM BY THE TIME HIS NORMAL DISTANCE FROM THE bank IS 20m?
148 Points
13 years ago
Dear vardaan
V=z1/ 2
velocity of man =2 m/sec
time taken by man to swim 20 m in normal direction t= 20/2 =10 sec
and time taken be man to swim z distance in normal direction t= z/2............1
now consider motion from z to z+dz distance time taken by man to travel z to z+dz postion is dt
and for this small distance dz ,velocity of river assumed to be constant V =z1/ 2
so distance moved in downstreem direction in time dt is
ds =z1/ 2 dt
ds =(2t)1/2 dt from equation 1
ds =(2t)1/2 dt
S =√2 t3/2 /3/2 limit o to 10
=2√2/3 t3/2 limit o to 10
=40√5/3
Please feel free to post as many doubts on our discussion forum as you can.
If you find any question Difficult to understand - post it here and we will get you
the answer and detailed solution very quickly.
We are all IITians and here to help you in your IIT JEE preparation.
All the best.
Regards,
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<MASK>
A Rule of Thumb for Pricing
Note that the extra revenue from an incremental unit of quantity, Δ(PQ)/ΔQ, has two components: 1. Producing one extra unit and selling it at price P brings in revenue (1)(P) = P. 2. But because the firm faces a downward-sloping demand curve, producing and selling this extra unit also results in a small drop in price ΔP/ΔQ, which reduces the revenue from all units sold (i.e., a change in revenue Q[ΔP/ΔQ]). Thus,
<MASK>
Monopoly Monopoly pricing compared to perfect competition pricing:
P > MC Price is larger than MC by an amount that depends inversely on the elasticity of demand Perfect Competition P = MC Demand is perfectly elastic so P=MC
<MASK>
Effect of Excise Tax on Monopolist
\$/Q D = AR MR Increase in P: P0 to P1 > tax Q1 P1 Q0 P0 MC + tax t MC Quantity
<MASK>
Monopsony Competitive Buyer Price taker
P = Marginal expenditure = Average expenditure D = Marginal value Graphically can compare competitive buyer to competitive seller
<MASK>
Monopsony Power The degree of monopsony power depends on three factors. Number of buyers The fewer the number of buyers, the less elastic the supply and the greater the monopsony power. Interaction Among Buyers The less the buyers compete, the greater the monopsony power. Elasticity of market supply Extent to which price is marked down below MV depends on elasticity of supply facing buyer If supply is very elastic, markdown will be small The more inelastic the supply the more monopsony power
Download ppt "Market Power: Monopoly and Monopsony"
<MASK>
<UNMASK>
<MASK>
## Presentation on theme: "Market Power: Monopoly and Monopsony"— Presentation transcript:
<MASK>
Average and Marginal Revenue
1 2 3 \$ per unit of output 4 5 6 7 Marginal Revenue Average Revenue (Demand) Output 1 2 3 4 5 6 7
<MASK>
Monopoly: An Example
<MASK>
A Rule of Thumb for Pricing
Note that the extra revenue from an incremental unit of quantity, Δ(PQ)/ΔQ, has two components: 1. Producing one extra unit and selling it at price P brings in revenue (1)(P) = P. 2. But because the firm faces a downward-sloping demand curve, producing and selling this extra unit also results in a small drop in price ΔP/ΔQ, which reduces the revenue from all units sold (i.e., a change in revenue Q[ΔP/ΔQ]). Thus,
(Q/P)(ΔP/ΔQ) is the reciprocal of the elasticity of demand, 1/Ed, measured at the profit-maximizing output, and Now, because the firm’s objective is to maximize profit, we can set marginal revenue equal to marginal cost: which can be rearranged to give us Equivalently, we can rearrange this equation to express price directly as a markup over marginal cost:
<MASK>
A Rule of Thumb for Pricing
<MASK>
Monopoly Monopoly pricing compared to perfect competition pricing:
P > MC Price is larger than MC by an amount that depends inversely on the elasticity of demand Perfect Competition P = MC Demand is perfectly elastic so P=MC
<MASK>
Monopoly Shifts in demand usually cause a change in both price and quantity. Example show how monopolistic market differs from perfectly competitive market Competitive market supplies specific quantity a every price This relationship does not exist for a monopolistic market
<MASK>
Effect of Excise Tax on Monopolist
\$/Q D = AR MR Increase in P: P0 to P1 > tax Q1 P1 Q0 P0 MC + tax t MC Quantity
<MASK>
Monopsony Competitive Buyer Price taker
P = Marginal expenditure = Average expenditure D = Marginal value Graphically can compare competitive buyer to competitive seller
<MASK>
Monopoly and Monopsony
Monopsony is easier to understand if we compare to monopoly We can see this graphically Monopolist Can charge price above MC because faces downward sloping demand (average revenue) MR < AR MR=MC gives quantity less than competitive market and price that is higher
<MASK>
Monopoly and Monopsony
MR < P P > MC Qm < QC Pm > PC Monopsony ME > P P < MV Qm < QC Pm < PC
Monopsony Power The degree of monopsony power depends on three factors. Number of buyers The fewer the number of buyers, the less elastic the supply and the greater the monopsony power. Interaction Among Buyers The less the buyers compete, the greater the monopsony power. Elasticity of market supply Extent to which price is marked down below MV depends on elasticity of supply facing buyer If supply is very elastic, markdown will be small The more inelastic the supply the more monopsony power
Download ppt "Market Power: Monopoly and Monopsony"
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<MASK>
## Presentation on theme: "Market Power: Monopoly and Monopsony"— Presentation transcript:
<MASK>
Average and Marginal Revenue
1 2 3 \$ per unit of output 4 5 6 7 Marginal Revenue Average Revenue (Demand) Output 1 2 3 4 5 6 7
<MASK>
Monopoly: An Example
<MASK>
A Rule of Thumb for Pricing
Note that the extra revenue from an incremental unit of quantity, Δ(PQ)/ΔQ, has two components: 1. Producing one extra unit and selling it at price P brings in revenue (1)(P) = P. 2. But because the firm faces a downward-sloping demand curve, producing and selling this extra unit also results in a small drop in price ΔP/ΔQ, which reduces the revenue from all units sold (i.e., a change in revenue Q[ΔP/ΔQ]). Thus,
(Q/P)(ΔP/ΔQ) is the reciprocal of the elasticity of demand, 1/Ed, measured at the profit-maximizing output, and Now, because the firm’s objective is to maximize profit, we can set marginal revenue equal to marginal cost: which can be rearranged to give us Equivalently, we can rearrange this equation to express price directly as a markup over marginal cost:
<MASK>
A Rule of Thumb for Pricing
<MASK>
Monopoly Monopoly pricing compared to perfect competition pricing:
P > MC Price is larger than MC by an amount that depends inversely on the elasticity of demand Perfect Competition P = MC Demand is perfectly elastic so P=MC
<MASK>
Monopoly Shifts in demand usually cause a change in both price and quantity. Example show how monopolistic market differs from perfectly competitive market Competitive market supplies specific quantity a every price This relationship does not exist for a monopolistic market
<MASK>
Effect of Excise Tax on Monopolist
\$/Q D = AR MR Increase in P: P0 to P1 > tax Q1 P1 Q0 P0 MC + tax t MC Quantity
<MASK>
Monopsony Competitive Buyer Price taker
P = Marginal expenditure = Average expenditure D = Marginal value Graphically can compare competitive buyer to competitive seller
<MASK>
Monopoly and Monopsony
Monopsony is easier to understand if we compare to monopoly We can see this graphically Monopolist Can charge price above MC because faces downward sloping demand (average revenue) MR < AR MR=MC gives quantity less than competitive market and price that is higher
<MASK>
Monopoly and Monopsony
MR < P P > MC Qm < QC Pm > PC Monopsony ME > P P < MV Qm < QC Pm < PC
Monopsony Power The degree of monopsony power depends on three factors. Number of buyers The fewer the number of buyers, the less elastic the supply and the greater the monopsony power. Interaction Among Buyers The less the buyers compete, the greater the monopsony power. Elasticity of market supply Extent to which price is marked down below MV depends on elasticity of supply facing buyer If supply is very elastic, markdown will be small The more inelastic the supply the more monopsony power
Download ppt "Market Power: Monopoly and Monopsony"
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<UNMASK>
<MASK>
## Presentation on theme: "Market Power: Monopoly and Monopsony"— Presentation transcript:
<MASK>
Review of Perfect Competition
Q P Market D S Q0 P0 Q P Individual Firm P0 D = MR = P q0 LRAC LMC
<MASK>
Average and Marginal Revenue
1 2 3 \$ per unit of output 4 5 6 7 Marginal Revenue Average Revenue (Demand) Output 1 2 3 4 5 6 7
<MASK>
Monopoly: An Example
<MASK>
A Rule of Thumb for Pricing
Note that the extra revenue from an incremental unit of quantity, Δ(PQ)/ΔQ, has two components: 1. Producing one extra unit and selling it at price P brings in revenue (1)(P) = P. 2. But because the firm faces a downward-sloping demand curve, producing and selling this extra unit also results in a small drop in price ΔP/ΔQ, which reduces the revenue from all units sold (i.e., a change in revenue Q[ΔP/ΔQ]). Thus,
(Q/P)(ΔP/ΔQ) is the reciprocal of the elasticity of demand, 1/Ed, measured at the profit-maximizing output, and Now, because the firm’s objective is to maximize profit, we can set marginal revenue equal to marginal cost: which can be rearranged to give us Equivalently, we can rearrange this equation to express price directly as a markup over marginal cost:
<MASK>
A Rule of Thumb for Pricing
Produce one more unit brings in revenue (1)(P) = P With downward sloping demand, producing and selling one more unit results in small drop in price P/Q. Reduces revenue from all units sold, change in revenue: Q(P/Q)
<MASK>
A Rule of Thumb for Pricing
<MASK>
Monopoly Monopoly pricing compared to perfect competition pricing:
P > MC Price is larger than MC by an amount that depends inversely on the elasticity of demand Perfect Competition P = MC Demand is perfectly elastic so P=MC
Shifts in Demand In perfect competition, the market supply curve is determined by marginal cost. For a monopoly, output is determined by marginal cost and the shape of the demand curve. There is no supply curve for monopolistic market Shifts in demand do not trace out price and quantity changes corresponding to a supply curve Shifts in demand lead to Changes in price with no change in output Changes in output with no change in price Changes in both price and quantity
Monopoly Shifts in demand usually cause a change in both price and quantity. Example show how monopolistic market differs from perfectly competitive market Competitive market supplies specific quantity a every price This relationship does not exist for a monopolistic market
<MASK>
Effect of Excise Tax on Monopolist
\$/Q D = AR MR Increase in P: P0 to P1 > tax Q1 P1 Q0 P0 MC + tax t MC Quantity
<MASK>
Deadweight Loss from Monopoly Power
\$/Q Because of the higher price, consumers lose A+B and producer gains A-C. Lost Consumer Surplus MC Deadweight Loss B A QC PC C C Qm Quantity
<MASK>
Monopsony Competitive Buyer Price taker
P = Marginal expenditure = Average expenditure D = Marginal value Graphically can compare competitive buyer to competitive seller
<MASK>
Monopoly and Monopsony
Monopsony is easier to understand if we compare to monopoly We can see this graphically Monopolist Can charge price above MC because faces downward sloping demand (average revenue) MR < AR MR=MC gives quantity less than competitive market and price that is higher
Monopoly and Monopsony
\$/Q MV ME S = AE Q* P* PC QC Monopsony Note: ME = MV; ME > AE; MV > P
Monopoly and Monopsony
MR < P P > MC Qm < QC Pm > PC Monopsony ME > P P < MV Qm < QC Pm < PC
Monopsony Power The degree of monopsony power depends on three factors. Number of buyers The fewer the number of buyers, the less elastic the supply and the greater the monopsony power. Interaction Among Buyers The less the buyers compete, the greater the monopsony power. Elasticity of market supply Extent to which price is marked down below MV depends on elasticity of supply facing buyer If supply is very elastic, markdown will be small The more inelastic the supply the more monopsony power
Download ppt "Market Power: Monopoly and Monopsony"
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<MASK>
## Presentation on theme: "Market Power: Monopoly and Monopsony"— Presentation transcript:
<MASK>
Review of Perfect Competition
Q P Market D S Q0 P0 Q P Individual Firm P0 D = MR = P q0 LRAC LMC
<MASK>
Average and Marginal Revenue
1 2 3 \$ per unit of output 4 5 6 7 Marginal Revenue Average Revenue (Demand) Output 1 2 3 4 5 6 7
<MASK>
Monopoly: An Example
<MASK>
A Rule of Thumb for Pricing
Note that the extra revenue from an incremental unit of quantity, Δ(PQ)/ΔQ, has two components: 1. Producing one extra unit and selling it at price P brings in revenue (1)(P) = P. 2. But because the firm faces a downward-sloping demand curve, producing and selling this extra unit also results in a small drop in price ΔP/ΔQ, which reduces the revenue from all units sold (i.e., a change in revenue Q[ΔP/ΔQ]). Thus,
(Q/P)(ΔP/ΔQ) is the reciprocal of the elasticity of demand, 1/Ed, measured at the profit-maximizing output, and Now, because the firm’s objective is to maximize profit, we can set marginal revenue equal to marginal cost: which can be rearranged to give us Equivalently, we can rearrange this equation to express price directly as a markup over marginal cost:
<MASK>
A Rule of Thumb for Pricing
Produce one more unit brings in revenue (1)(P) = P With downward sloping demand, producing and selling one more unit results in small drop in price P/Q. Reduces revenue from all units sold, change in revenue: Q(P/Q)
<MASK>
A Rule of Thumb for Pricing
<MASK>
Monopoly Monopoly pricing compared to perfect competition pricing:
P > MC Price is larger than MC by an amount that depends inversely on the elasticity of demand Perfect Competition P = MC Demand is perfectly elastic so P=MC
Shifts in Demand In perfect competition, the market supply curve is determined by marginal cost. For a monopoly, output is determined by marginal cost and the shape of the demand curve. There is no supply curve for monopolistic market Shifts in demand do not trace out price and quantity changes corresponding to a supply curve Shifts in demand lead to Changes in price with no change in output Changes in output with no change in price Changes in both price and quantity
Monopoly Shifts in demand usually cause a change in both price and quantity. Example show how monopolistic market differs from perfectly competitive market Competitive market supplies specific quantity a every price This relationship does not exist for a monopolistic market
<MASK>
Effect of Excise Tax on Monopolist
\$/Q D = AR MR Increase in P: P0 to P1 > tax Q1 P1 Q0 P0 MC + tax t MC Quantity
<MASK>
Deadweight Loss from Monopoly Power
\$/Q Because of the higher price, consumers lose A+B and producer gains A-C. Lost Consumer Surplus MC Deadweight Loss B A QC PC C C Qm Quantity
<MASK>
Monopsony Competitive Buyer Price taker
P = Marginal expenditure = Average expenditure D = Marginal value Graphically can compare competitive buyer to competitive seller
<MASK>
Monopoly and Monopsony
Monopsony is easier to understand if we compare to monopoly We can see this graphically Monopolist Can charge price above MC because faces downward sloping demand (average revenue) MR < AR MR=MC gives quantity less than competitive market and price that is higher
Monopoly and Monopsony
\$/Q MV ME S = AE Q* P* PC QC Monopsony Note: ME = MV; ME > AE; MV > P
Monopoly and Monopsony
MR < P P > MC Qm < QC Pm > PC Monopsony ME > P P < MV Qm < QC Pm < PC
Monopsony Power The degree of monopsony power depends on three factors. Number of buyers The fewer the number of buyers, the less elastic the supply and the greater the monopsony power. Interaction Among Buyers The less the buyers compete, the greater the monopsony power. Elasticity of market supply Extent to which price is marked down below MV depends on elasticity of supply facing buyer If supply is very elastic, markdown will be small The more inelastic the supply the more monopsony power
Download ppt "Market Power: Monopoly and Monopsony"
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<UNMASK>
# Market Power: Monopoly and Monopsony
## Presentation on theme: "Market Power: Monopoly and Monopsony"— Presentation transcript:
<MASK>
Review of Perfect Competition
Q P Market D S Q0 P0 Q P Individual Firm P0 D = MR = P q0 LRAC LMC
Monopoly Monopoly One seller - many buyers
One product (no good substitutes) Barriers to entry Price Maker The monopolist is the supply-side of the market and has complete control over the amount offered for sale. Monopolist controls price but must consider consumer demand Profits will be maximized at the level of output where marginal revenue equals marginal cost.
<MASK>
Average and Marginal Revenue
1 2 3 \$ per unit of output 4 5 6 7 Marginal Revenue Average Revenue (Demand) Output 1 2 3 4 5 6 7
We can also see algebraically that Q. maximizes profit
We can also see algebraically that Q* maximizes profit. Profit π is the difference between revenue and cost, both of which depend on Q: As Q is increased from zero, profit will increase until it reaches a maximum and then begin to decrease. Thus the profit-maximizing Q is such that the incremental profit resulting from a small increase in Q is just zero (i.e., Δπ /ΔQ = 0). Then But ΔR/ΔQ is marginal revenue and ΔC/ΔQ is marginal cost. Thus the profit-maximizing condition is that , or
Monopolist’s Output Decision
\$ per unit of output D = AR MR MC AC P1 Q1 P* Q* P2 Q2 Lost profit Lost profit Quantity
Monopoly: An Example
Monopoly: An Example
A Rule of Thumb for Pricing
Note that the extra revenue from an incremental unit of quantity, Δ(PQ)/ΔQ, has two components: 1. Producing one extra unit and selling it at price P brings in revenue (1)(P) = P. 2. But because the firm faces a downward-sloping demand curve, producing and selling this extra unit also results in a small drop in price ΔP/ΔQ, which reduces the revenue from all units sold (i.e., a change in revenue Q[ΔP/ΔQ]). Thus,
(Q/P)(ΔP/ΔQ) is the reciprocal of the elasticity of demand, 1/Ed, measured at the profit-maximizing output, and Now, because the firm’s objective is to maximize profit, we can set marginal revenue equal to marginal cost: which can be rearranged to give us Equivalently, we can rearrange this equation to express price directly as a markup over marginal cost:
Example of Profit Maximization
\$/Q 10 20 40 Profit = (P - AC) x Q = (\$30 - \$15)(10) = \$150 MC AC P=30 Profit AR AC=15 MR 5 10 15 20 Quantity
A Rule of Thumb for Pricing
Produce one more unit brings in revenue (1)(P) = P With downward sloping demand, producing and selling one more unit results in small drop in price P/Q. Reduces revenue from all units sold, change in revenue: Q(P/Q)
A Rule of Thumb for Pricing
<MASK>
A Rule of Thumb for Pricing
A Rule of Thumb for Pricing
Monopoly Monopoly pricing compared to perfect competition pricing:
P > MC Price is larger than MC by an amount that depends inversely on the elasticity of demand Perfect Competition P = MC Demand is perfectly elastic so P=MC
Shifts in Demand In perfect competition, the market supply curve is determined by marginal cost. For a monopoly, output is determined by marginal cost and the shape of the demand curve. There is no supply curve for monopolistic market Shifts in demand do not trace out price and quantity changes corresponding to a supply curve Shifts in demand lead to Changes in price with no change in output Changes in output with no change in price Changes in both price and quantity
Monopoly Shifts in demand usually cause a change in both price and quantity. Example show how monopolistic market differs from perfectly competitive market Competitive market supplies specific quantity a every price This relationship does not exist for a monopolistic market
The Effect of a Tax In competitive market, a per-unit tax causes price to rise by less than tax: burden shared by producers and consumers Under monopoly, price can sometimes rise by more than the amount of the tax. To determine the impact of a tax: t = specific tax MC = MC + t
Effect of Excise Tax on Monopolist
\$/Q D = AR MR Increase in P: P0 to P1 > tax Q1 P1 Q0 P0 MC + tax t MC Quantity
<MASK>
The Multi-plant Firm Firm should increase output from each plant until the additional profit from last unit produced at Plant 1 equals 0
Deadweight Loss from Monopoly Power
\$/Q Because of the higher price, consumers lose A+B and producer gains A-C. Lost Consumer Surplus MC Deadweight Loss B A QC PC C C Qm Quantity
Monopsony A monopsony is a market in which there is a single buyer.
An oligopsony is a market with only a few buyers. Monopsony power is the ability of the buyer to affect the price of the good and pay less than the price that would exist in a competitive market. Typically choose to buy until the benefit from last unit equals that unit’s cost Marginal value is the additional benefit derived from purchasing one more unit of a good Demand curve – downward sloping Marginal expenditure is the additional cost of buying one more unit of a good Depends on buying power
Monopsony Competitive Buyer Price taker
P = Marginal expenditure = Average expenditure D = Marginal value Graphically can compare competitive buyer to competitive seller
Monopsonist Buyer ME S = AE PC P*m D = MV Q*m QC \$/Q Monopsony
ME above S Quantity where ME = MV: Qm Price from Supply curve: Pm ME D = MV S = AE Q*m P*m Competitive P = PC Q = Q+C PC QC Quantity
Monopoly and Monopsony
Monopsony is easier to understand if we compare to monopoly We can see this graphically Monopolist Can charge price above MC because faces downward sloping demand (average revenue) MR < AR MR=MC gives quantity less than competitive market and price that is higher
Monopoly and Monopsony
\$/Q MV ME S = AE Q* P* PC QC Monopsony Note: ME = MV; ME > AE; MV > P
Monopoly and Monopsony
MR < P P > MC Qm < QC Pm > PC Monopsony ME > P P < MV Qm < QC Pm < PC
Monopsony Power The degree of monopsony power depends on three factors. Number of buyers The fewer the number of buyers, the less elastic the supply and the greater the monopsony power. Interaction Among Buyers The less the buyers compete, the greater the monopsony power. Elasticity of market supply Extent to which price is marked down below MV depends on elasticity of supply facing buyer If supply is very elastic, markdown will be small The more inelastic the supply the more monopsony power
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# Market Power: Monopoly and Monopsony
## Presentation on theme: "Market Power: Monopoly and Monopsony"— Presentation transcript:
<MASK>
Review of Perfect Competition
Q P Market D S Q0 P0 Q P Individual Firm P0 D = MR = P q0 LRAC LMC
Monopoly Monopoly One seller - many buyers
One product (no good substitutes) Barriers to entry Price Maker The monopolist is the supply-side of the market and has complete control over the amount offered for sale. Monopolist controls price but must consider consumer demand Profits will be maximized at the level of output where marginal revenue equals marginal cost.
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Average and Marginal Revenue
1 2 3 \$ per unit of output 4 5 6 7 Marginal Revenue Average Revenue (Demand) Output 1 2 3 4 5 6 7
We can also see algebraically that Q. maximizes profit
We can also see algebraically that Q* maximizes profit. Profit π is the difference between revenue and cost, both of which depend on Q: As Q is increased from zero, profit will increase until it reaches a maximum and then begin to decrease. Thus the profit-maximizing Q is such that the incremental profit resulting from a small increase in Q is just zero (i.e., Δπ /ΔQ = 0). Then But ΔR/ΔQ is marginal revenue and ΔC/ΔQ is marginal cost. Thus the profit-maximizing condition is that , or
Monopolist’s Output Decision
\$ per unit of output D = AR MR MC AC P1 Q1 P* Q* P2 Q2 Lost profit Lost profit Quantity
Monopoly: An Example
Monopoly: An Example
A Rule of Thumb for Pricing
Note that the extra revenue from an incremental unit of quantity, Δ(PQ)/ΔQ, has two components: 1. Producing one extra unit and selling it at price P brings in revenue (1)(P) = P. 2. But because the firm faces a downward-sloping demand curve, producing and selling this extra unit also results in a small drop in price ΔP/ΔQ, which reduces the revenue from all units sold (i.e., a change in revenue Q[ΔP/ΔQ]). Thus,
(Q/P)(ΔP/ΔQ) is the reciprocal of the elasticity of demand, 1/Ed, measured at the profit-maximizing output, and Now, because the firm’s objective is to maximize profit, we can set marginal revenue equal to marginal cost: which can be rearranged to give us Equivalently, we can rearrange this equation to express price directly as a markup over marginal cost:
Example of Profit Maximization
\$/Q 10 20 40 Profit = (P - AC) x Q = (\$30 - \$15)(10) = \$150 MC AC P=30 Profit AR AC=15 MR 5 10 15 20 Quantity
A Rule of Thumb for Pricing
Produce one more unit brings in revenue (1)(P) = P With downward sloping demand, producing and selling one more unit results in small drop in price P/Q. Reduces revenue from all units sold, change in revenue: Q(P/Q)
A Rule of Thumb for Pricing
<MASK>
A Rule of Thumb for Pricing
A Rule of Thumb for Pricing
Monopoly Monopoly pricing compared to perfect competition pricing:
P > MC Price is larger than MC by an amount that depends inversely on the elasticity of demand Perfect Competition P = MC Demand is perfectly elastic so P=MC
Shifts in Demand In perfect competition, the market supply curve is determined by marginal cost. For a monopoly, output is determined by marginal cost and the shape of the demand curve. There is no supply curve for monopolistic market Shifts in demand do not trace out price and quantity changes corresponding to a supply curve Shifts in demand lead to Changes in price with no change in output Changes in output with no change in price Changes in both price and quantity
Monopoly Shifts in demand usually cause a change in both price and quantity. Example show how monopolistic market differs from perfectly competitive market Competitive market supplies specific quantity a every price This relationship does not exist for a monopolistic market
The Effect of a Tax In competitive market, a per-unit tax causes price to rise by less than tax: burden shared by producers and consumers Under monopoly, price can sometimes rise by more than the amount of the tax. To determine the impact of a tax: t = specific tax MC = MC + t
Effect of Excise Tax on Monopolist
\$/Q D = AR MR Increase in P: P0 to P1 > tax Q1 P1 Q0 P0 MC + tax t MC Quantity
<MASK>
The Multi-plant Firm Firm should increase output from each plant until the additional profit from last unit produced at Plant 1 equals 0
Deadweight Loss from Monopoly Power
\$/Q Because of the higher price, consumers lose A+B and producer gains A-C. Lost Consumer Surplus MC Deadweight Loss B A QC PC C C Qm Quantity
Monopsony A monopsony is a market in which there is a single buyer.
An oligopsony is a market with only a few buyers. Monopsony power is the ability of the buyer to affect the price of the good and pay less than the price that would exist in a competitive market. Typically choose to buy until the benefit from last unit equals that unit’s cost Marginal value is the additional benefit derived from purchasing one more unit of a good Demand curve – downward sloping Marginal expenditure is the additional cost of buying one more unit of a good Depends on buying power
Monopsony Competitive Buyer Price taker
P = Marginal expenditure = Average expenditure D = Marginal value Graphically can compare competitive buyer to competitive seller
Monopsonist Buyer ME S = AE PC P*m D = MV Q*m QC \$/Q Monopsony
ME above S Quantity where ME = MV: Qm Price from Supply curve: Pm ME D = MV S = AE Q*m P*m Competitive P = PC Q = Q+C PC QC Quantity
Monopoly and Monopsony
Monopsony is easier to understand if we compare to monopoly We can see this graphically Monopolist Can charge price above MC because faces downward sloping demand (average revenue) MR < AR MR=MC gives quantity less than competitive market and price that is higher
Monopoly and Monopsony
\$/Q MV ME S = AE Q* P* PC QC Monopsony Note: ME = MV; ME > AE; MV > P
Monopoly and Monopsony
MR < P P > MC Qm < QC Pm > PC Monopsony ME > P P < MV Qm < QC Pm < PC
Monopsony Power The degree of monopsony power depends on three factors. Number of buyers The fewer the number of buyers, the less elastic the supply and the greater the monopsony power. Interaction Among Buyers The less the buyers compete, the greater the monopsony power. Elasticity of market supply Extent to which price is marked down below MV depends on elasticity of supply facing buyer If supply is very elastic, markdown will be small The more inelastic the supply the more monopsony power
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# Market Power: Monopoly and Monopsony
## Presentation on theme: "Market Power: Monopoly and Monopsony"— Presentation transcript:
Market Power: Monopoly and Monopsony
Chapter 9
Review of Perfect Competition
P = LMC = LRAC Normal profits or zero economic profits in the long run Large number of buyers and sellers Homogenous product Perfect information Firm is a price taker
Review of Perfect Competition
Q P Market D S Q0 P0 Q P Individual Firm P0 D = MR = P q0 LRAC LMC
Monopoly Monopoly One seller - many buyers
One product (no good substitutes) Barriers to entry Price Maker The monopolist is the supply-side of the market and has complete control over the amount offered for sale. Monopolist controls price but must consider consumer demand Profits will be maximized at the level of output where marginal revenue equals marginal cost.
Average & Marginal Revenue
The monopolist’s average revenue, price received per unit sold, is the market demand curve. Monopolist also needs to find marginal revenue, change in revenue resulting from a unit change in output. Finding Marginal Revenue As the sole producer, the monopolist works with the market demand to determine output and price. An example can be used to show the relationship between average and marginal revenue Assume a monopolist with demand: P = 6 - Q
Average and Marginal Revenue
1 2 3 \$ per unit of output 4 5 6 7 Marginal Revenue Average Revenue (Demand) Output 1 2 3 4 5 6 7
We can also see algebraically that Q. maximizes profit
We can also see algebraically that Q* maximizes profit. Profit π is the difference between revenue and cost, both of which depend on Q: As Q is increased from zero, profit will increase until it reaches a maximum and then begin to decrease. Thus the profit-maximizing Q is such that the incremental profit resulting from a small increase in Q is just zero (i.e., Δπ /ΔQ = 0). Then But ΔR/ΔQ is marginal revenue and ΔC/ΔQ is marginal cost. Thus the profit-maximizing condition is that , or
Monopolist’s Output Decision
\$ per unit of output D = AR MR MC AC P1 Q1 P* Q* P2 Q2 Lost profit Lost profit Quantity
Monopoly: An Example
Monopoly: An Example
A Rule of Thumb for Pricing
Note that the extra revenue from an incremental unit of quantity, Δ(PQ)/ΔQ, has two components: 1. Producing one extra unit and selling it at price P brings in revenue (1)(P) = P. 2. But because the firm faces a downward-sloping demand curve, producing and selling this extra unit also results in a small drop in price ΔP/ΔQ, which reduces the revenue from all units sold (i.e., a change in revenue Q[ΔP/ΔQ]). Thus,
(Q/P)(ΔP/ΔQ) is the reciprocal of the elasticity of demand, 1/Ed, measured at the profit-maximizing output, and Now, because the firm’s objective is to maximize profit, we can set marginal revenue equal to marginal cost: which can be rearranged to give us Equivalently, we can rearrange this equation to express price directly as a markup over marginal cost:
Example of Profit Maximization
\$/Q 10 20 40 Profit = (P - AC) x Q = (\$30 - \$15)(10) = \$150 MC AC P=30 Profit AR AC=15 MR 5 10 15 20 Quantity
A Rule of Thumb for Pricing
Produce one more unit brings in revenue (1)(P) = P With downward sloping demand, producing and selling one more unit results in small drop in price P/Q. Reduces revenue from all units sold, change in revenue: Q(P/Q)
A Rule of Thumb for Pricing
A Rule of Thumb for Pricing
A Rule of Thumb for Pricing
A Rule of Thumb for Pricing
Monopoly Monopoly pricing compared to perfect competition pricing:
P > MC Price is larger than MC by an amount that depends inversely on the elasticity of demand Perfect Competition P = MC Demand is perfectly elastic so P=MC
Shifts in Demand In perfect competition, the market supply curve is determined by marginal cost. For a monopoly, output is determined by marginal cost and the shape of the demand curve. There is no supply curve for monopolistic market Shifts in demand do not trace out price and quantity changes corresponding to a supply curve Shifts in demand lead to Changes in price with no change in output Changes in output with no change in price Changes in both price and quantity
Monopoly Shifts in demand usually cause a change in both price and quantity. Example show how monopolistic market differs from perfectly competitive market Competitive market supplies specific quantity a every price This relationship does not exist for a monopolistic market
The Effect of a Tax In competitive market, a per-unit tax causes price to rise by less than tax: burden shared by producers and consumers Under monopoly, price can sometimes rise by more than the amount of the tax. To determine the impact of a tax: t = specific tax MC = MC + t
Effect of Excise Tax on Monopolist
\$/Q D = AR MR Increase in P: P0 to P1 > tax Q1 P1 Q0 P0 MC + tax t MC Quantity
The Multi-plant Firm For some firms, production takes place in more than one plant each with different costs Firm must determine how to distribute production between both plants Production should be split so that the MC in the plants is the same Output is chosen where MR=MC. Profits is therefore maximized when MR=MC at each plant We can show this algebraically: Q1 and C1 is output and cost of production for Plant 1 Q2 and C2 is output and cost of production for Plant 2 QT = Q1 + Q2 is total output Profit is then: = PQT – C1(Q1) – C2(Q2)
The Multi-plant Firm Firm should increase output from each plant until the additional profit from last unit produced at Plant 1 equals 0
Deadweight Loss from Monopoly Power
\$/Q Because of the higher price, consumers lose A+B and producer gains A-C. Lost Consumer Surplus MC Deadweight Loss B A QC PC C C Qm Quantity
Monopsony A monopsony is a market in which there is a single buyer.
An oligopsony is a market with only a few buyers. Monopsony power is the ability of the buyer to affect the price of the good and pay less than the price that would exist in a competitive market. Typically choose to buy until the benefit from last unit equals that unit’s cost Marginal value is the additional benefit derived from purchasing one more unit of a good Demand curve – downward sloping Marginal expenditure is the additional cost of buying one more unit of a good Depends on buying power
Monopsony Competitive Buyer Price taker
P = Marginal expenditure = Average expenditure D = Marginal value Graphically can compare competitive buyer to competitive seller
Monopsonist Buyer ME S = AE PC P*m D = MV Q*m QC \$/Q Monopsony
ME above S Quantity where ME = MV: Qm Price from Supply curve: Pm ME D = MV S = AE Q*m P*m Competitive P = PC Q = Q+C PC QC Quantity
Monopoly and Monopsony
Monopsony is easier to understand if we compare to monopoly We can see this graphically Monopolist Can charge price above MC because faces downward sloping demand (average revenue) MR < AR MR=MC gives quantity less than competitive market and price that is higher
Monopoly and Monopsony
\$/Q MV ME S = AE Q* P* PC QC Monopsony Note: ME = MV; ME > AE; MV > P
Monopoly and Monopsony
MR < P P > MC Qm < QC Pm > PC Monopsony ME > P P < MV Qm < QC Pm < PC
Monopsony Power The degree of monopsony power depends on three factors. Number of buyers The fewer the number of buyers, the less elastic the supply and the greater the monopsony power. Interaction Among Buyers The less the buyers compete, the greater the monopsony power. Elasticity of market supply Extent to which price is marked down below MV depends on elasticity of supply facing buyer If supply is very elastic, markdown will be small The more inelastic the supply the more monopsony power
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Student Entries
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Article posted June 2, 2012 at 07:33 AM GMT-5 • comment • Reads 539 Today is Sunday, June 10, and I'm doing this blog a week late because the website crashed last week. It also means that there are exactly 2 school days until Finals start! I'm starting to worry it will be really hard to study without being distracted by the nice weather. To get in the spirit of studying, I'm going to post a review question from one of my past tests today on this blog. This is it: Determine which three numbers could be the sides of a right triangle. A. 64,73,98 B. 64,72,96 C. 65,72,97 The answer is C because of the Pythagorean theorem. When you square each number, the first two must add up to the third one's square. If that happens, then it is a right triangle. Choice C is the only one that happens for, so it is the correct answer. Until Next week, Article posted June 2, 2012 at 07:33 AM GMT-5 • comment • Reads 539
Article posted June 10, 2012 at 07:03 AM GMT-5 • comment • Reads 309 I'm doing the second blog for this week today, Sunday, June 10 because the website crashed last week. As I said earlier, there are exactly two school days until finals start! While this may seem terrible, I'm actually very happy because it means that we are in the home stretch before summer! I'm posting another review question this week: this will be about lines in triangles. Which line in triangles meet at the orthocenter when all three are drawn in a triangle? The answer to this question is altitudes. An altitude, also known as the height of a triangle, is a perpendicular line to the base drawn from a vertex. School may be almost done, but I can't give up now. This is the last blog of this year, and I can say that's a good thing! Article posted June 10, 2012 at 07:03 AM GMT-5 • comment • Reads 309
Article posted June 10, 2012 at 07:40 PM GMT-5 • comment • Reads 291 For this weeks blog I was asked to post two review questions for the final exam. The final exam is this week and I need to start studying, so I am going to start by making these review questions. My first question is: What does MAPA COCI stand for? A hint for this question is to remember lines in triangles and their concurrent points! My next question is: How do you find the circumference of a circle with a radius of 10 cm? Answers: Question 1: Median, Altitude, Perpendicular Bisector, Angle Bisector. Centroid, Orthocenter, Circumcenter, Incenter. Question 2: 20pi cm. Article posted June 10, 2012 at 07:40 PM GMT-5 • comment • Reads 291
Article posted June 7, 2012 at 04:31 PM GMT-5 • comment • Reads 279 On Sunday I watched a lot of TV because of all the crappy weather that we had. The best thing I watched was the golf touroment where Tiger Woods hit an unbelievable shot to clinch the victory. The problem I am working on for this blog is # 12 from the green version on test # 7. It asks for you to find the area of a parallelogram that has sides of 8 and 6. On the test, I got the problem wrong because I didn't know the area formula for parallelograms but when I went back and re-did the problem while studying for the retake I realized that I had to find the height by making a 45-45-90 triangle because the formula is a=bh. I found that the height was 3 root 2 and 8 time 3 root 2 is 24 root 2. Article posted June 7, 2012 at 04:31 PM GMT-5 • comment • Reads 279
Article posted June 7, 2012 at 04:49 PM GMT-5 • comment • Reads 356 I am blogging on a thursday for probably the first time, it seems wierd. I am getting ready to do my geometry homework. I am also going to watch the Miami Heat vs the Boston Celtics later, it is must win game for the Heat. The problem that I am going to work on in this blog is # 17 from the green version of test # 7. It asks you find the area of a deck that surrounds a hot tub if the hot tub has a diameter of 6 meters and the deck is 2 meters wide. When I took the test I knew how to do the problem but I just had a brain cramp and screwed up the final answer. The way to figure out this problem is to find the area of the hot tub and then subtract that from the area of the deck. The only formula that you need to use on this problem is a=pie(r) squared. The first thing to do is sub the number 3 into that formula and you end up with 9 pie. You then plug the number 5 into that formula because 6+ 2+2 equals 10 as a diameter and half of that equals 5. When plugging that into the equation you end up with 25 pie. The problem that I made on the test was that I forgot to subtract 9 pie from 25 pie so that is the last step in this problem. Your final answer should be 16 pie. Article posted June 7, 2012 at 04:49 PM GMT-5 • comment • Reads 356
Article posted June 10, 2012 at 10:26 PM GMT-5 • comment • Reads 224 This blog is the twenty-third of my personal weekly blog This blog was supposed to be posted last week but blogmeister was down! Oh no! We got everything squared away though, so yay! Last week was pretty gloomy, the weather hasn't been that nice until this weekend! Last week in geometry we started learning more in trigonometry and then started learning about law of sines and law of cosines! It's been pretty easy, but some things trip me up easily. Anyways the year is winding down and I can not wait until SUMMER! Midterms start this week! Here is a review question on what we have learned in Geometry! Ms. J told us that studying our tests and quizzes would be most helpful so that is exactly what I will be doing! A particular unit that I didn't fair so well on was working with the lines of triangles! Here is a review question from the Lines of Triangles Quiz that we took on February 7th. Which of the following are the slopes of two perpendicular lines? a. 3 and -3 b. 5 and 1/5 c. no slope and undefined d. -2/3 and 3/2 If you recall that the slope of any line perpendicular to another is the negative reciprocal, then this problem is easy to solve! A is not the answer because the negative reciprocal or 3 would be -1/3. B does not work because the negative reciprocal of 5 would be -1/5. c does not work because they aren't specific lines! THEREFORE D IS THE ANSWER! and to prove it.. the negative reciprocal of -2/3 would be - ( -3/2) or 3/2! TADA! Article posted June 10, 2012 at 10:26 PM GMT-5 • comment • Reads 224
Article posted June 10, 2012 at 10:34 PM GMT-5 • comment • Reads 419 This is the twenty-fourth of my personal weekly blog. Alas I have reached my last blog. Farewell classblogmeister... I won't miss you too much I promise! This week in Geometry we continued working with the law of sines and the law of cosines! We have a test this week before our final so that means a lot of studying on our parts! We pretty much did a lot of reviewing on the all the parts of trigonometry, which was helpful! This review question will be coming from the Quadrilaterals test we took on March 29th. I GOT THIS QUESTION WRONG! I honestly have no idea how I could've made the mistake, but I did, so make sure you don't jump to conclusions! Consecutive sides of a rectangle are congruent. a. sometimes b. always c. never the answer is SOMETIMES because a square (which is a rectangle) has congruent consecutive sides! YAY GEOMETRY!(: Article posted June 10, 2012 at 10:34 PM GMT-5 • comment • Reads 419
Article posted June 10, 2012 at 10:32 AM GMT-5 • comment • Reads 224 For Ch 12, the work was on transformations. The review question will be for rotations. The question as what is the point of rotation, and what is the angle of rotation. When given a pre-image and an image, to find the point of rotation, it will either be where the two images intersect or a point not connected to either image. Then to find the angle of rotation, you will need a protractor. Put your protractor on the point of rotation and pick a point you want to measure from. You always measure from the pre-image and counterclockwise unless stated to go clockwise. After you pick the point, find the point on the image that is the same, and measure. That will give you the angle of rotation Stay classy bloggers. Article posted June 10, 2012 at 10:32 AM GMT-5 • comment • Reads 224
Article posted June 10, 2012 at 11:24 AM GMT-5 • comment • Reads 217 Our final exam is Wednesday! Ahhh! And in preparation to that I am going to be posting a review question from one of the chapters we learned about this semester. In chapter 7 we learned more about areas of formulas. My review question today, will be finding the area of a rhombus (remember that the diagonals of a rhombus bisect each other). The formula for an area of a rhombus is 1/2(d1+d2), d stands for diagonal. If your given information was that one diagonal was 7cm and half of the other diagonal was 5.5cm, this is how you would solve it: We know that the other half of the diagonal is 5.5cm, because it gets bisected from the other diagonal. Now we plug in our given information to come up with 1/2(7+11) = area. Simplify to get 1/2(18) = area. Now simply multiply by 1/2 or divide by two, and we get 9cm squared = area Hopefully this helped you! I will be posting another review question soon! Goodbye! Article posted June 10, 2012 at 11:24 AM GMT-5 • comment • Reads 217
Article posted June 10, 2012 at 12:01 PM GMT-5 • comment • Reads 233 Hello again! Today, I am posting another review question. This comes from chapter 5. Question: Which of they following is the concurrent point of the altitudes? Answer: Orthocenter. I always remember which concurrent point goes with which line of a triangle because of MAPA COCI. MAPA COCI stands for: Median --> Centroid Altitude --> Orthocenter Perpendicular Bisector -->Circumcenter Angle Bisector --> Incenter This helped me remember and hopefully it will help you too! :) Now down to the sad stuff. :( I won't be saying 'see you next week' anymore..... because this is my last blog! *Gasp* I hope my blogs have helped you in some way or another, but right now I have to say goodbye for good. I know you'll miss me though! Farewell, Livy Article posted June 10, 2012 at 12:01 PM GMT-5 • comment • Reads 233
Article posted June 11, 2012 at 02:06 PM GMT-5 • comment • Reads 235 Question Given: Circle O with Diamter CD, AB is parrallel to CD, and arc AB=80 degrees Find arc CA [1] 50 [2] 60 [3] 80 [4] 100 Answer 50 Article posted June 11, 2012 at 02:06 PM GMT-5 • comment • Reads 235
Article posted June 11, 2012 at 02:10 PM GMT-5 • comment • Reads 381 Question Given: In triangle ABC, B=120, c=15, and a=15 Find C Answer 30 degrees Article posted June 11, 2012 at 02:10 PM GMT-5 • comment • Reads 381
Article posted June 11, 2012 at 09:04 PM GMT-5 • comment • Reads 241 The midterm is in two days and I'm getting a little nervous. We also have a test tomorrow that isn't helping with the nerves! Hopefully the test tomorrow will make get me feeling completely confident with the trigonometry unit, so it'll be one less thing to study for the final! My first review question that I want to go over is number 13 from practice 58. It is working with vectors which I needed a slight refresher on. The question: Homing pigeons have the ability or instinct to find their way home when released hundreds of miles away from home. Homing pigeons carried news of Olympic victories to various cities in ancient Greece. Suppose one such pigeon took off from Athens and landed in Sparta, which is 73 miles west and 64 miles south of Athens. Find the distance and its direction of flight. The first step you would need to take is plugging the coordinates into the distance formula. It comes out with 97, so Athens and Sparta are 97 miles apart. Then, you need to use the tangent ratio. This is the opposite side over the adjacent side. Therefore, tan(x) = 64/73 --> x = tan-1( 64/73) --> x = 41°, so the direction of flight is 41°. Article posted June 11, 2012 at 09:04 PM GMT-5 • comment • Reads 241
Article posted June 11, 2012 at 09:16 PM GMT-5 • comment • Reads 229 We were required two review questions, so I think I'll choose one from the beginning of the semester just as a refresher. Number 25 on the quadrilaterals test was asking us to label the coordinates of parallelogram LAST using only three variables. I chose d, c, and b (along with 0). Point 'a' was on the y-axis so the x coordinate was 0. Then, I chose 'b' to act as the height, or y coordinate. Point 'l' is the lower, left-hand point, and it is on the x-axis, making the y coordinate 0. Then, I made the x coordinate '-c' because it is to the left of the y-axis. Point 't' is the lower right-hand point, and it also lies on the x-axis, making the y coordinate 0. Then, I chose 'd' to act as the x coordinate. Point 's' was the upper right point of the parallelogram, so the height was also 'b', making the y coordinate automatically 'b'. The x coordinate is going to be 'd+c', because 'c' is the length that the segment goes on longer past 'd', which is where you get the '+c'. Hopefully that all made sense and was (slightly) helpful! Article posted June 11, 2012 at 09:16 PM GMT-5 • comment • Reads 229
Article posted June 10, 2012 at 09:00 PM GMT-5 • comment • Reads 222 This blog is supposed to be for two weeks ago but due to problems with the website the due date was postponed. This blog I will be explaining how to do a problem from one of the previous tests that I have taken. I have chosen to explain number 4 from test 7. The problem gives you a rhombus and tells you to find the area using the diagonals of 18 and 21. To find the area you must use the formula A=.5(d1)(d2). You would use substitution of make the equation A=.5(18)(21). If you do the math the area comes out to be 189ft squared. That is how you would do a problem like that... Article posted June 10, 2012 at 09:00 PM GMT-5 • comment • Reads 222
Article posted June 10, 2012 at 09:23 PM GMT-5 • comment • Reads 311 This week in geometry I am doing the same thing as last week. I have to pick a problem from any of my previous tests and I must explain how to do it.On problem 4 from the chapter 7 retest, the problem gives you a parallelogram and tells you to find the area using a height of 24 and a base of 10. The formula you must use is A= base times height. You would set up the equation as A= 24 times 10. The answer to the problem is 240 meters squared. That is how you do an problem like that... Article posted June 10, 2012 at 09:23 PM GMT-5 • comment • Reads 311
Article posted June 12, 2012 at 05:29 PM GMT-5 • comment • Reads 234 Question: The diagonals of a square bisect all angles Answer: Always Article posted June 12, 2012 at 05:29 PM GMT-5 • comment • Reads 234
Article posted June 12, 2012 at 05:31 PM GMT-5 • comment • Reads 312 Question: A triangle has side lengths of 8 cm, 14 cm, and 11 cm. Classify the triangle. Answer:Obtuse Article posted June 12, 2012 at 05:31 PM GMT-5 • comment • Reads 312
Article posted June 9, 2012 at 10:48 AM GMT-5 • comment • Reads 303 Hey guys so the school year's almost over. Yeah, summer. Unfortunately to get to summer we have to go through probably the wort week of the school year. Finals. Yeah you all know what I'm talking about. So here is one of the review questions that I'm using to study for my geometry final. This question is on right triangle trig. Good luck! Find X and H Answer: X=13 H=8.1 Article posted June 9, 2012 at 10:48 AM GMT-5 • comment • Reads 303
Article posted June 4, 2012 at 06:43 AM GMT-5 • comment • Reads 261 Hey guys so as I said in my last blog we are learning about SOH, CAH, TOA. Well now we are learning about the law of sines. Here is a review question and its answer about the law of sines. If m Answer: Measure of angle B is 59.4. Article posted June 4, 2012 at 06:43 AM GMT-5 • comment • Reads 261
Article posted June 7, 2012 at 09:17 PM GMT-5 • comment • Reads 224 We have finals in two weeks, and I don't really have any specific emotions for the end of the year. It was an awesome freshman year and a good first highschool experience. I hope next year will be as good, but then again I won't get to have Ms. J anymore:( To study for the final we must post a review question. Mine is- If a plane takes off at 25 degrees and flies 1600 ft, what is its altitude. SPOILER BELOW To do this problem, you must first write out the formula for sine. Using SOH, it would come out to be sin25=X/1600. Then you would multiply by 1600 on both sides to isolate X. This would make it 1600*sin(25)=X. Enter 1600*sin(25)into your calculator and the resulting number is the altitude-676.18 ft Article posted June 7, 2012 at 09:17 PM GMT-5 • comment • Reads 224
Article posted June 10, 2012 at 10:45 PM GMT-5 • comment • Reads 240 Hello everyone, I'm sad to tell you that this will be my last geometry blog :( The year is finally coming to an end. Six more days of school until summer! Here is my final exam review question: A triangle has three sides with the lengths of 6:8:10 What type of triangle is it? A.) acute B.)obtuse C.) right Answer: C. Right Article posted June 10, 2012 at 10:45 PM GMT-5 • comment • Reads 240
Article posted June 10, 2012 at 11:25 AM GMT-5 • comment • Reads 208 Hello Bloggers!!! This week I will be giving you a practice problem for you to solve!! At the bottom of the blog I will show you the work to get to the answer!! Question: You know two sides and an angle (6cm, 10cm, and 56 degrees) find the missing side. Answer: X2(squared)=6(squared)+10(squared) - 2(6)(10) - cos(56) by plugging this equation: 6(squared) + 10(squared) -2(6)(10) - cos(56) into your calculator, you will come up with X2(squared)= 15.4408071 you need to find X, so you have to find the square root. Type this into your calculator and X= 3.92947924 Great Job!!!! I will be blogging soon!!!! Article posted June 10, 2012 at 11:25 AM GMT-5 • comment • Reads 208
Article posted June 9, 2012 at 02:17 PM GMT-5 • comment • Reads 217 Hey bloggers! Just as I promised on Wednesday, here is another final review question! "In parallelogram BARK, m first draw the paralleglogram: 4x = xsquared - 60 xsqaured - 4x - 60 = 0 (x + 6)(x - 10) xsquared - 10x + 6x - 60 xsquared - 4x - 60 x + 6 = 0 or x - 10 = 0 x = -6 or x = 10 7x + m 70 + m m m< ARK = 110 dgrees Wish me luck on my final! Grace Article posted June 9, 2012 at 02:17 PM GMT-5 • comment • Reads 217
Article posted June 6, 2012 at 04:34 PM GMT-5 • comment • Reads 234 Hey bloggers! I am going to show you a good review question for our Geometry final next week! "Q is the intersection of AC and BD and ABCD. Find the area of kite ABCD if AB= 10, BC= 20, AC = 30, and BQ = 5." A= 1/2 x d1 x d2 d1= DB = 2BQ = 10. d2 = AC = 30 A = 1/2 x 10 x 30 Area = 150 squared units I will be posting another question later in the week due to the website being down last weekend. Later, Grace Article posted June 6, 2012 at 04:34 PM GMT-5 • comment • Reads 234
Article posted June 10, 2012 at 08:47 PM GMT-5 • comment • Reads 219 Welcome back everyone, since we are nearing the end of the school year, I was thinking that we could all start posting review questions to prepare ourselves for the upcoming final! What is a perpendicular bisector? And, how do you find it for any triangle? A perpendicular bisector is a line that divides a side of a triangle directly in the middle, but it bisects it at a 90˚ angle. Each side of the triangle has it’s own perpendicular bisector. In a triangle, there are 3 different sides each that have a perpendicular bisector. Since there are three different types of triangles (acute, right, and obtuse), each location of the circumcenter is different. For any acute triangle, the circumcenter is inside of the triangle. The point where all three of the Perpendicular bisectors intersect is at the circumcenter. But, like always there are 2 special cases. For a right triangle, the circumcenter of a right triangle is on the hypotenuse. But for an obtuse triangle: the circumcenter of an obtuse triangle is outside of the triangle. A circumscribed circle is a circle that is drawn from the circumcenter that hits all of the vertices of the triangle. The circumcenter is always equidistant to all vertices or any other point on the circumscribed circle. If you need anymore help with perpendicular bisectors reply to this post! I hope you all begin to review now! See you all next week! -Kathleen Article posted June 10, 2012 at 08:47 PM GMT-5 • comment • Reads 219
Article posted June 10, 2012 at 10:17 AM GMT-5 • comment • Reads 204 Hello again! As we are approaching our final exam next week, we are continuing to post review questions! These are aimed to help us remember the concepts we have learned this semester! The question I'm going to ask takes us back a chapter, and deals with triangles! Be sure to blog back with any questions. Q: A triangle has side lengths of 8cm, 11cm, and 14cm. Classify this triangle as acute, obtuse, or right. A: OBTUSE To find this answer, we first begin with the Pythagorean theorem (A squared + B squared = c squared). So, when we plug in what is known, our formula becomes 64 + 121 = 196. Here's the rule to classify: IF (A squared + b squared > C squared) then the triangle is acute. IF (A squared + b squared < C squared) then the triangle is obtuse. IF (A squared + b squared = C squared) then the triangle is right. In our formula, 64 + 121 = 185. Because 185 < 196, then the triangle is obtuse! Hope that helped and you learned something knew! This is my last weekly blog... but I'll be writing occasionally until the end of the year. Talk to you then! Emma Article posted June 10, 2012 at 10:17 AM GMT-5 • comment • Reads 204
Article posted June 14, 2012 at 12:40 PM GMT-5 • comment • Reads 240 We finally made it! Finals are around the corner, and so is summer! I can’t wait to ditch my backpack and head to the beach. But first, I have to survive finals. To review for Geometry, I’ve posted one of my favorite homework problems from a past unit below. KL⎮⎮JM in isosceles trapezoid JKLM. Find the values of x and y if m⦟J=(23x-8)º, m⦟K=(12y-13)º, and m⦟M=(17x+10)º. m⦟J=m⦟M 23x-8=17x+10 6x-8=10 6x=18 x=3 m⦟J+m⦟K=180 23x-8+12y-13=180 61+12y-13=180 12y=132 y=11 I chose this problem because I was from one of the first units of the semester, and I didn’t remember that unit at all. I thought that it would be nice to review since I had forgotten about it. Article posted June 14, 2012 at 12:40 PM GMT-5 • comment • Reads 240
Article posted June 14, 2012 at 12:40 PM GMT-5 • comment • Reads 244 We finally made it! Finals are around the corner, and so is summer! I can’t wait to ditch my backpack and head to the beach. But first, I have to survive finals. To review for Geometry, I’ve posted one of my favorite homework problems from a past unit below. KL⎮⎮JM in isosceles trapezoid JKLM. Find the values of x and y if m⦟J=(23x-8)º, m⦟K=(12y-13)º, and m⦟M=(17x+10)º. m⦟J=m⦟M 23x-8=17x+10 6x-8=10 6x=18 x=3 m⦟J+m⦟K=180 23x-8+12y-13=180 61+12y-13=180 12y=132 y=11 I chose this problem because I was from one of the first units of the semester, and I didn’t remember that unit at all. I thought that it would be nice to review since I had forgotten about it. Article posted June 14, 2012 at 12:40 PM GMT-5 • comment • Reads 244
Article posted June 6, 2012 at 04:33 PM GMT-5 • comment • Reads 257 Hello fellow blog readers! Sorry that this blog is so late compared to my others. Blogmeister, as I'm sure you know, has been down for the past couple of days. I am sorry to keep you waiting for so long. Now to the important stuff. In geometry, I have a question about the latest test. My question is how to find the length of an arc. I do not remember and was wondering if you bloggers could help me out. If you comment with the correct answer, I will mention you in my next blog. Well that is all for now blog-readers! Stay tuned for this Sunday's blog and question. Stay classy and thanks for reading! Article posted June 6, 2012 at 04:33 PM GMT-5 • comment • Reads 257
Article posted June 10, 2012 at 04:52 PM GMT-5 • comment • Reads 215 The year is almost over! We have two tests this next week in geometry. One tomorrow on chapter 9 and one on June 18th on everything we've learned this semester (our final). To help review for this final I have been told to blog a review question with the answer. Do you remember the distance formula? Well for coordinate proofs we had to use the distance formula to show how one side was of equal length to the other side. To show this you use the formula d = the square root of (y2-y1)squared + (x2-x1)squared. If you plug in the coordinates from #26 on the quadrilaterals test you get AS = the square root of (b-b)squared + (a+c-0)squared. This simplifies to the square root of (a+c)squared which equals a + c. Hope this was helpful, good luck on the test and the final! Article posted June 10, 2012 at 04:52 PM GMT-5 • comment • Reads 215
Article posted June 10, 2012 at 04:11 PM GMT-5 • comment • Reads 218 AHHHH! Three more days until final exams! Before I show you the review question take a deep breath... Inhale, now exhale. Feel better? Good. Let's begin! This is from chapter 9. This chapter is on SOH CAH TOA, the law of sines, the law of cosines, and vectors. Say you had a triangle where the length of the hypotenuse (AC) is 12cm, and the short side (AB) and the long side (BC) is unknown. Angle C is given as 23 degrees and angle B is 90 degrees. To solve this problem you use CAH because you are given a right triangle, an angle (23 degrees), the hypotenuse to that angle (12cm), and the long side is present. To set it up you write; cos23 = X/12. To solve you multiple cos23 by 12, so it's 12cos23 = X. X = 11cm. I hope this was helpful! Article posted June 10, 2012 at 04:11 PM GMT-5 • comment • Reads 218
Article posted June 10, 2012 at 08:54 PM GMT-5 • comment • Reads 216 Hello everyone! Hope your week has gone nicely! Mine has been pretty interesting, because my brother had an English exchange student arrive on Thursday. He will be staying with us for the week, and I am really excited! English accents are so cool! This exchange trip has also reminded me that the end of the school year is really close! In order to prepare for the geometry final, I am going to tell you about a question we had to do for homework as a way to review. This problem was on a worksheet I did recently involving the Law of Sines: sinA = sinB a b A and B stand for angles of an oblique triangle, while a and b stand for the 2 sides opposite them. Basically, you can use this law to find one of these four measures, as long as you have an angle and the side opposite. Basically, the problem was a triangle with angles A, B, and C, and sides a, b, and c. The measure of angle A was 35º, the measure of side a was 8 cm, and the measure of side b was 12 cm. The problem asked you to find the measure of angle B. Through substitution, I came out with this equation: sin(35) = sin(B) 8 12 By multiplying both sides by 12 and then by sin to the (-1), I was able to then solve for B: sin-1(12sin(35)) = sin-1(sin(B)) 8 m This process took some time to get used to, but I've found I actually like this sorts of problems in trigonometry. Let me know if you have any questions! Have a great week! (: Article posted June 10, 2012 at 08:54 PM GMT-5 • comment • Reads 216
Article posted June 10, 2012 at 09:02 PM GMT-5 • comment • Reads 207 Hello everyone! Firstly, I would just like to explain that last week Blogmeister was experiencing some technical difficulties, so I was unable to post at that time. The post from both then and the most recent one should be uploaded now. Anyways, this week is the beginning of finals for my school! I am very nervous because I've never taken finals before, and I'm not sure what to expect. Wish me luck! In order to review once again for geometry, I am posting another review questions below. I chose this particular one because I find this area formula hard to remember sometimes, because it doesn't involve side lengths, rather, diagonal lengths. For this problem, there was a rhombus shown with diagonals of 21 ft and 18 ft. The instructions were to find the area, and the possible answers were as shown: a) 378 sq. ft b) 189 sq. ft c) 162 sq. ft d) 27(square root of)85 sq. ft. I knew that the formula for the area of a rhombus is 1/2d1*d2, so I substituted in the diameter lengths: A = 1/2 * 21 * 18 = 189 With that equation, I was able to find that the area was 189 sq. ft., or answer b. I hope this has been a helpful review question! Have a great week! (: Article posted June 10, 2012 at 09:02 PM GMT-5 • comment • Reads 207
Article posted June 11, 2012 at 08:18 AM GMT-5 • comment • Reads 256 Well, the final is coming up for geometry this. Just one more test that I need to study for. I can't say that I enjoy these types of things, but I still need to do them. To help out everybody, though, here are two problems that were on homework from this term. There are two since the blog wasn't working last week when I was supposed to post one and one from this week, so I'm just combining the two blogs into this week's. Here they are: The lengths of the diagonals of a rhombus are 2 in. and 5 in. Find the measures of the angles of the rhombus to the nearest degree. Describe each translation using an ordered pair. 2 units to the left, 1 unit down. Good luck to everyone on the final! Article posted June 11, 2012 at 08:18 AM GMT-5 • comment • Reads 256
Article posted June 10, 2012 at 07:04 PM GMT-5 • comment • Reads 190 It seems like just yesterday that we were beginning our weekly blogs, and now we are writing our last ones. This is the week of finals, and after this week, we only have one day left of school. I cannot believe that school went by this fast. I would have to say that this school year was filled with learning, and fun! On another note, we still must remain serious. We still have our final exam on Monday, and that requires a gracious amount of studying. To study for the exam, I have come up with a few review guides that will help me go over anything that I have been struggling with. One thing that I have had a bit of trouble with is vectors. Therefore, I picked out a problem about vectors from my book and worked out the answer to it. The problem was "describe each vector as an ordered pair". To do so, you had to first figure out the sine ratio as well as the cosine ratio. Then, you would order them so that they describe the vector. It took some practice to understand, but eventually, I got it! I now feel prepared for this final exam! Article posted June 10, 2012 at 07:04 PM GMT-5 • comment • Reads 190
Article posted June 4, 2012 at 09:38 AM GMT-5 • comment • Reads 251 Salutations fellow bloggers! This week in geometry class we learned about the law of sine and cosine. It isn't to difficult but I don't really understand how it saves you time. My question for the final review is: What quadrilaterals have diagonals that bisect each other? Answer: Parallelogram, rhombus, rectangle, and square Have a great week everyone! Article posted June 4, 2012 at 09:38 AM GMT-5 • comment • Reads 251
Article posted June 5, 2012 at 07:53 PM GMT-5 • comment • Reads 237 Hi everyone! Just writing because before I could add my review question, Blogmeister shut itself down! NOOOOOO!!!! Well, anyway, here’s my Trigonometry Review question: Kevin and Zane have just built a tree house. It is 20 ft off the ground, and the tree is perpendicular to the ground. When Kevin’s Mom goes out to look at the tree house, the angle of elevation from her feet to the tree house is 50*. Zane wants to know how far the tree that has the tree house in it is from the house, but he has lost the tape measure. How far away is the tree from the house? ***SPOILER ALERT*** Answer: Roughly 17 Feet --Joe Article posted June 5, 2012 at 07:53 PM GMT-5 • comment • Reads 237
Article posted June 10, 2012 at 07:42 AM GMT-5 • comment • Reads 198 HELLO! WELCOME TO THE FINAL COMPLETE WEEK OF SCHOOL! Wow, the year has gone by so fast! Did I mention that this is our final weekly blog? I can’t believe how close we are to the end of the year! It seems like just yesterday we walked into Geometry for the first time. I have had a great weekend, capped of by a RockEucharist at my church! This week in Geometry, we continued with our trigonometry unit, and we learned the law of cosines. At first I didn’t understand it, but once I got it, I had it for good. I think that on our trig. test this week, I will not have that much trouble, as this unit has been about mostly algebra and I do not have much trouble with algebra. But you never know-- it may be very hard. Our final is also coming up. Wow, that snuck up on me. Here is another review question: Ben and Griffin love modern houses. They want to make a museum about the evolution of houses. While searching for a place to put the museum, they find a modern building that is a rhombus. The sides of the house are 50 feet long. The angles formed by two intersecting lines are 120* and 60*. How much area is in the museum? ANSWER: 4330 ft. squared Have a great last week! --Joe Article posted June 10, 2012 at 07:42 AM GMT-5 • comment • Reads 198
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Article posted June 2, 2012 at 07:33 AM GMT-5 • comment • Reads 539 Today is Sunday, June 10, and I'm doing this blog a week late because the website crashed last week. It also means that there are exactly 2 school days until Finals start! I'm starting to worry it will be really hard to study without being distracted by the nice weather. To get in the spirit of studying, I'm going to post a review question from one of my past tests today on this blog. This is it: Determine which three numbers could be the sides of a right triangle. A. 64,73,98 B. 64,72,96 C. 65,72,97 The answer is C because of the Pythagorean theorem. When you square each number, the first two must add up to the third one's square. If that happens, then it is a right triangle. Choice C is the only one that happens for, so it is the correct answer. Until Next week, Article posted June 2, 2012 at 07:33 AM GMT-5 • comment • Reads 539
Article posted June 10, 2012 at 07:03 AM GMT-5 • comment • Reads 309 I'm doing the second blog for this week today, Sunday, June 10 because the website crashed last week. As I said earlier, there are exactly two school days until finals start! While this may seem terrible, I'm actually very happy because it means that we are in the home stretch before summer! I'm posting another review question this week: this will be about lines in triangles. Which line in triangles meet at the orthocenter when all three are drawn in a triangle? The answer to this question is altitudes. An altitude, also known as the height of a triangle, is a perpendicular line to the base drawn from a vertex. School may be almost done, but I can't give up now. This is the last blog of this year, and I can say that's a good thing! Article posted June 10, 2012 at 07:03 AM GMT-5 • comment • Reads 309
Article posted June 10, 2012 at 07:40 PM GMT-5 • comment • Reads 291 For this weeks blog I was asked to post two review questions for the final exam. The final exam is this week and I need to start studying, so I am going to start by making these review questions. My first question is: What does MAPA COCI stand for? A hint for this question is to remember lines in triangles and their concurrent points! My next question is: How do you find the circumference of a circle with a radius of 10 cm? Answers: Question 1: Median, Altitude, Perpendicular Bisector, Angle Bisector. Centroid, Orthocenter, Circumcenter, Incenter. Question 2: 20pi cm. Article posted June 10, 2012 at 07:40 PM GMT-5 • comment • Reads 291
Article posted June 7, 2012 at 04:31 PM GMT-5 • comment • Reads 279 On Sunday I watched a lot of TV because of all the crappy weather that we had. The best thing I watched was the golf touroment where Tiger Woods hit an unbelievable shot to clinch the victory. The problem I am working on for this blog is # 12 from the green version on test # 7. It asks for you to find the area of a parallelogram that has sides of 8 and 6. On the test, I got the problem wrong because I didn't know the area formula for parallelograms but when I went back and re-did the problem while studying for the retake I realized that I had to find the height by making a 45-45-90 triangle because the formula is a=bh. I found that the height was 3 root 2 and 8 time 3 root 2 is 24 root 2. Article posted June 7, 2012 at 04:31 PM GMT-5 • comment • Reads 279
Article posted June 7, 2012 at 04:49 PM GMT-5 • comment • Reads 356 I am blogging on a thursday for probably the first time, it seems wierd. I am getting ready to do my geometry homework. I am also going to watch the Miami Heat vs the Boston Celtics later, it is must win game for the Heat. The problem that I am going to work on in this blog is # 17 from the green version of test # 7. It asks you find the area of a deck that surrounds a hot tub if the hot tub has a diameter of 6 meters and the deck is 2 meters wide. When I took the test I knew how to do the problem but I just had a brain cramp and screwed up the final answer. The way to figure out this problem is to find the area of the hot tub and then subtract that from the area of the deck. The only formula that you need to use on this problem is a=pie(r) squared. The first thing to do is sub the number 3 into that formula and you end up with 9 pie. You then plug the number 5 into that formula because 6+ 2+2 equals 10 as a diameter and half of that equals 5. When plugging that into the equation you end up with 25 pie. The problem that I made on the test was that I forgot to subtract 9 pie from 25 pie so that is the last step in this problem. Your final answer should be 16 pie. Article posted June 7, 2012 at 04:49 PM GMT-5 • comment • Reads 356
Article posted June 10, 2012 at 10:26 PM GMT-5 • comment • Reads 224 This blog is the twenty-third of my personal weekly blog This blog was supposed to be posted last week but blogmeister was down! Oh no! We got everything squared away though, so yay! Last week was pretty gloomy, the weather hasn't been that nice until this weekend! Last week in geometry we started learning more in trigonometry and then started learning about law of sines and law of cosines! It's been pretty easy, but some things trip me up easily. Anyways the year is winding down and I can not wait until SUMMER! Midterms start this week! Here is a review question on what we have learned in Geometry! Ms. J told us that studying our tests and quizzes would be most helpful so that is exactly what I will be doing! A particular unit that I didn't fair so well on was working with the lines of triangles! Here is a review question from the Lines of Triangles Quiz that we took on February 7th. Which of the following are the slopes of two perpendicular lines? a. 3 and -3 b. 5 and 1/5 c. no slope and undefined d. -2/3 and 3/2 If you recall that the slope of any line perpendicular to another is the negative reciprocal, then this problem is easy to solve! A is not the answer because the negative reciprocal or 3 would be -1/3. B does not work because the negative reciprocal of 5 would be -1/5. c does not work because they aren't specific lines! THEREFORE D IS THE ANSWER! and to prove it.. the negative reciprocal of -2/3 would be - ( -3/2) or 3/2! TADA! Article posted June 10, 2012 at 10:26 PM GMT-5 • comment • Reads 224
Article posted June 10, 2012 at 10:34 PM GMT-5 • comment • Reads 419 This is the twenty-fourth of my personal weekly blog. Alas I have reached my last blog. Farewell classblogmeister... I won't miss you too much I promise! This week in Geometry we continued working with the law of sines and the law of cosines! We have a test this week before our final so that means a lot of studying on our parts! We pretty much did a lot of reviewing on the all the parts of trigonometry, which was helpful! This review question will be coming from the Quadrilaterals test we took on March 29th. I GOT THIS QUESTION WRONG! I honestly have no idea how I could've made the mistake, but I did, so make sure you don't jump to conclusions! Consecutive sides of a rectangle are congruent. a. sometimes b. always c. never the answer is SOMETIMES because a square (which is a rectangle) has congruent consecutive sides! YAY GEOMETRY!(: Article posted June 10, 2012 at 10:34 PM GMT-5 • comment • Reads 419
Article posted June 10, 2012 at 10:32 AM GMT-5 • comment • Reads 224 For Ch 12, the work was on transformations. The review question will be for rotations. The question as what is the point of rotation, and what is the angle of rotation. When given a pre-image and an image, to find the point of rotation, it will either be where the two images intersect or a point not connected to either image. Then to find the angle of rotation, you will need a protractor. Put your protractor on the point of rotation and pick a point you want to measure from. You always measure from the pre-image and counterclockwise unless stated to go clockwise. After you pick the point, find the point on the image that is the same, and measure. That will give you the angle of rotation Stay classy bloggers. Article posted June 10, 2012 at 10:32 AM GMT-5 • comment • Reads 224
Article posted June 10, 2012 at 11:24 AM GMT-5 • comment • Reads 217 Our final exam is Wednesday! Ahhh! And in preparation to that I am going to be posting a review question from one of the chapters we learned about this semester. In chapter 7 we learned more about areas of formulas. My review question today, will be finding the area of a rhombus (remember that the diagonals of a rhombus bisect each other). The formula for an area of a rhombus is 1/2(d1+d2), d stands for diagonal. If your given information was that one diagonal was 7cm and half of the other diagonal was 5.5cm, this is how you would solve it: We know that the other half of the diagonal is 5.5cm, because it gets bisected from the other diagonal. Now we plug in our given information to come up with 1/2(7+11) = area. Simplify to get 1/2(18) = area. Now simply multiply by 1/2 or divide by two, and we get 9cm squared = area Hopefully this helped you! I will be posting another review question soon! Goodbye! Article posted June 10, 2012 at 11:24 AM GMT-5 • comment • Reads 217
Article posted June 10, 2012 at 12:01 PM GMT-5 • comment • Reads 233 Hello again! Today, I am posting another review question. This comes from chapter 5. Question: Which of they following is the concurrent point of the altitudes? Answer: Orthocenter. I always remember which concurrent point goes with which line of a triangle because of MAPA COCI. MAPA COCI stands for: Median --> Centroid Altitude --> Orthocenter Perpendicular Bisector -->Circumcenter Angle Bisector --> Incenter This helped me remember and hopefully it will help you too! :) Now down to the sad stuff. :( I won't be saying 'see you next week' anymore..... because this is my last blog! *Gasp* I hope my blogs have helped you in some way or another, but right now I have to say goodbye for good. I know you'll miss me though! Farewell, Livy Article posted June 10, 2012 at 12:01 PM GMT-5 • comment • Reads 233
Article posted June 11, 2012 at 02:06 PM GMT-5 • comment • Reads 235 Question Given: Circle O with Diamter CD, AB is parrallel to CD, and arc AB=80 degrees Find arc CA [1] 50 [2] 60 [3] 80 [4] 100 Answer 50 Article posted June 11, 2012 at 02:06 PM GMT-5 • comment • Reads 235
Article posted June 11, 2012 at 02:10 PM GMT-5 • comment • Reads 381 Question Given: In triangle ABC, B=120, c=15, and a=15 Find C Answer 30 degrees Article posted June 11, 2012 at 02:10 PM GMT-5 • comment • Reads 381
Article posted June 11, 2012 at 09:04 PM GMT-5 • comment • Reads 241 The midterm is in two days and I'm getting a little nervous. We also have a test tomorrow that isn't helping with the nerves! Hopefully the test tomorrow will make get me feeling completely confident with the trigonometry unit, so it'll be one less thing to study for the final! My first review question that I want to go over is number 13 from practice 58. It is working with vectors which I needed a slight refresher on. The question: Homing pigeons have the ability or instinct to find their way home when released hundreds of miles away from home. Homing pigeons carried news of Olympic victories to various cities in ancient Greece. Suppose one such pigeon took off from Athens and landed in Sparta, which is 73 miles west and 64 miles south of Athens. Find the distance and its direction of flight. The first step you would need to take is plugging the coordinates into the distance formula. It comes out with 97, so Athens and Sparta are 97 miles apart. Then, you need to use the tangent ratio. This is the opposite side over the adjacent side. Therefore, tan(x) = 64/73 --> x = tan-1( 64/73) --> x = 41°, so the direction of flight is 41°. Article posted June 11, 2012 at 09:04 PM GMT-5 • comment • Reads 241
Article posted June 11, 2012 at 09:16 PM GMT-5 • comment • Reads 229 We were required two review questions, so I think I'll choose one from the beginning of the semester just as a refresher. Number 25 on the quadrilaterals test was asking us to label the coordinates of parallelogram LAST using only three variables. I chose d, c, and b (along with 0). Point 'a' was on the y-axis so the x coordinate was 0. Then, I chose 'b' to act as the height, or y coordinate. Point 'l' is the lower, left-hand point, and it is on the x-axis, making the y coordinate 0. Then, I made the x coordinate '-c' because it is to the left of the y-axis. Point 't' is the lower right-hand point, and it also lies on the x-axis, making the y coordinate 0. Then, I chose 'd' to act as the x coordinate. Point 's' was the upper right point of the parallelogram, so the height was also 'b', making the y coordinate automatically 'b'. The x coordinate is going to be 'd+c', because 'c' is the length that the segment goes on longer past 'd', which is where you get the '+c'. Hopefully that all made sense and was (slightly) helpful! Article posted June 11, 2012 at 09:16 PM GMT-5 • comment • Reads 229
Article posted June 10, 2012 at 09:00 PM GMT-5 • comment • Reads 222 This blog is supposed to be for two weeks ago but due to problems with the website the due date was postponed. This blog I will be explaining how to do a problem from one of the previous tests that I have taken. I have chosen to explain number 4 from test 7. The problem gives you a rhombus and tells you to find the area using the diagonals of 18 and 21. To find the area you must use the formula A=.5(d1)(d2). You would use substitution of make the equation A=.5(18)(21). If you do the math the area comes out to be 189ft squared. That is how you would do a problem like that... Article posted June 10, 2012 at 09:00 PM GMT-5 • comment • Reads 222
Article posted June 10, 2012 at 09:23 PM GMT-5 • comment • Reads 311 This week in geometry I am doing the same thing as last week. I have to pick a problem from any of my previous tests and I must explain how to do it.On problem 4 from the chapter 7 retest, the problem gives you a parallelogram and tells you to find the area using a height of 24 and a base of 10. The formula you must use is A= base times height. You would set up the equation as A= 24 times 10. The answer to the problem is 240 meters squared. That is how you do an problem like that... Article posted June 10, 2012 at 09:23 PM GMT-5 • comment • Reads 311
Article posted June 12, 2012 at 05:29 PM GMT-5 • comment • Reads 234 Question: The diagonals of a square bisect all angles Answer: Always Article posted June 12, 2012 at 05:29 PM GMT-5 • comment • Reads 234
Article posted June 12, 2012 at 05:31 PM GMT-5 • comment • Reads 312 Question: A triangle has side lengths of 8 cm, 14 cm, and 11 cm. Classify the triangle. Answer:Obtuse Article posted June 12, 2012 at 05:31 PM GMT-5 • comment • Reads 312
Article posted June 9, 2012 at 10:48 AM GMT-5 • comment • Reads 303 Hey guys so the school year's almost over. Yeah, summer. Unfortunately to get to summer we have to go through probably the wort week of the school year. Finals. Yeah you all know what I'm talking about. So here is one of the review questions that I'm using to study for my geometry final. This question is on right triangle trig. Good luck! Find X and H Answer: X=13 H=8.1 Article posted June 9, 2012 at 10:48 AM GMT-5 • comment • Reads 303
Article posted June 4, 2012 at 06:43 AM GMT-5 • comment • Reads 261 Hey guys so as I said in my last blog we are learning about SOH, CAH, TOA. Well now we are learning about the law of sines. Here is a review question and its answer about the law of sines. If m Answer: Measure of angle B is 59.4. Article posted June 4, 2012 at 06:43 AM GMT-5 • comment • Reads 261
Article posted June 7, 2012 at 09:17 PM GMT-5 • comment • Reads 224 We have finals in two weeks, and I don't really have any specific emotions for the end of the year. It was an awesome freshman year and a good first highschool experience. I hope next year will be as good, but then again I won't get to have Ms. J anymore:( To study for the final we must post a review question. Mine is- If a plane takes off at 25 degrees and flies 1600 ft, what is its altitude. SPOILER BELOW To do this problem, you must first write out the formula for sine. Using SOH, it would come out to be sin25=X/1600. Then you would multiply by 1600 on both sides to isolate X. This would make it 1600*sin(25)=X. Enter 1600*sin(25)into your calculator and the resulting number is the altitude-676.18 ft Article posted June 7, 2012 at 09:17 PM GMT-5 • comment • Reads 224
Article posted June 10, 2012 at 10:45 PM GMT-5 • comment • Reads 240 Hello everyone, I'm sad to tell you that this will be my last geometry blog :( The year is finally coming to an end. Six more days of school until summer! Here is my final exam review question: A triangle has three sides with the lengths of 6:8:10 What type of triangle is it? A.) acute B.)obtuse C.) right Answer: C. Right Article posted June 10, 2012 at 10:45 PM GMT-5 • comment • Reads 240
Article posted June 10, 2012 at 11:25 AM GMT-5 • comment • Reads 208 Hello Bloggers!!! This week I will be giving you a practice problem for you to solve!! At the bottom of the blog I will show you the work to get to the answer!! Question: You know two sides and an angle (6cm, 10cm, and 56 degrees) find the missing side. Answer: X2(squared)=6(squared)+10(squared) - 2(6)(10) - cos(56) by plugging this equation: 6(squared) + 10(squared) -2(6)(10) - cos(56) into your calculator, you will come up with X2(squared)= 15.4408071 you need to find X, so you have to find the square root. Type this into your calculator and X= 3.92947924 Great Job!!!! I will be blogging soon!!!! Article posted June 10, 2012 at 11:25 AM GMT-5 • comment • Reads 208
Article posted June 9, 2012 at 02:17 PM GMT-5 • comment • Reads 217 Hey bloggers! Just as I promised on Wednesday, here is another final review question! "In parallelogram BARK, m first draw the paralleglogram: 4x = xsquared - 60 xsqaured - 4x - 60 = 0 (x + 6)(x - 10) xsquared - 10x + 6x - 60 xsquared - 4x - 60 x + 6 = 0 or x - 10 = 0 x = -6 or x = 10 7x + m 70 + m m m< ARK = 110 dgrees Wish me luck on my final! Grace Article posted June 9, 2012 at 02:17 PM GMT-5 • comment • Reads 217
Article posted June 6, 2012 at 04:34 PM GMT-5 • comment • Reads 234 Hey bloggers! I am going to show you a good review question for our Geometry final next week! "Q is the intersection of AC and BD and ABCD. Find the area of kite ABCD if AB= 10, BC= 20, AC = 30, and BQ = 5." A= 1/2 x d1 x d2 d1= DB = 2BQ = 10. d2 = AC = 30 A = 1/2 x 10 x 30 Area = 150 squared units I will be posting another question later in the week due to the website being down last weekend. Later, Grace Article posted June 6, 2012 at 04:34 PM GMT-5 • comment • Reads 234
Article posted June 10, 2012 at 08:47 PM GMT-5 • comment • Reads 219 Welcome back everyone, since we are nearing the end of the school year, I was thinking that we could all start posting review questions to prepare ourselves for the upcoming final! What is a perpendicular bisector? And, how do you find it for any triangle? A perpendicular bisector is a line that divides a side of a triangle directly in the middle, but it bisects it at a 90˚ angle. Each side of the triangle has it’s own perpendicular bisector. In a triangle, there are 3 different sides each that have a perpendicular bisector. Since there are three different types of triangles (acute, right, and obtuse), each location of the circumcenter is different. For any acute triangle, the circumcenter is inside of the triangle. The point where all three of the Perpendicular bisectors intersect is at the circumcenter. But, like always there are 2 special cases. For a right triangle, the circumcenter of a right triangle is on the hypotenuse. But for an obtuse triangle: the circumcenter of an obtuse triangle is outside of the triangle. A circumscribed circle is a circle that is drawn from the circumcenter that hits all of the vertices of the triangle. The circumcenter is always equidistant to all vertices or any other point on the circumscribed circle. If you need anymore help with perpendicular bisectors reply to this post! I hope you all begin to review now! See you all next week! -Kathleen Article posted June 10, 2012 at 08:47 PM GMT-5 • comment • Reads 219
Article posted June 10, 2012 at 10:17 AM GMT-5 • comment • Reads 204 Hello again! As we are approaching our final exam next week, we are continuing to post review questions! These are aimed to help us remember the concepts we have learned this semester! The question I'm going to ask takes us back a chapter, and deals with triangles! Be sure to blog back with any questions. Q: A triangle has side lengths of 8cm, 11cm, and 14cm. Classify this triangle as acute, obtuse, or right. A: OBTUSE To find this answer, we first begin with the Pythagorean theorem (A squared + B squared = c squared). So, when we plug in what is known, our formula becomes 64 + 121 = 196. Here's the rule to classify: IF (A squared + b squared > C squared) then the triangle is acute. IF (A squared + b squared < C squared) then the triangle is obtuse. IF (A squared + b squared = C squared) then the triangle is right. In our formula, 64 + 121 = 185. Because 185 < 196, then the triangle is obtuse! Hope that helped and you learned something knew! This is my last weekly blog... but I'll be writing occasionally until the end of the year. Talk to you then! Emma Article posted June 10, 2012 at 10:17 AM GMT-5 • comment • Reads 204
Article posted June 14, 2012 at 12:40 PM GMT-5 • comment • Reads 240 We finally made it! Finals are around the corner, and so is summer! I can’t wait to ditch my backpack and head to the beach. But first, I have to survive finals. To review for Geometry, I’ve posted one of my favorite homework problems from a past unit below. KL⎮⎮JM in isosceles trapezoid JKLM. Find the values of x and y if m⦟J=(23x-8)º, m⦟K=(12y-13)º, and m⦟M=(17x+10)º. m⦟J=m⦟M 23x-8=17x+10 6x-8=10 6x=18 x=3 m⦟J+m⦟K=180 23x-8+12y-13=180 61+12y-13=180 12y=132 y=11 I chose this problem because I was from one of the first units of the semester, and I didn’t remember that unit at all. I thought that it would be nice to review since I had forgotten about it. Article posted June 14, 2012 at 12:40 PM GMT-5 • comment • Reads 240
Article posted June 14, 2012 at 12:40 PM GMT-5 • comment • Reads 244 We finally made it! Finals are around the corner, and so is summer! I can’t wait to ditch my backpack and head to the beach. But first, I have to survive finals. To review for Geometry, I’ve posted one of my favorite homework problems from a past unit below. KL⎮⎮JM in isosceles trapezoid JKLM. Find the values of x and y if m⦟J=(23x-8)º, m⦟K=(12y-13)º, and m⦟M=(17x+10)º. m⦟J=m⦟M 23x-8=17x+10 6x-8=10 6x=18 x=3 m⦟J+m⦟K=180 23x-8+12y-13=180 61+12y-13=180 12y=132 y=11 I chose this problem because I was from one of the first units of the semester, and I didn’t remember that unit at all. I thought that it would be nice to review since I had forgotten about it. Article posted June 14, 2012 at 12:40 PM GMT-5 • comment • Reads 244
Article posted June 6, 2012 at 04:33 PM GMT-5 • comment • Reads 257 Hello fellow blog readers! Sorry that this blog is so late compared to my others. Blogmeister, as I'm sure you know, has been down for the past couple of days. I am sorry to keep you waiting for so long. Now to the important stuff. In geometry, I have a question about the latest test. My question is how to find the length of an arc. I do not remember and was wondering if you bloggers could help me out. If you comment with the correct answer, I will mention you in my next blog. Well that is all for now blog-readers! Stay tuned for this Sunday's blog and question. Stay classy and thanks for reading! Article posted June 6, 2012 at 04:33 PM GMT-5 • comment • Reads 257
Article posted June 10, 2012 at 04:52 PM GMT-5 • comment • Reads 215 The year is almost over! We have two tests this next week in geometry. One tomorrow on chapter 9 and one on June 18th on everything we've learned this semester (our final). To help review for this final I have been told to blog a review question with the answer. Do you remember the distance formula? Well for coordinate proofs we had to use the distance formula to show how one side was of equal length to the other side. To show this you use the formula d = the square root of (y2-y1)squared + (x2-x1)squared. If you plug in the coordinates from #26 on the quadrilaterals test you get AS = the square root of (b-b)squared + (a+c-0)squared. This simplifies to the square root of (a+c)squared which equals a + c. Hope this was helpful, good luck on the test and the final! Article posted June 10, 2012 at 04:52 PM GMT-5 • comment • Reads 215
Article posted June 10, 2012 at 04:11 PM GMT-5 • comment • Reads 218 AHHHH! Three more days until final exams! Before I show you the review question take a deep breath... Inhale, now exhale. Feel better? Good. Let's begin! This is from chapter 9. This chapter is on SOH CAH TOA, the law of sines, the law of cosines, and vectors. Say you had a triangle where the length of the hypotenuse (AC) is 12cm, and the short side (AB) and the long side (BC) is unknown. Angle C is given as 23 degrees and angle B is 90 degrees. To solve this problem you use CAH because you are given a right triangle, an angle (23 degrees), the hypotenuse to that angle (12cm), and the long side is present. To set it up you write; cos23 = X/12. To solve you multiple cos23 by 12, so it's 12cos23 = X. X = 11cm. I hope this was helpful! Article posted June 10, 2012 at 04:11 PM GMT-5 • comment • Reads 218
Article posted June 10, 2012 at 08:54 PM GMT-5 • comment • Reads 216 Hello everyone! Hope your week has gone nicely! Mine has been pretty interesting, because my brother had an English exchange student arrive on Thursday. He will be staying with us for the week, and I am really excited! English accents are so cool! This exchange trip has also reminded me that the end of the school year is really close! In order to prepare for the geometry final, I am going to tell you about a question we had to do for homework as a way to review. This problem was on a worksheet I did recently involving the Law of Sines: sinA = sinB a b A and B stand for angles of an oblique triangle, while a and b stand for the 2 sides opposite them. Basically, you can use this law to find one of these four measures, as long as you have an angle and the side opposite. Basically, the problem was a triangle with angles A, B, and C, and sides a, b, and c. The measure of angle A was 35º, the measure of side a was 8 cm, and the measure of side b was 12 cm. The problem asked you to find the measure of angle B. Through substitution, I came out with this equation: sin(35) = sin(B) 8 12 By multiplying both sides by 12 and then by sin to the (-1), I was able to then solve for B: sin-1(12sin(35)) = sin-1(sin(B)) 8 m This process took some time to get used to, but I've found I actually like this sorts of problems in trigonometry. Let me know if you have any questions! Have a great week! (: Article posted June 10, 2012 at 08:54 PM GMT-5 • comment • Reads 216
Article posted June 10, 2012 at 09:02 PM GMT-5 • comment • Reads 207 Hello everyone! Firstly, I would just like to explain that last week Blogmeister was experiencing some technical difficulties, so I was unable to post at that time. The post from both then and the most recent one should be uploaded now. Anyways, this week is the beginning of finals for my school! I am very nervous because I've never taken finals before, and I'm not sure what to expect. Wish me luck! In order to review once again for geometry, I am posting another review questions below. I chose this particular one because I find this area formula hard to remember sometimes, because it doesn't involve side lengths, rather, diagonal lengths. For this problem, there was a rhombus shown with diagonals of 21 ft and 18 ft. The instructions were to find the area, and the possible answers were as shown: a) 378 sq. ft b) 189 sq. ft c) 162 sq. ft d) 27(square root of)85 sq. ft. I knew that the formula for the area of a rhombus is 1/2d1*d2, so I substituted in the diameter lengths: A = 1/2 * 21 * 18 = 189 With that equation, I was able to find that the area was 189 sq. ft., or answer b. I hope this has been a helpful review question! Have a great week! (: Article posted June 10, 2012 at 09:02 PM GMT-5 • comment • Reads 207
Article posted June 11, 2012 at 08:18 AM GMT-5 • comment • Reads 256 Well, the final is coming up for geometry this. Just one more test that I need to study for. I can't say that I enjoy these types of things, but I still need to do them. To help out everybody, though, here are two problems that were on homework from this term. There are two since the blog wasn't working last week when I was supposed to post one and one from this week, so I'm just combining the two blogs into this week's. Here they are: The lengths of the diagonals of a rhombus are 2 in. and 5 in. Find the measures of the angles of the rhombus to the nearest degree. Describe each translation using an ordered pair. 2 units to the left, 1 unit down. Good luck to everyone on the final! Article posted June 11, 2012 at 08:18 AM GMT-5 • comment • Reads 256
Article posted June 10, 2012 at 07:04 PM GMT-5 • comment • Reads 190 It seems like just yesterday that we were beginning our weekly blogs, and now we are writing our last ones. This is the week of finals, and after this week, we only have one day left of school. I cannot believe that school went by this fast. I would have to say that this school year was filled with learning, and fun! On another note, we still must remain serious. We still have our final exam on Monday, and that requires a gracious amount of studying. To study for the exam, I have come up with a few review guides that will help me go over anything that I have been struggling with. One thing that I have had a bit of trouble with is vectors. Therefore, I picked out a problem about vectors from my book and worked out the answer to it. The problem was "describe each vector as an ordered pair". To do so, you had to first figure out the sine ratio as well as the cosine ratio. Then, you would order them so that they describe the vector. It took some practice to understand, but eventually, I got it! I now feel prepared for this final exam! Article posted June 10, 2012 at 07:04 PM GMT-5 • comment • Reads 190
Article posted June 4, 2012 at 09:38 AM GMT-5 • comment • Reads 251 Salutations fellow bloggers! This week in geometry class we learned about the law of sine and cosine. It isn't to difficult but I don't really understand how it saves you time. My question for the final review is: What quadrilaterals have diagonals that bisect each other? Answer: Parallelogram, rhombus, rectangle, and square Have a great week everyone! Article posted June 4, 2012 at 09:38 AM GMT-5 • comment • Reads 251
Article posted June 5, 2012 at 07:53 PM GMT-5 • comment • Reads 237 Hi everyone! Just writing because before I could add my review question, Blogmeister shut itself down! NOOOOOO!!!! Well, anyway, here’s my Trigonometry Review question: Kevin and Zane have just built a tree house. It is 20 ft off the ground, and the tree is perpendicular to the ground. When Kevin’s Mom goes out to look at the tree house, the angle of elevation from her feet to the tree house is 50*. Zane wants to know how far the tree that has the tree house in it is from the house, but he has lost the tape measure. How far away is the tree from the house? ***SPOILER ALERT*** Answer: Roughly 17 Feet --Joe Article posted June 5, 2012 at 07:53 PM GMT-5 • comment • Reads 237
Article posted June 10, 2012 at 07:42 AM GMT-5 • comment • Reads 198 HELLO! WELCOME TO THE FINAL COMPLETE WEEK OF SCHOOL! Wow, the year has gone by so fast! Did I mention that this is our final weekly blog? I can’t believe how close we are to the end of the year! It seems like just yesterday we walked into Geometry for the first time. I have had a great weekend, capped of by a RockEucharist at my church! This week in Geometry, we continued with our trigonometry unit, and we learned the law of cosines. At first I didn’t understand it, but once I got it, I had it for good. I think that on our trig. test this week, I will not have that much trouble, as this unit has been about mostly algebra and I do not have much trouble with algebra. But you never know-- it may be very hard. Our final is also coming up. Wow, that snuck up on me. Here is another review question: Ben and Griffin love modern houses. They want to make a museum about the evolution of houses. While searching for a place to put the museum, they find a modern building that is a rhombus. The sides of the house are 50 feet long. The angles formed by two intersecting lines are 120* and 60*. How much area is in the museum? ANSWER: 4330 ft. squared Have a great last week! --Joe Article posted June 10, 2012 at 07:42 AM GMT-5 • comment • Reads 198
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Article posted June 2, 2012 at 07:33 AM GMT-5 • comment • Reads 539 Today is Sunday, June 10, and I'm doing this blog a week late because the website crashed last week. It also means that there are exactly 2 school days until Finals start! I'm starting to worry it will be really hard to study without being distracted by the nice weather. To get in the spirit of studying, I'm going to post a review question from one of my past tests today on this blog. This is it: Determine which three numbers could be the sides of a right triangle. A. 64,73,98 B. 64,72,96 C. 65,72,97 The answer is C because of the Pythagorean theorem. When you square each number, the first two must add up to the third one's square. If that happens, then it is a right triangle. Choice C is the only one that happens for, so it is the correct answer. Until Next week, Article posted June 2, 2012 at 07:33 AM GMT-5 • comment • Reads 539
Article posted June 10, 2012 at 07:03 AM GMT-5 • comment • Reads 309 I'm doing the second blog for this week today, Sunday, June 10 because the website crashed last week. As I said earlier, there are exactly two school days until finals start! While this may seem terrible, I'm actually very happy because it means that we are in the home stretch before summer! I'm posting another review question this week: this will be about lines in triangles. Which line in triangles meet at the orthocenter when all three are drawn in a triangle? The answer to this question is altitudes. An altitude, also known as the height of a triangle, is a perpendicular line to the base drawn from a vertex. School may be almost done, but I can't give up now. This is the last blog of this year, and I can say that's a good thing! Article posted June 10, 2012 at 07:03 AM GMT-5 • comment • Reads 309
Article posted June 10, 2012 at 07:40 PM GMT-5 • comment • Reads 291 For this weeks blog I was asked to post two review questions for the final exam. The final exam is this week and I need to start studying, so I am going to start by making these review questions. My first question is: What does MAPA COCI stand for? A hint for this question is to remember lines in triangles and their concurrent points! My next question is: How do you find the circumference of a circle with a radius of 10 cm? Answers: Question 1: Median, Altitude, Perpendicular Bisector, Angle Bisector. Centroid, Orthocenter, Circumcenter, Incenter. Question 2: 20pi cm. Article posted June 10, 2012 at 07:40 PM GMT-5 • comment • Reads 291
Article posted June 7, 2012 at 04:31 PM GMT-5 • comment • Reads 279 On Sunday I watched a lot of TV because of all the crappy weather that we had. The best thing I watched was the golf touroment where Tiger Woods hit an unbelievable shot to clinch the victory. The problem I am working on for this blog is # 12 from the green version on test # 7. It asks for you to find the area of a parallelogram that has sides of 8 and 6. On the test, I got the problem wrong because I didn't know the area formula for parallelograms but when I went back and re-did the problem while studying for the retake I realized that I had to find the height by making a 45-45-90 triangle because the formula is a=bh. I found that the height was 3 root 2 and 8 time 3 root 2 is 24 root 2. Article posted June 7, 2012 at 04:31 PM GMT-5 • comment • Reads 279
Article posted June 7, 2012 at 04:49 PM GMT-5 • comment • Reads 356 I am blogging on a thursday for probably the first time, it seems wierd. I am getting ready to do my geometry homework. I am also going to watch the Miami Heat vs the Boston Celtics later, it is must win game for the Heat. The problem that I am going to work on in this blog is # 17 from the green version of test # 7. It asks you find the area of a deck that surrounds a hot tub if the hot tub has a diameter of 6 meters and the deck is 2 meters wide. When I took the test I knew how to do the problem but I just had a brain cramp and screwed up the final answer. The way to figure out this problem is to find the area of the hot tub and then subtract that from the area of the deck. The only formula that you need to use on this problem is a=pie(r) squared. The first thing to do is sub the number 3 into that formula and you end up with 9 pie. You then plug the number 5 into that formula because 6+ 2+2 equals 10 as a diameter and half of that equals 5. When plugging that into the equation you end up with 25 pie. The problem that I made on the test was that I forgot to subtract 9 pie from 25 pie so that is the last step in this problem. Your final answer should be 16 pie. Article posted June 7, 2012 at 04:49 PM GMT-5 • comment • Reads 356
Article posted June 10, 2012 at 10:26 PM GMT-5 • comment • Reads 224 This blog is the twenty-third of my personal weekly blog This blog was supposed to be posted last week but blogmeister was down! Oh no! We got everything squared away though, so yay! Last week was pretty gloomy, the weather hasn't been that nice until this weekend! Last week in geometry we started learning more in trigonometry and then started learning about law of sines and law of cosines! It's been pretty easy, but some things trip me up easily. Anyways the year is winding down and I can not wait until SUMMER! Midterms start this week! Here is a review question on what we have learned in Geometry! Ms. J told us that studying our tests and quizzes would be most helpful so that is exactly what I will be doing! A particular unit that I didn't fair so well on was working with the lines of triangles! Here is a review question from the Lines of Triangles Quiz that we took on February 7th. Which of the following are the slopes of two perpendicular lines? a. 3 and -3 b. 5 and 1/5 c. no slope and undefined d. -2/3 and 3/2 If you recall that the slope of any line perpendicular to another is the negative reciprocal, then this problem is easy to solve! A is not the answer because the negative reciprocal or 3 would be -1/3. B does not work because the negative reciprocal of 5 would be -1/5. c does not work because they aren't specific lines! THEREFORE D IS THE ANSWER! and to prove it.. the negative reciprocal of -2/3 would be - ( -3/2) or 3/2! TADA! Article posted June 10, 2012 at 10:26 PM GMT-5 • comment • Reads 224
Article posted June 10, 2012 at 10:34 PM GMT-5 • comment • Reads 419 This is the twenty-fourth of my personal weekly blog. Alas I have reached my last blog. Farewell classblogmeister... I won't miss you too much I promise! This week in Geometry we continued working with the law of sines and the law of cosines! We have a test this week before our final so that means a lot of studying on our parts! We pretty much did a lot of reviewing on the all the parts of trigonometry, which was helpful! This review question will be coming from the Quadrilaterals test we took on March 29th. I GOT THIS QUESTION WRONG! I honestly have no idea how I could've made the mistake, but I did, so make sure you don't jump to conclusions! Consecutive sides of a rectangle are congruent. a. sometimes b. always c. never the answer is SOMETIMES because a square (which is a rectangle) has congruent consecutive sides! YAY GEOMETRY!(: Article posted June 10, 2012 at 10:34 PM GMT-5 • comment • Reads 419
Article posted June 10, 2012 at 10:32 AM GMT-5 • comment • Reads 224 For Ch 12, the work was on transformations. The review question will be for rotations. The question as what is the point of rotation, and what is the angle of rotation. When given a pre-image and an image, to find the point of rotation, it will either be where the two images intersect or a point not connected to either image. Then to find the angle of rotation, you will need a protractor. Put your protractor on the point of rotation and pick a point you want to measure from. You always measure from the pre-image and counterclockwise unless stated to go clockwise. After you pick the point, find the point on the image that is the same, and measure. That will give you the angle of rotation Stay classy bloggers. Article posted June 10, 2012 at 10:32 AM GMT-5 • comment • Reads 224
Article posted June 10, 2012 at 11:24 AM GMT-5 • comment • Reads 217 Our final exam is Wednesday! Ahhh! And in preparation to that I am going to be posting a review question from one of the chapters we learned about this semester. In chapter 7 we learned more about areas of formulas. My review question today, will be finding the area of a rhombus (remember that the diagonals of a rhombus bisect each other). The formula for an area of a rhombus is 1/2(d1+d2), d stands for diagonal. If your given information was that one diagonal was 7cm and half of the other diagonal was 5.5cm, this is how you would solve it: We know that the other half of the diagonal is 5.5cm, because it gets bisected from the other diagonal. Now we plug in our given information to come up with 1/2(7+11) = area. Simplify to get 1/2(18) = area. Now simply multiply by 1/2 or divide by two, and we get 9cm squared = area Hopefully this helped you! I will be posting another review question soon! Goodbye! Article posted June 10, 2012 at 11:24 AM GMT-5 • comment • Reads 217
Article posted June 10, 2012 at 12:01 PM GMT-5 • comment • Reads 233 Hello again! Today, I am posting another review question. This comes from chapter 5. Question: Which of they following is the concurrent point of the altitudes? Answer: Orthocenter. I always remember which concurrent point goes with which line of a triangle because of MAPA COCI. MAPA COCI stands for: Median --> Centroid Altitude --> Orthocenter Perpendicular Bisector -->Circumcenter Angle Bisector --> Incenter This helped me remember and hopefully it will help you too! :) Now down to the sad stuff. :( I won't be saying 'see you next week' anymore..... because this is my last blog! *Gasp* I hope my blogs have helped you in some way or another, but right now I have to say goodbye for good. I know you'll miss me though! Farewell, Livy Article posted June 10, 2012 at 12:01 PM GMT-5 • comment • Reads 233
Article posted June 11, 2012 at 02:06 PM GMT-5 • comment • Reads 235 Question Given: Circle O with Diamter CD, AB is parrallel to CD, and arc AB=80 degrees Find arc CA [1] 50 [2] 60 [3] 80 [4] 100 Answer 50 Article posted June 11, 2012 at 02:06 PM GMT-5 • comment • Reads 235
Article posted June 11, 2012 at 02:10 PM GMT-5 • comment • Reads 381 Question Given: In triangle ABC, B=120, c=15, and a=15 Find C Answer 30 degrees Article posted June 11, 2012 at 02:10 PM GMT-5 • comment • Reads 381
Article posted June 11, 2012 at 09:04 PM GMT-5 • comment • Reads 241 The midterm is in two days and I'm getting a little nervous. We also have a test tomorrow that isn't helping with the nerves! Hopefully the test tomorrow will make get me feeling completely confident with the trigonometry unit, so it'll be one less thing to study for the final! My first review question that I want to go over is number 13 from practice 58. It is working with vectors which I needed a slight refresher on. The question: Homing pigeons have the ability or instinct to find their way home when released hundreds of miles away from home. Homing pigeons carried news of Olympic victories to various cities in ancient Greece. Suppose one such pigeon took off from Athens and landed in Sparta, which is 73 miles west and 64 miles south of Athens. Find the distance and its direction of flight. The first step you would need to take is plugging the coordinates into the distance formula. It comes out with 97, so Athens and Sparta are 97 miles apart. Then, you need to use the tangent ratio. This is the opposite side over the adjacent side. Therefore, tan(x) = 64/73 --> x = tan-1( 64/73) --> x = 41°, so the direction of flight is 41°. Article posted June 11, 2012 at 09:04 PM GMT-5 • comment • Reads 241
Article posted June 11, 2012 at 09:16 PM GMT-5 • comment • Reads 229 We were required two review questions, so I think I'll choose one from the beginning of the semester just as a refresher. Number 25 on the quadrilaterals test was asking us to label the coordinates of parallelogram LAST using only three variables. I chose d, c, and b (along with 0). Point 'a' was on the y-axis so the x coordinate was 0. Then, I chose 'b' to act as the height, or y coordinate. Point 'l' is the lower, left-hand point, and it is on the x-axis, making the y coordinate 0. Then, I made the x coordinate '-c' because it is to the left of the y-axis. Point 't' is the lower right-hand point, and it also lies on the x-axis, making the y coordinate 0. Then, I chose 'd' to act as the x coordinate. Point 's' was the upper right point of the parallelogram, so the height was also 'b', making the y coordinate automatically 'b'. The x coordinate is going to be 'd+c', because 'c' is the length that the segment goes on longer past 'd', which is where you get the '+c'. Hopefully that all made sense and was (slightly) helpful! Article posted June 11, 2012 at 09:16 PM GMT-5 • comment • Reads 229
Article posted June 10, 2012 at 09:00 PM GMT-5 • comment • Reads 222 This blog is supposed to be for two weeks ago but due to problems with the website the due date was postponed. This blog I will be explaining how to do a problem from one of the previous tests that I have taken. I have chosen to explain number 4 from test 7. The problem gives you a rhombus and tells you to find the area using the diagonals of 18 and 21. To find the area you must use the formula A=.5(d1)(d2). You would use substitution of make the equation A=.5(18)(21). If you do the math the area comes out to be 189ft squared. That is how you would do a problem like that... Article posted June 10, 2012 at 09:00 PM GMT-5 • comment • Reads 222
Article posted June 10, 2012 at 09:23 PM GMT-5 • comment • Reads 311 This week in geometry I am doing the same thing as last week. I have to pick a problem from any of my previous tests and I must explain how to do it.On problem 4 from the chapter 7 retest, the problem gives you a parallelogram and tells you to find the area using a height of 24 and a base of 10. The formula you must use is A= base times height. You would set up the equation as A= 24 times 10. The answer to the problem is 240 meters squared. That is how you do an problem like that... Article posted June 10, 2012 at 09:23 PM GMT-5 • comment • Reads 311
Article posted June 12, 2012 at 05:29 PM GMT-5 • comment • Reads 234 Question: The diagonals of a square bisect all angles Answer: Always Article posted June 12, 2012 at 05:29 PM GMT-5 • comment • Reads 234
Article posted June 12, 2012 at 05:31 PM GMT-5 • comment • Reads 312 Question: A triangle has side lengths of 8 cm, 14 cm, and 11 cm. Classify the triangle. Answer:Obtuse Article posted June 12, 2012 at 05:31 PM GMT-5 • comment • Reads 312
Article posted June 9, 2012 at 10:48 AM GMT-5 • comment • Reads 303 Hey guys so the school year's almost over. Yeah, summer. Unfortunately to get to summer we have to go through probably the wort week of the school year. Finals. Yeah you all know what I'm talking about. So here is one of the review questions that I'm using to study for my geometry final. This question is on right triangle trig. Good luck! Find X and H Answer: X=13 H=8.1 Article posted June 9, 2012 at 10:48 AM GMT-5 • comment • Reads 303
Article posted June 4, 2012 at 06:43 AM GMT-5 • comment • Reads 261 Hey guys so as I said in my last blog we are learning about SOH, CAH, TOA. Well now we are learning about the law of sines. Here is a review question and its answer about the law of sines. If m Answer: Measure of angle B is 59.4. Article posted June 4, 2012 at 06:43 AM GMT-5 • comment • Reads 261
Article posted June 7, 2012 at 09:17 PM GMT-5 • comment • Reads 224 We have finals in two weeks, and I don't really have any specific emotions for the end of the year. It was an awesome freshman year and a good first highschool experience. I hope next year will be as good, but then again I won't get to have Ms. J anymore:( To study for the final we must post a review question. Mine is- If a plane takes off at 25 degrees and flies 1600 ft, what is its altitude. SPOILER BELOW To do this problem, you must first write out the formula for sine. Using SOH, it would come out to be sin25=X/1600. Then you would multiply by 1600 on both sides to isolate X. This would make it 1600*sin(25)=X. Enter 1600*sin(25)into your calculator and the resulting number is the altitude-676.18 ft Article posted June 7, 2012 at 09:17 PM GMT-5 • comment • Reads 224
Article posted June 10, 2012 at 10:45 PM GMT-5 • comment • Reads 240 Hello everyone, I'm sad to tell you that this will be my last geometry blog :( The year is finally coming to an end. Six more days of school until summer! Here is my final exam review question: A triangle has three sides with the lengths of 6:8:10 What type of triangle is it? A.) acute B.)obtuse C.) right Answer: C. Right Article posted June 10, 2012 at 10:45 PM GMT-5 • comment • Reads 240
Article posted June 10, 2012 at 11:25 AM GMT-5 • comment • Reads 208 Hello Bloggers!!! This week I will be giving you a practice problem for you to solve!! At the bottom of the blog I will show you the work to get to the answer!! Question: You know two sides and an angle (6cm, 10cm, and 56 degrees) find the missing side. Answer: X2(squared)=6(squared)+10(squared) - 2(6)(10) - cos(56) by plugging this equation: 6(squared) + 10(squared) -2(6)(10) - cos(56) into your calculator, you will come up with X2(squared)= 15.4408071 you need to find X, so you have to find the square root. Type this into your calculator and X= 3.92947924 Great Job!!!! I will be blogging soon!!!! Article posted June 10, 2012 at 11:25 AM GMT-5 • comment • Reads 208
Article posted June 9, 2012 at 02:17 PM GMT-5 • comment • Reads 217 Hey bloggers! Just as I promised on Wednesday, here is another final review question! "In parallelogram BARK, m first draw the paralleglogram: 4x = xsquared - 60 xsqaured - 4x - 60 = 0 (x + 6)(x - 10) xsquared - 10x + 6x - 60 xsquared - 4x - 60 x + 6 = 0 or x - 10 = 0 x = -6 or x = 10 7x + m 70 + m m m< ARK = 110 dgrees Wish me luck on my final! Grace Article posted June 9, 2012 at 02:17 PM GMT-5 • comment • Reads 217
Article posted June 6, 2012 at 04:34 PM GMT-5 • comment • Reads 234 Hey bloggers! I am going to show you a good review question for our Geometry final next week! "Q is the intersection of AC and BD and ABCD. Find the area of kite ABCD if AB= 10, BC= 20, AC = 30, and BQ = 5." A= 1/2 x d1 x d2 d1= DB = 2BQ = 10. d2 = AC = 30 A = 1/2 x 10 x 30 Area = 150 squared units I will be posting another question later in the week due to the website being down last weekend. Later, Grace Article posted June 6, 2012 at 04:34 PM GMT-5 • comment • Reads 234
Article posted June 10, 2012 at 08:47 PM GMT-5 • comment • Reads 219 Welcome back everyone, since we are nearing the end of the school year, I was thinking that we could all start posting review questions to prepare ourselves for the upcoming final! What is a perpendicular bisector? And, how do you find it for any triangle? A perpendicular bisector is a line that divides a side of a triangle directly in the middle, but it bisects it at a 90˚ angle. Each side of the triangle has it’s own perpendicular bisector. In a triangle, there are 3 different sides each that have a perpendicular bisector. Since there are three different types of triangles (acute, right, and obtuse), each location of the circumcenter is different. For any acute triangle, the circumcenter is inside of the triangle. The point where all three of the Perpendicular bisectors intersect is at the circumcenter. But, like always there are 2 special cases. For a right triangle, the circumcenter of a right triangle is on the hypotenuse. But for an obtuse triangle: the circumcenter of an obtuse triangle is outside of the triangle. A circumscribed circle is a circle that is drawn from the circumcenter that hits all of the vertices of the triangle. The circumcenter is always equidistant to all vertices or any other point on the circumscribed circle. If you need anymore help with perpendicular bisectors reply to this post! I hope you all begin to review now! See you all next week! -Kathleen Article posted June 10, 2012 at 08:47 PM GMT-5 • comment • Reads 219
Article posted June 10, 2012 at 10:17 AM GMT-5 • comment • Reads 204 Hello again! As we are approaching our final exam next week, we are continuing to post review questions! These are aimed to help us remember the concepts we have learned this semester! The question I'm going to ask takes us back a chapter, and deals with triangles! Be sure to blog back with any questions. Q: A triangle has side lengths of 8cm, 11cm, and 14cm. Classify this triangle as acute, obtuse, or right. A: OBTUSE To find this answer, we first begin with the Pythagorean theorem (A squared + B squared = c squared). So, when we plug in what is known, our formula becomes 64 + 121 = 196. Here's the rule to classify: IF (A squared + b squared > C squared) then the triangle is acute. IF (A squared + b squared < C squared) then the triangle is obtuse. IF (A squared + b squared = C squared) then the triangle is right. In our formula, 64 + 121 = 185. Because 185 < 196, then the triangle is obtuse! Hope that helped and you learned something knew! This is my last weekly blog... but I'll be writing occasionally until the end of the year. Talk to you then! Emma Article posted June 10, 2012 at 10:17 AM GMT-5 • comment • Reads 204
Article posted June 14, 2012 at 12:40 PM GMT-5 • comment • Reads 240 We finally made it! Finals are around the corner, and so is summer! I can’t wait to ditch my backpack and head to the beach. But first, I have to survive finals. To review for Geometry, I’ve posted one of my favorite homework problems from a past unit below. KL⎮⎮JM in isosceles trapezoid JKLM. Find the values of x and y if m⦟J=(23x-8)º, m⦟K=(12y-13)º, and m⦟M=(17x+10)º. m⦟J=m⦟M 23x-8=17x+10 6x-8=10 6x=18 x=3 m⦟J+m⦟K=180 23x-8+12y-13=180 61+12y-13=180 12y=132 y=11 I chose this problem because I was from one of the first units of the semester, and I didn’t remember that unit at all. I thought that it would be nice to review since I had forgotten about it. Article posted June 14, 2012 at 12:40 PM GMT-5 • comment • Reads 240
Article posted June 14, 2012 at 12:40 PM GMT-5 • comment • Reads 244 We finally made it! Finals are around the corner, and so is summer! I can’t wait to ditch my backpack and head to the beach. But first, I have to survive finals. To review for Geometry, I’ve posted one of my favorite homework problems from a past unit below. KL⎮⎮JM in isosceles trapezoid JKLM. Find the values of x and y if m⦟J=(23x-8)º, m⦟K=(12y-13)º, and m⦟M=(17x+10)º. m⦟J=m⦟M 23x-8=17x+10 6x-8=10 6x=18 x=3 m⦟J+m⦟K=180 23x-8+12y-13=180 61+12y-13=180 12y=132 y=11 I chose this problem because I was from one of the first units of the semester, and I didn’t remember that unit at all. I thought that it would be nice to review since I had forgotten about it. Article posted June 14, 2012 at 12:40 PM GMT-5 • comment • Reads 244
Article posted June 6, 2012 at 04:33 PM GMT-5 • comment • Reads 257 Hello fellow blog readers! Sorry that this blog is so late compared to my others. Blogmeister, as I'm sure you know, has been down for the past couple of days. I am sorry to keep you waiting for so long. Now to the important stuff. In geometry, I have a question about the latest test. My question is how to find the length of an arc. I do not remember and was wondering if you bloggers could help me out. If you comment with the correct answer, I will mention you in my next blog. Well that is all for now blog-readers! Stay tuned for this Sunday's blog and question. Stay classy and thanks for reading! Article posted June 6, 2012 at 04:33 PM GMT-5 • comment • Reads 257
Article posted June 10, 2012 at 04:52 PM GMT-5 • comment • Reads 215 The year is almost over! We have two tests this next week in geometry. One tomorrow on chapter 9 and one on June 18th on everything we've learned this semester (our final). To help review for this final I have been told to blog a review question with the answer. Do you remember the distance formula? Well for coordinate proofs we had to use the distance formula to show how one side was of equal length to the other side. To show this you use the formula d = the square root of (y2-y1)squared + (x2-x1)squared. If you plug in the coordinates from #26 on the quadrilaterals test you get AS = the square root of (b-b)squared + (a+c-0)squared. This simplifies to the square root of (a+c)squared which equals a + c. Hope this was helpful, good luck on the test and the final! Article posted June 10, 2012 at 04:52 PM GMT-5 • comment • Reads 215
Article posted June 10, 2012 at 04:11 PM GMT-5 • comment • Reads 218 AHHHH! Three more days until final exams! Before I show you the review question take a deep breath... Inhale, now exhale. Feel better? Good. Let's begin! This is from chapter 9. This chapter is on SOH CAH TOA, the law of sines, the law of cosines, and vectors. Say you had a triangle where the length of the hypotenuse (AC) is 12cm, and the short side (AB) and the long side (BC) is unknown. Angle C is given as 23 degrees and angle B is 90 degrees. To solve this problem you use CAH because you are given a right triangle, an angle (23 degrees), the hypotenuse to that angle (12cm), and the long side is present. To set it up you write; cos23 = X/12. To solve you multiple cos23 by 12, so it's 12cos23 = X. X = 11cm. I hope this was helpful! Article posted June 10, 2012 at 04:11 PM GMT-5 • comment • Reads 218
Article posted June 10, 2012 at 08:54 PM GMT-5 • comment • Reads 216 Hello everyone! Hope your week has gone nicely! Mine has been pretty interesting, because my brother had an English exchange student arrive on Thursday. He will be staying with us for the week, and I am really excited! English accents are so cool! This exchange trip has also reminded me that the end of the school year is really close! In order to prepare for the geometry final, I am going to tell you about a question we had to do for homework as a way to review. This problem was on a worksheet I did recently involving the Law of Sines: sinA = sinB a b A and B stand for angles of an oblique triangle, while a and b stand for the 2 sides opposite them. Basically, you can use this law to find one of these four measures, as long as you have an angle and the side opposite. Basically, the problem was a triangle with angles A, B, and C, and sides a, b, and c. The measure of angle A was 35º, the measure of side a was 8 cm, and the measure of side b was 12 cm. The problem asked you to find the measure of angle B. Through substitution, I came out with this equation: sin(35) = sin(B) 8 12 By multiplying both sides by 12 and then by sin to the (-1), I was able to then solve for B: sin-1(12sin(35)) = sin-1(sin(B)) 8 m This process took some time to get used to, but I've found I actually like this sorts of problems in trigonometry. Let me know if you have any questions! Have a great week! (: Article posted June 10, 2012 at 08:54 PM GMT-5 • comment • Reads 216
Article posted June 10, 2012 at 09:02 PM GMT-5 • comment • Reads 207 Hello everyone! Firstly, I would just like to explain that last week Blogmeister was experiencing some technical difficulties, so I was unable to post at that time. The post from both then and the most recent one should be uploaded now. Anyways, this week is the beginning of finals for my school! I am very nervous because I've never taken finals before, and I'm not sure what to expect. Wish me luck! In order to review once again for geometry, I am posting another review questions below. I chose this particular one because I find this area formula hard to remember sometimes, because it doesn't involve side lengths, rather, diagonal lengths. For this problem, there was a rhombus shown with diagonals of 21 ft and 18 ft. The instructions were to find the area, and the possible answers were as shown: a) 378 sq. ft b) 189 sq. ft c) 162 sq. ft d) 27(square root of)85 sq. ft. I knew that the formula for the area of a rhombus is 1/2d1*d2, so I substituted in the diameter lengths: A = 1/2 * 21 * 18 = 189 With that equation, I was able to find that the area was 189 sq. ft., or answer b. I hope this has been a helpful review question! Have a great week! (: Article posted June 10, 2012 at 09:02 PM GMT-5 • comment • Reads 207
Article posted June 11, 2012 at 08:18 AM GMT-5 • comment • Reads 256 Well, the final is coming up for geometry this. Just one more test that I need to study for. I can't say that I enjoy these types of things, but I still need to do them. To help out everybody, though, here are two problems that were on homework from this term. There are two since the blog wasn't working last week when I was supposed to post one and one from this week, so I'm just combining the two blogs into this week's. Here they are: The lengths of the diagonals of a rhombus are 2 in. and 5 in. Find the measures of the angles of the rhombus to the nearest degree. Describe each translation using an ordered pair. 2 units to the left, 1 unit down. Good luck to everyone on the final! Article posted June 11, 2012 at 08:18 AM GMT-5 • comment • Reads 256
Article posted June 10, 2012 at 07:04 PM GMT-5 • comment • Reads 190 It seems like just yesterday that we were beginning our weekly blogs, and now we are writing our last ones. This is the week of finals, and after this week, we only have one day left of school. I cannot believe that school went by this fast. I would have to say that this school year was filled with learning, and fun! On another note, we still must remain serious. We still have our final exam on Monday, and that requires a gracious amount of studying. To study for the exam, I have come up with a few review guides that will help me go over anything that I have been struggling with. One thing that I have had a bit of trouble with is vectors. Therefore, I picked out a problem about vectors from my book and worked out the answer to it. The problem was "describe each vector as an ordered pair". To do so, you had to first figure out the sine ratio as well as the cosine ratio. Then, you would order them so that they describe the vector. It took some practice to understand, but eventually, I got it! I now feel prepared for this final exam! Article posted June 10, 2012 at 07:04 PM GMT-5 • comment • Reads 190
Article posted June 4, 2012 at 09:38 AM GMT-5 • comment • Reads 251 Salutations fellow bloggers! This week in geometry class we learned about the law of sine and cosine. It isn't to difficult but I don't really understand how it saves you time. My question for the final review is: What quadrilaterals have diagonals that bisect each other? Answer: Parallelogram, rhombus, rectangle, and square Have a great week everyone! Article posted June 4, 2012 at 09:38 AM GMT-5 • comment • Reads 251
Article posted June 5, 2012 at 07:53 PM GMT-5 • comment • Reads 237 Hi everyone! Just writing because before I could add my review question, Blogmeister shut itself down! NOOOOOO!!!! Well, anyway, here’s my Trigonometry Review question: Kevin and Zane have just built a tree house. It is 20 ft off the ground, and the tree is perpendicular to the ground. When Kevin’s Mom goes out to look at the tree house, the angle of elevation from her feet to the tree house is 50*. Zane wants to know how far the tree that has the tree house in it is from the house, but he has lost the tape measure. How far away is the tree from the house? ***SPOILER ALERT*** Answer: Roughly 17 Feet --Joe Article posted June 5, 2012 at 07:53 PM GMT-5 • comment • Reads 237
Article posted June 10, 2012 at 07:42 AM GMT-5 • comment • Reads 198 HELLO! WELCOME TO THE FINAL COMPLETE WEEK OF SCHOOL! Wow, the year has gone by so fast! Did I mention that this is our final weekly blog? I can’t believe how close we are to the end of the year! It seems like just yesterday we walked into Geometry for the first time. I have had a great weekend, capped of by a RockEucharist at my church! This week in Geometry, we continued with our trigonometry unit, and we learned the law of cosines. At first I didn’t understand it, but once I got it, I had it for good. I think that on our trig. test this week, I will not have that much trouble, as this unit has been about mostly algebra and I do not have much trouble with algebra. But you never know-- it may be very hard. Our final is also coming up. Wow, that snuck up on me. Here is another review question: Ben and Griffin love modern houses. They want to make a museum about the evolution of houses. While searching for a place to put the museum, they find a modern building that is a rhombus. The sides of the house are 50 feet long. The angles formed by two intersecting lines are 120* and 60*. How much area is in the museum? ANSWER: 4330 ft. squared Have a great last week! --Joe Article posted June 10, 2012 at 07:42 AM GMT-5 • comment • Reads 198
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Article posted June 2, 2012 at 07:33 AM GMT-5 • comment • Reads 539 Today is Sunday, June 10, and I'm doing this blog a week late because the website crashed last week. It also means that there are exactly 2 school days until Finals start! I'm starting to worry it will be really hard to study without being distracted by the nice weather. To get in the spirit of studying, I'm going to post a review question from one of my past tests today on this blog. This is it: Determine which three numbers could be the sides of a right triangle. A. 64,73,98 B. 64,72,96 C. 65,72,97 The answer is C because of the Pythagorean theorem. When you square each number, the first two must add up to the third one's square. If that happens, then it is a right triangle. Choice C is the only one that happens for, so it is the correct answer. Until Next week, Article posted June 2, 2012 at 07:33 AM GMT-5 • comment • Reads 539
Article posted June 10, 2012 at 07:03 AM GMT-5 • comment • Reads 309 I'm doing the second blog for this week today, Sunday, June 10 because the website crashed last week. As I said earlier, there are exactly two school days until finals start! While this may seem terrible, I'm actually very happy because it means that we are in the home stretch before summer! I'm posting another review question this week: this will be about lines in triangles. Which line in triangles meet at the orthocenter when all three are drawn in a triangle? The answer to this question is altitudes. An altitude, also known as the height of a triangle, is a perpendicular line to the base drawn from a vertex. School may be almost done, but I can't give up now. This is the last blog of this year, and I can say that's a good thing! Article posted June 10, 2012 at 07:03 AM GMT-5 • comment • Reads 309
Article posted June 10, 2012 at 07:40 PM GMT-5 • comment • Reads 291 For this weeks blog I was asked to post two review questions for the final exam. The final exam is this week and I need to start studying, so I am going to start by making these review questions. My first question is: What does MAPA COCI stand for? A hint for this question is to remember lines in triangles and their concurrent points! My next question is: How do you find the circumference of a circle with a radius of 10 cm? Answers: Question 1: Median, Altitude, Perpendicular Bisector, Angle Bisector. Centroid, Orthocenter, Circumcenter, Incenter. Question 2: 20pi cm. Article posted June 10, 2012 at 07:40 PM GMT-5 • comment • Reads 291
Article posted June 7, 2012 at 04:31 PM GMT-5 • comment • Reads 279 On Sunday I watched a lot of TV because of all the crappy weather that we had. The best thing I watched was the golf touroment where Tiger Woods hit an unbelievable shot to clinch the victory. The problem I am working on for this blog is # 12 from the green version on test # 7. It asks for you to find the area of a parallelogram that has sides of 8 and 6. On the test, I got the problem wrong because I didn't know the area formula for parallelograms but when I went back and re-did the problem while studying for the retake I realized that I had to find the height by making a 45-45-90 triangle because the formula is a=bh. I found that the height was 3 root 2 and 8 time 3 root 2 is 24 root 2. Article posted June 7, 2012 at 04:31 PM GMT-5 • comment • Reads 279
Article posted June 7, 2012 at 04:49 PM GMT-5 • comment • Reads 356 I am blogging on a thursday for probably the first time, it seems wierd. I am getting ready to do my geometry homework. I am also going to watch the Miami Heat vs the Boston Celtics later, it is must win game for the Heat. The problem that I am going to work on in this blog is # 17 from the green version of test # 7. It asks you find the area of a deck that surrounds a hot tub if the hot tub has a diameter of 6 meters and the deck is 2 meters wide. When I took the test I knew how to do the problem but I just had a brain cramp and screwed up the final answer. The way to figure out this problem is to find the area of the hot tub and then subtract that from the area of the deck. The only formula that you need to use on this problem is a=pie(r) squared. The first thing to do is sub the number 3 into that formula and you end up with 9 pie. You then plug the number 5 into that formula because 6+ 2+2 equals 10 as a diameter and half of that equals 5. When plugging that into the equation you end up with 25 pie. The problem that I made on the test was that I forgot to subtract 9 pie from 25 pie so that is the last step in this problem. Your final answer should be 16 pie. Article posted June 7, 2012 at 04:49 PM GMT-5 • comment • Reads 356
Article posted June 10, 2012 at 10:26 PM GMT-5 • comment • Reads 224 This blog is the twenty-third of my personal weekly blog This blog was supposed to be posted last week but blogmeister was down! Oh no! We got everything squared away though, so yay! Last week was pretty gloomy, the weather hasn't been that nice until this weekend! Last week in geometry we started learning more in trigonometry and then started learning about law of sines and law of cosines! It's been pretty easy, but some things trip me up easily. Anyways the year is winding down and I can not wait until SUMMER! Midterms start this week! Here is a review question on what we have learned in Geometry! Ms. J told us that studying our tests and quizzes would be most helpful so that is exactly what I will be doing! A particular unit that I didn't fair so well on was working with the lines of triangles! Here is a review question from the Lines of Triangles Quiz that we took on February 7th. Which of the following are the slopes of two perpendicular lines? a. 3 and -3 b. 5 and 1/5 c. no slope and undefined d. -2/3 and 3/2 If you recall that the slope of any line perpendicular to another is the negative reciprocal, then this problem is easy to solve! A is not the answer because the negative reciprocal or 3 would be -1/3. B does not work because the negative reciprocal of 5 would be -1/5. c does not work because they aren't specific lines! THEREFORE D IS THE ANSWER! and to prove it.. the negative reciprocal of -2/3 would be - ( -3/2) or 3/2! TADA! Article posted June 10, 2012 at 10:26 PM GMT-5 • comment • Reads 224
Article posted June 10, 2012 at 10:34 PM GMT-5 • comment • Reads 419 This is the twenty-fourth of my personal weekly blog. Alas I have reached my last blog. Farewell classblogmeister... I won't miss you too much I promise! This week in Geometry we continued working with the law of sines and the law of cosines! We have a test this week before our final so that means a lot of studying on our parts! We pretty much did a lot of reviewing on the all the parts of trigonometry, which was helpful! This review question will be coming from the Quadrilaterals test we took on March 29th. I GOT THIS QUESTION WRONG! I honestly have no idea how I could've made the mistake, but I did, so make sure you don't jump to conclusions! Consecutive sides of a rectangle are congruent. a. sometimes b. always c. never the answer is SOMETIMES because a square (which is a rectangle) has congruent consecutive sides! YAY GEOMETRY!(: Article posted June 10, 2012 at 10:34 PM GMT-5 • comment • Reads 419
Article posted June 10, 2012 at 10:32 AM GMT-5 • comment • Reads 224 For Ch 12, the work was on transformations. The review question will be for rotations. The question as what is the point of rotation, and what is the angle of rotation. When given a pre-image and an image, to find the point of rotation, it will either be where the two images intersect or a point not connected to either image. Then to find the angle of rotation, you will need a protractor. Put your protractor on the point of rotation and pick a point you want to measure from. You always measure from the pre-image and counterclockwise unless stated to go clockwise. After you pick the point, find the point on the image that is the same, and measure. That will give you the angle of rotation Stay classy bloggers. Article posted June 10, 2012 at 10:32 AM GMT-5 • comment • Reads 224
Article posted June 10, 2012 at 11:24 AM GMT-5 • comment • Reads 217 Our final exam is Wednesday! Ahhh! And in preparation to that I am going to be posting a review question from one of the chapters we learned about this semester. In chapter 7 we learned more about areas of formulas. My review question today, will be finding the area of a rhombus (remember that the diagonals of a rhombus bisect each other). The formula for an area of a rhombus is 1/2(d1+d2), d stands for diagonal. If your given information was that one diagonal was 7cm and half of the other diagonal was 5.5cm, this is how you would solve it: We know that the other half of the diagonal is 5.5cm, because it gets bisected from the other diagonal. Now we plug in our given information to come up with 1/2(7+11) = area. Simplify to get 1/2(18) = area. Now simply multiply by 1/2 or divide by two, and we get 9cm squared = area Hopefully this helped you! I will be posting another review question soon! Goodbye! Article posted June 10, 2012 at 11:24 AM GMT-5 • comment • Reads 217
Article posted June 10, 2012 at 12:01 PM GMT-5 • comment • Reads 233 Hello again! Today, I am posting another review question. This comes from chapter 5. Question: Which of they following is the concurrent point of the altitudes? Answer: Orthocenter. I always remember which concurrent point goes with which line of a triangle because of MAPA COCI. MAPA COCI stands for: Median --> Centroid Altitude --> Orthocenter Perpendicular Bisector -->Circumcenter Angle Bisector --> Incenter This helped me remember and hopefully it will help you too! :) Now down to the sad stuff. :( I won't be saying 'see you next week' anymore..... because this is my last blog! *Gasp* I hope my blogs have helped you in some way or another, but right now I have to say goodbye for good. I know you'll miss me though! Farewell, Livy Article posted June 10, 2012 at 12:01 PM GMT-5 • comment • Reads 233
Article posted June 11, 2012 at 02:06 PM GMT-5 • comment • Reads 235 Question Given: Circle O with Diamter CD, AB is parrallel to CD, and arc AB=80 degrees Find arc CA [1] 50 [2] 60 [3] 80 [4] 100 Answer 50 Article posted June 11, 2012 at 02:06 PM GMT-5 • comment • Reads 235
Article posted June 11, 2012 at 02:10 PM GMT-5 • comment • Reads 381 Question Given: In triangle ABC, B=120, c=15, and a=15 Find C Answer 30 degrees Article posted June 11, 2012 at 02:10 PM GMT-5 • comment • Reads 381
Article posted June 11, 2012 at 09:04 PM GMT-5 • comment • Reads 241 The midterm is in two days and I'm getting a little nervous. We also have a test tomorrow that isn't helping with the nerves! Hopefully the test tomorrow will make get me feeling completely confident with the trigonometry unit, so it'll be one less thing to study for the final! My first review question that I want to go over is number 13 from practice 58. It is working with vectors which I needed a slight refresher on. The question: Homing pigeons have the ability or instinct to find their way home when released hundreds of miles away from home. Homing pigeons carried news of Olympic victories to various cities in ancient Greece. Suppose one such pigeon took off from Athens and landed in Sparta, which is 73 miles west and 64 miles south of Athens. Find the distance and its direction of flight. The first step you would need to take is plugging the coordinates into the distance formula. It comes out with 97, so Athens and Sparta are 97 miles apart. Then, you need to use the tangent ratio. This is the opposite side over the adjacent side. Therefore, tan(x) = 64/73 --> x = tan-1( 64/73) --> x = 41°, so the direction of flight is 41°. Article posted June 11, 2012 at 09:04 PM GMT-5 • comment • Reads 241
Article posted June 11, 2012 at 09:16 PM GMT-5 • comment • Reads 229 We were required two review questions, so I think I'll choose one from the beginning of the semester just as a refresher. Number 25 on the quadrilaterals test was asking us to label the coordinates of parallelogram LAST using only three variables. I chose d, c, and b (along with 0). Point 'a' was on the y-axis so the x coordinate was 0. Then, I chose 'b' to act as the height, or y coordinate. Point 'l' is the lower, left-hand point, and it is on the x-axis, making the y coordinate 0. Then, I made the x coordinate '-c' because it is to the left of the y-axis. Point 't' is the lower right-hand point, and it also lies on the x-axis, making the y coordinate 0. Then, I chose 'd' to act as the x coordinate. Point 's' was the upper right point of the parallelogram, so the height was also 'b', making the y coordinate automatically 'b'. The x coordinate is going to be 'd+c', because 'c' is the length that the segment goes on longer past 'd', which is where you get the '+c'. Hopefully that all made sense and was (slightly) helpful! Article posted June 11, 2012 at 09:16 PM GMT-5 • comment • Reads 229
Article posted June 10, 2012 at 09:00 PM GMT-5 • comment • Reads 222 This blog is supposed to be for two weeks ago but due to problems with the website the due date was postponed. This blog I will be explaining how to do a problem from one of the previous tests that I have taken. I have chosen to explain number 4 from test 7. The problem gives you a rhombus and tells you to find the area using the diagonals of 18 and 21. To find the area you must use the formula A=.5(d1)(d2). You would use substitution of make the equation A=.5(18)(21). If you do the math the area comes out to be 189ft squared. That is how you would do a problem like that... Article posted June 10, 2012 at 09:00 PM GMT-5 • comment • Reads 222
Article posted June 10, 2012 at 09:23 PM GMT-5 • comment • Reads 311 This week in geometry I am doing the same thing as last week. I have to pick a problem from any of my previous tests and I must explain how to do it.On problem 4 from the chapter 7 retest, the problem gives you a parallelogram and tells you to find the area using a height of 24 and a base of 10. The formula you must use is A= base times height. You would set up the equation as A= 24 times 10. The answer to the problem is 240 meters squared. That is how you do an problem like that... Article posted June 10, 2012 at 09:23 PM GMT-5 • comment • Reads 311
Article posted June 12, 2012 at 05:29 PM GMT-5 • comment • Reads 234 Question: The diagonals of a square bisect all angles Answer: Always Article posted June 12, 2012 at 05:29 PM GMT-5 • comment • Reads 234
Article posted June 12, 2012 at 05:31 PM GMT-5 • comment • Reads 312 Question: A triangle has side lengths of 8 cm, 14 cm, and 11 cm. Classify the triangle. Answer:Obtuse Article posted June 12, 2012 at 05:31 PM GMT-5 • comment • Reads 312
Article posted June 9, 2012 at 10:48 AM GMT-5 • comment • Reads 303 Hey guys so the school year's almost over. Yeah, summer. Unfortunately to get to summer we have to go through probably the wort week of the school year. Finals. Yeah you all know what I'm talking about. So here is one of the review questions that I'm using to study for my geometry final. This question is on right triangle trig. Good luck! Find X and H Answer: X=13 H=8.1 Article posted June 9, 2012 at 10:48 AM GMT-5 • comment • Reads 303
Article posted June 4, 2012 at 06:43 AM GMT-5 • comment • Reads 261 Hey guys so as I said in my last blog we are learning about SOH, CAH, TOA. Well now we are learning about the law of sines. Here is a review question and its answer about the law of sines. If m Answer: Measure of angle B is 59.4. Article posted June 4, 2012 at 06:43 AM GMT-5 • comment • Reads 261
Article posted June 7, 2012 at 09:17 PM GMT-5 • comment • Reads 224 We have finals in two weeks, and I don't really have any specific emotions for the end of the year. It was an awesome freshman year and a good first highschool experience. I hope next year will be as good, but then again I won't get to have Ms. J anymore:( To study for the final we must post a review question. Mine is- If a plane takes off at 25 degrees and flies 1600 ft, what is its altitude. SPOILER BELOW To do this problem, you must first write out the formula for sine. Using SOH, it would come out to be sin25=X/1600. Then you would multiply by 1600 on both sides to isolate X. This would make it 1600*sin(25)=X. Enter 1600*sin(25)into your calculator and the resulting number is the altitude-676.18 ft Article posted June 7, 2012 at 09:17 PM GMT-5 • comment • Reads 224
Article posted June 10, 2012 at 10:45 PM GMT-5 • comment • Reads 240 Hello everyone, I'm sad to tell you that this will be my last geometry blog :( The year is finally coming to an end. Six more days of school until summer! Here is my final exam review question: A triangle has three sides with the lengths of 6:8:10 What type of triangle is it? A.) acute B.)obtuse C.) right Answer: C. Right Article posted June 10, 2012 at 10:45 PM GMT-5 • comment • Reads 240
Article posted June 10, 2012 at 11:25 AM GMT-5 • comment • Reads 208 Hello Bloggers!!! This week I will be giving you a practice problem for you to solve!! At the bottom of the blog I will show you the work to get to the answer!! Question: You know two sides and an angle (6cm, 10cm, and 56 degrees) find the missing side. Answer: X2(squared)=6(squared)+10(squared) - 2(6)(10) - cos(56) by plugging this equation: 6(squared) + 10(squared) -2(6)(10) - cos(56) into your calculator, you will come up with X2(squared)= 15.4408071 you need to find X, so you have to find the square root. Type this into your calculator and X= 3.92947924 Great Job!!!! I will be blogging soon!!!! Article posted June 10, 2012 at 11:25 AM GMT-5 • comment • Reads 208
Article posted June 9, 2012 at 02:17 PM GMT-5 • comment • Reads 217 Hey bloggers! Just as I promised on Wednesday, here is another final review question! "In parallelogram BARK, m first draw the paralleglogram: 4x = xsquared - 60 xsqaured - 4x - 60 = 0 (x + 6)(x - 10) xsquared - 10x + 6x - 60 xsquared - 4x - 60 x + 6 = 0 or x - 10 = 0 x = -6 or x = 10 7x + m 70 + m m m< ARK = 110 dgrees Wish me luck on my final! Grace Article posted June 9, 2012 at 02:17 PM GMT-5 • comment • Reads 217
Article posted June 6, 2012 at 04:34 PM GMT-5 • comment • Reads 234 Hey bloggers! I am going to show you a good review question for our Geometry final next week! "Q is the intersection of AC and BD and ABCD. Find the area of kite ABCD if AB= 10, BC= 20, AC = 30, and BQ = 5." A= 1/2 x d1 x d2 d1= DB = 2BQ = 10. d2 = AC = 30 A = 1/2 x 10 x 30 Area = 150 squared units I will be posting another question later in the week due to the website being down last weekend. Later, Grace Article posted June 6, 2012 at 04:34 PM GMT-5 • comment • Reads 234
Article posted June 10, 2012 at 08:47 PM GMT-5 • comment • Reads 219 Welcome back everyone, since we are nearing the end of the school year, I was thinking that we could all start posting review questions to prepare ourselves for the upcoming final! What is a perpendicular bisector? And, how do you find it for any triangle? A perpendicular bisector is a line that divides a side of a triangle directly in the middle, but it bisects it at a 90˚ angle. Each side of the triangle has it’s own perpendicular bisector. In a triangle, there are 3 different sides each that have a perpendicular bisector. Since there are three different types of triangles (acute, right, and obtuse), each location of the circumcenter is different. For any acute triangle, the circumcenter is inside of the triangle. The point where all three of the Perpendicular bisectors intersect is at the circumcenter. But, like always there are 2 special cases. For a right triangle, the circumcenter of a right triangle is on the hypotenuse. But for an obtuse triangle: the circumcenter of an obtuse triangle is outside of the triangle. A circumscribed circle is a circle that is drawn from the circumcenter that hits all of the vertices of the triangle. The circumcenter is always equidistant to all vertices or any other point on the circumscribed circle. If you need anymore help with perpendicular bisectors reply to this post! I hope you all begin to review now! See you all next week! -Kathleen Article posted June 10, 2012 at 08:47 PM GMT-5 • comment • Reads 219
Article posted June 10, 2012 at 10:17 AM GMT-5 • comment • Reads 204 Hello again! As we are approaching our final exam next week, we are continuing to post review questions! These are aimed to help us remember the concepts we have learned this semester! The question I'm going to ask takes us back a chapter, and deals with triangles! Be sure to blog back with any questions. Q: A triangle has side lengths of 8cm, 11cm, and 14cm. Classify this triangle as acute, obtuse, or right. A: OBTUSE To find this answer, we first begin with the Pythagorean theorem (A squared + B squared = c squared). So, when we plug in what is known, our formula becomes 64 + 121 = 196. Here's the rule to classify: IF (A squared + b squared > C squared) then the triangle is acute. IF (A squared + b squared < C squared) then the triangle is obtuse. IF (A squared + b squared = C squared) then the triangle is right. In our formula, 64 + 121 = 185. Because 185 < 196, then the triangle is obtuse! Hope that helped and you learned something knew! This is my last weekly blog... but I'll be writing occasionally until the end of the year. Talk to you then! Emma Article posted June 10, 2012 at 10:17 AM GMT-5 • comment • Reads 204
Article posted June 14, 2012 at 12:40 PM GMT-5 • comment • Reads 240 We finally made it! Finals are around the corner, and so is summer! I can’t wait to ditch my backpack and head to the beach. But first, I have to survive finals. To review for Geometry, I’ve posted one of my favorite homework problems from a past unit below. KL⎮⎮JM in isosceles trapezoid JKLM. Find the values of x and y if m⦟J=(23x-8)º, m⦟K=(12y-13)º, and m⦟M=(17x+10)º. m⦟J=m⦟M 23x-8=17x+10 6x-8=10 6x=18 x=3 m⦟J+m⦟K=180 23x-8+12y-13=180 61+12y-13=180 12y=132 y=11 I chose this problem because I was from one of the first units of the semester, and I didn’t remember that unit at all. I thought that it would be nice to review since I had forgotten about it. Article posted June 14, 2012 at 12:40 PM GMT-5 • comment • Reads 240
Article posted June 14, 2012 at 12:40 PM GMT-5 • comment • Reads 244 We finally made it! Finals are around the corner, and so is summer! I can’t wait to ditch my backpack and head to the beach. But first, I have to survive finals. To review for Geometry, I’ve posted one of my favorite homework problems from a past unit below. KL⎮⎮JM in isosceles trapezoid JKLM. Find the values of x and y if m⦟J=(23x-8)º, m⦟K=(12y-13)º, and m⦟M=(17x+10)º. m⦟J=m⦟M 23x-8=17x+10 6x-8=10 6x=18 x=3 m⦟J+m⦟K=180 23x-8+12y-13=180 61+12y-13=180 12y=132 y=11 I chose this problem because I was from one of the first units of the semester, and I didn’t remember that unit at all. I thought that it would be nice to review since I had forgotten about it. Article posted June 14, 2012 at 12:40 PM GMT-5 • comment • Reads 244
Article posted June 6, 2012 at 04:33 PM GMT-5 • comment • Reads 257 Hello fellow blog readers! Sorry that this blog is so late compared to my others. Blogmeister, as I'm sure you know, has been down for the past couple of days. I am sorry to keep you waiting for so long. Now to the important stuff. In geometry, I have a question about the latest test. My question is how to find the length of an arc. I do not remember and was wondering if you bloggers could help me out. If you comment with the correct answer, I will mention you in my next blog. Well that is all for now blog-readers! Stay tuned for this Sunday's blog and question. Stay classy and thanks for reading! Article posted June 6, 2012 at 04:33 PM GMT-5 • comment • Reads 257
Article posted June 10, 2012 at 04:52 PM GMT-5 • comment • Reads 215 The year is almost over! We have two tests this next week in geometry. One tomorrow on chapter 9 and one on June 18th on everything we've learned this semester (our final). To help review for this final I have been told to blog a review question with the answer. Do you remember the distance formula? Well for coordinate proofs we had to use the distance formula to show how one side was of equal length to the other side. To show this you use the formula d = the square root of (y2-y1)squared + (x2-x1)squared. If you plug in the coordinates from #26 on the quadrilaterals test you get AS = the square root of (b-b)squared + (a+c-0)squared. This simplifies to the square root of (a+c)squared which equals a + c. Hope this was helpful, good luck on the test and the final! Article posted June 10, 2012 at 04:52 PM GMT-5 • comment • Reads 215
Article posted June 10, 2012 at 04:11 PM GMT-5 • comment • Reads 218 AHHHH! Three more days until final exams! Before I show you the review question take a deep breath... Inhale, now exhale. Feel better? Good. Let's begin! This is from chapter 9. This chapter is on SOH CAH TOA, the law of sines, the law of cosines, and vectors. Say you had a triangle where the length of the hypotenuse (AC) is 12cm, and the short side (AB) and the long side (BC) is unknown. Angle C is given as 23 degrees and angle B is 90 degrees. To solve this problem you use CAH because you are given a right triangle, an angle (23 degrees), the hypotenuse to that angle (12cm), and the long side is present. To set it up you write; cos23 = X/12. To solve you multiple cos23 by 12, so it's 12cos23 = X. X = 11cm. I hope this was helpful! Article posted June 10, 2012 at 04:11 PM GMT-5 • comment • Reads 218
Article posted June 10, 2012 at 08:54 PM GMT-5 • comment • Reads 216 Hello everyone! Hope your week has gone nicely! Mine has been pretty interesting, because my brother had an English exchange student arrive on Thursday. He will be staying with us for the week, and I am really excited! English accents are so cool! This exchange trip has also reminded me that the end of the school year is really close! In order to prepare for the geometry final, I am going to tell you about a question we had to do for homework as a way to review. This problem was on a worksheet I did recently involving the Law of Sines: sinA = sinB a b A and B stand for angles of an oblique triangle, while a and b stand for the 2 sides opposite them. Basically, you can use this law to find one of these four measures, as long as you have an angle and the side opposite. Basically, the problem was a triangle with angles A, B, and C, and sides a, b, and c. The measure of angle A was 35º, the measure of side a was 8 cm, and the measure of side b was 12 cm. The problem asked you to find the measure of angle B. Through substitution, I came out with this equation: sin(35) = sin(B) 8 12 By multiplying both sides by 12 and then by sin to the (-1), I was able to then solve for B: sin-1(12sin(35)) = sin-1(sin(B)) 8 m This process took some time to get used to, but I've found I actually like this sorts of problems in trigonometry. Let me know if you have any questions! Have a great week! (: Article posted June 10, 2012 at 08:54 PM GMT-5 • comment • Reads 216
Article posted June 10, 2012 at 09:02 PM GMT-5 • comment • Reads 207 Hello everyone! Firstly, I would just like to explain that last week Blogmeister was experiencing some technical difficulties, so I was unable to post at that time. The post from both then and the most recent one should be uploaded now. Anyways, this week is the beginning of finals for my school! I am very nervous because I've never taken finals before, and I'm not sure what to expect. Wish me luck! In order to review once again for geometry, I am posting another review questions below. I chose this particular one because I find this area formula hard to remember sometimes, because it doesn't involve side lengths, rather, diagonal lengths. For this problem, there was a rhombus shown with diagonals of 21 ft and 18 ft. The instructions were to find the area, and the possible answers were as shown: a) 378 sq. ft b) 189 sq. ft c) 162 sq. ft d) 27(square root of)85 sq. ft. I knew that the formula for the area of a rhombus is 1/2d1*d2, so I substituted in the diameter lengths: A = 1/2 * 21 * 18 = 189 With that equation, I was able to find that the area was 189 sq. ft., or answer b. I hope this has been a helpful review question! Have a great week! (: Article posted June 10, 2012 at 09:02 PM GMT-5 • comment • Reads 207
Article posted June 11, 2012 at 08:18 AM GMT-5 • comment • Reads 256 Well, the final is coming up for geometry this. Just one more test that I need to study for. I can't say that I enjoy these types of things, but I still need to do them. To help out everybody, though, here are two problems that were on homework from this term. There are two since the blog wasn't working last week when I was supposed to post one and one from this week, so I'm just combining the two blogs into this week's. Here they are: The lengths of the diagonals of a rhombus are 2 in. and 5 in. Find the measures of the angles of the rhombus to the nearest degree. Describe each translation using an ordered pair. 2 units to the left, 1 unit down. Good luck to everyone on the final! Article posted June 11, 2012 at 08:18 AM GMT-5 • comment • Reads 256
Article posted June 10, 2012 at 07:04 PM GMT-5 • comment • Reads 190 It seems like just yesterday that we were beginning our weekly blogs, and now we are writing our last ones. This is the week of finals, and after this week, we only have one day left of school. I cannot believe that school went by this fast. I would have to say that this school year was filled with learning, and fun! On another note, we still must remain serious. We still have our final exam on Monday, and that requires a gracious amount of studying. To study for the exam, I have come up with a few review guides that will help me go over anything that I have been struggling with. One thing that I have had a bit of trouble with is vectors. Therefore, I picked out a problem about vectors from my book and worked out the answer to it. The problem was "describe each vector as an ordered pair". To do so, you had to first figure out the sine ratio as well as the cosine ratio. Then, you would order them so that they describe the vector. It took some practice to understand, but eventually, I got it! I now feel prepared for this final exam! Article posted June 10, 2012 at 07:04 PM GMT-5 • comment • Reads 190
Article posted June 4, 2012 at 09:38 AM GMT-5 • comment • Reads 251 Salutations fellow bloggers! This week in geometry class we learned about the law of sine and cosine. It isn't to difficult but I don't really understand how it saves you time. My question for the final review is: What quadrilaterals have diagonals that bisect each other? Answer: Parallelogram, rhombus, rectangle, and square Have a great week everyone! Article posted June 4, 2012 at 09:38 AM GMT-5 • comment • Reads 251
Article posted June 5, 2012 at 07:53 PM GMT-5 • comment • Reads 237 Hi everyone! Just writing because before I could add my review question, Blogmeister shut itself down! NOOOOOO!!!! Well, anyway, here’s my Trigonometry Review question: Kevin and Zane have just built a tree house. It is 20 ft off the ground, and the tree is perpendicular to the ground. When Kevin’s Mom goes out to look at the tree house, the angle of elevation from her feet to the tree house is 50*. Zane wants to know how far the tree that has the tree house in it is from the house, but he has lost the tape measure. How far away is the tree from the house? ***SPOILER ALERT*** Answer: Roughly 17 Feet --Joe Article posted June 5, 2012 at 07:53 PM GMT-5 • comment • Reads 237
Article posted June 10, 2012 at 07:42 AM GMT-5 • comment • Reads 198 HELLO! WELCOME TO THE FINAL COMPLETE WEEK OF SCHOOL! Wow, the year has gone by so fast! Did I mention that this is our final weekly blog? I can’t believe how close we are to the end of the year! It seems like just yesterday we walked into Geometry for the first time. I have had a great weekend, capped of by a RockEucharist at my church! This week in Geometry, we continued with our trigonometry unit, and we learned the law of cosines. At first I didn’t understand it, but once I got it, I had it for good. I think that on our trig. test this week, I will not have that much trouble, as this unit has been about mostly algebra and I do not have much trouble with algebra. But you never know-- it may be very hard. Our final is also coming up. Wow, that snuck up on me. Here is another review question: Ben and Griffin love modern houses. They want to make a museum about the evolution of houses. While searching for a place to put the museum, they find a modern building that is a rhombus. The sides of the house are 50 feet long. The angles formed by two intersecting lines are 120* and 60*. How much area is in the museum? ANSWER: 4330 ft. squared Have a great last week! --Joe Article posted June 10, 2012 at 07:42 AM GMT-5 • comment • Reads 198
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Article posted June 2, 2012 at 07:33 AM GMT-5 • comment • Reads 539 Today is Sunday, June 10, and I'm doing this blog a week late because the website crashed last week. It also means that there are exactly 2 school days until Finals start! I'm starting to worry it will be really hard to study without being distracted by the nice weather. To get in the spirit of studying, I'm going to post a review question from one of my past tests today on this blog. This is it: Determine which three numbers could be the sides of a right triangle. A. 64,73,98 B. 64,72,96 C. 65,72,97 The answer is C because of the Pythagorean theorem. When you square each number, the first two must add up to the third one's square. If that happens, then it is a right triangle. Choice C is the only one that happens for, so it is the correct answer. Until Next week, Article posted June 2, 2012 at 07:33 AM GMT-5 • comment • Reads 539
Article posted June 10, 2012 at 07:03 AM GMT-5 • comment • Reads 309 I'm doing the second blog for this week today, Sunday, June 10 because the website crashed last week. As I said earlier, there are exactly two school days until finals start! While this may seem terrible, I'm actually very happy because it means that we are in the home stretch before summer! I'm posting another review question this week: this will be about lines in triangles. Which line in triangles meet at the orthocenter when all three are drawn in a triangle? The answer to this question is altitudes. An altitude, also known as the height of a triangle, is a perpendicular line to the base drawn from a vertex. School may be almost done, but I can't give up now. This is the last blog of this year, and I can say that's a good thing! Article posted June 10, 2012 at 07:03 AM GMT-5 • comment • Reads 309
Article posted June 10, 2012 at 07:40 PM GMT-5 • comment • Reads 291 For this weeks blog I was asked to post two review questions for the final exam. The final exam is this week and I need to start studying, so I am going to start by making these review questions. My first question is: What does MAPA COCI stand for? A hint for this question is to remember lines in triangles and their concurrent points! My next question is: How do you find the circumference of a circle with a radius of 10 cm? Answers: Question 1: Median, Altitude, Perpendicular Bisector, Angle Bisector. Centroid, Orthocenter, Circumcenter, Incenter. Question 2: 20pi cm. Article posted June 10, 2012 at 07:40 PM GMT-5 • comment • Reads 291
Article posted June 7, 2012 at 04:31 PM GMT-5 • comment • Reads 279 On Sunday I watched a lot of TV because of all the crappy weather that we had. The best thing I watched was the golf touroment where Tiger Woods hit an unbelievable shot to clinch the victory. The problem I am working on for this blog is # 12 from the green version on test # 7. It asks for you to find the area of a parallelogram that has sides of 8 and 6. On the test, I got the problem wrong because I didn't know the area formula for parallelograms but when I went back and re-did the problem while studying for the retake I realized that I had to find the height by making a 45-45-90 triangle because the formula is a=bh. I found that the height was 3 root 2 and 8 time 3 root 2 is 24 root 2. Article posted June 7, 2012 at 04:31 PM GMT-5 • comment • Reads 279
Article posted June 7, 2012 at 04:49 PM GMT-5 • comment • Reads 356 I am blogging on a thursday for probably the first time, it seems wierd. I am getting ready to do my geometry homework. I am also going to watch the Miami Heat vs the Boston Celtics later, it is must win game for the Heat. The problem that I am going to work on in this blog is # 17 from the green version of test # 7. It asks you find the area of a deck that surrounds a hot tub if the hot tub has a diameter of 6 meters and the deck is 2 meters wide. When I took the test I knew how to do the problem but I just had a brain cramp and screwed up the final answer. The way to figure out this problem is to find the area of the hot tub and then subtract that from the area of the deck. The only formula that you need to use on this problem is a=pie(r) squared. The first thing to do is sub the number 3 into that formula and you end up with 9 pie. You then plug the number 5 into that formula because 6+ 2+2 equals 10 as a diameter and half of that equals 5. When plugging that into the equation you end up with 25 pie. The problem that I made on the test was that I forgot to subtract 9 pie from 25 pie so that is the last step in this problem. Your final answer should be 16 pie. Article posted June 7, 2012 at 04:49 PM GMT-5 • comment • Reads 356
Article posted June 10, 2012 at 10:26 PM GMT-5 • comment • Reads 224 This blog is the twenty-third of my personal weekly blog This blog was supposed to be posted last week but blogmeister was down! Oh no! We got everything squared away though, so yay! Last week was pretty gloomy, the weather hasn't been that nice until this weekend! Last week in geometry we started learning more in trigonometry and then started learning about law of sines and law of cosines! It's been pretty easy, but some things trip me up easily. Anyways the year is winding down and I can not wait until SUMMER! Midterms start this week! Here is a review question on what we have learned in Geometry! Ms. J told us that studying our tests and quizzes would be most helpful so that is exactly what I will be doing! A particular unit that I didn't fair so well on was working with the lines of triangles! Here is a review question from the Lines of Triangles Quiz that we took on February 7th. Which of the following are the slopes of two perpendicular lines? a. 3 and -3 b. 5 and 1/5 c. no slope and undefined d. -2/3 and 3/2 If you recall that the slope of any line perpendicular to another is the negative reciprocal, then this problem is easy to solve! A is not the answer because the negative reciprocal or 3 would be -1/3. B does not work because the negative reciprocal of 5 would be -1/5. c does not work because they aren't specific lines! THEREFORE D IS THE ANSWER! and to prove it.. the negative reciprocal of -2/3 would be - ( -3/2) or 3/2! TADA! Article posted June 10, 2012 at 10:26 PM GMT-5 • comment • Reads 224
Article posted June 10, 2012 at 10:34 PM GMT-5 • comment • Reads 419 This is the twenty-fourth of my personal weekly blog. Alas I have reached my last blog. Farewell classblogmeister... I won't miss you too much I promise! This week in Geometry we continued working with the law of sines and the law of cosines! We have a test this week before our final so that means a lot of studying on our parts! We pretty much did a lot of reviewing on the all the parts of trigonometry, which was helpful! This review question will be coming from the Quadrilaterals test we took on March 29th. I GOT THIS QUESTION WRONG! I honestly have no idea how I could've made the mistake, but I did, so make sure you don't jump to conclusions! Consecutive sides of a rectangle are congruent. a. sometimes b. always c. never the answer is SOMETIMES because a square (which is a rectangle) has congruent consecutive sides! YAY GEOMETRY!(: Article posted June 10, 2012 at 10:34 PM GMT-5 • comment • Reads 419
Article posted June 10, 2012 at 10:32 AM GMT-5 • comment • Reads 224 For Ch 12, the work was on transformations. The review question will be for rotations. The question as what is the point of rotation, and what is the angle of rotation. When given a pre-image and an image, to find the point of rotation, it will either be where the two images intersect or a point not connected to either image. Then to find the angle of rotation, you will need a protractor. Put your protractor on the point of rotation and pick a point you want to measure from. You always measure from the pre-image and counterclockwise unless stated to go clockwise. After you pick the point, find the point on the image that is the same, and measure. That will give you the angle of rotation Stay classy bloggers. Article posted June 10, 2012 at 10:32 AM GMT-5 • comment • Reads 224
Article posted June 10, 2012 at 11:24 AM GMT-5 • comment • Reads 217 Our final exam is Wednesday! Ahhh! And in preparation to that I am going to be posting a review question from one of the chapters we learned about this semester. In chapter 7 we learned more about areas of formulas. My review question today, will be finding the area of a rhombus (remember that the diagonals of a rhombus bisect each other). The formula for an area of a rhombus is 1/2(d1+d2), d stands for diagonal. If your given information was that one diagonal was 7cm and half of the other diagonal was 5.5cm, this is how you would solve it: We know that the other half of the diagonal is 5.5cm, because it gets bisected from the other diagonal. Now we plug in our given information to come up with 1/2(7+11) = area. Simplify to get 1/2(18) = area. Now simply multiply by 1/2 or divide by two, and we get 9cm squared = area Hopefully this helped you! I will be posting another review question soon! Goodbye! Article posted June 10, 2012 at 11:24 AM GMT-5 • comment • Reads 217
Article posted June 10, 2012 at 12:01 PM GMT-5 • comment • Reads 233 Hello again! Today, I am posting another review question. This comes from chapter 5. Question: Which of they following is the concurrent point of the altitudes? Answer: Orthocenter. I always remember which concurrent point goes with which line of a triangle because of MAPA COCI. MAPA COCI stands for: Median --> Centroid Altitude --> Orthocenter Perpendicular Bisector -->Circumcenter Angle Bisector --> Incenter This helped me remember and hopefully it will help you too! :) Now down to the sad stuff. :( I won't be saying 'see you next week' anymore..... because this is my last blog! *Gasp* I hope my blogs have helped you in some way or another, but right now I have to say goodbye for good. I know you'll miss me though! Farewell, Livy Article posted June 10, 2012 at 12:01 PM GMT-5 • comment • Reads 233
Article posted June 11, 2012 at 02:06 PM GMT-5 • comment • Reads 235 Question Given: Circle O with Diamter CD, AB is parrallel to CD, and arc AB=80 degrees Find arc CA [1] 50 [2] 60 [3] 80 [4] 100 Answer 50 Article posted June 11, 2012 at 02:06 PM GMT-5 • comment • Reads 235
Article posted June 11, 2012 at 02:10 PM GMT-5 • comment • Reads 381 Question Given: In triangle ABC, B=120, c=15, and a=15 Find C Answer 30 degrees Article posted June 11, 2012 at 02:10 PM GMT-5 • comment • Reads 381
Article posted June 11, 2012 at 09:04 PM GMT-5 • comment • Reads 241 The midterm is in two days and I'm getting a little nervous. We also have a test tomorrow that isn't helping with the nerves! Hopefully the test tomorrow will make get me feeling completely confident with the trigonometry unit, so it'll be one less thing to study for the final! My first review question that I want to go over is number 13 from practice 58. It is working with vectors which I needed a slight refresher on. The question: Homing pigeons have the ability or instinct to find their way home when released hundreds of miles away from home. Homing pigeons carried news of Olympic victories to various cities in ancient Greece. Suppose one such pigeon took off from Athens and landed in Sparta, which is 73 miles west and 64 miles south of Athens. Find the distance and its direction of flight. The first step you would need to take is plugging the coordinates into the distance formula. It comes out with 97, so Athens and Sparta are 97 miles apart. Then, you need to use the tangent ratio. This is the opposite side over the adjacent side. Therefore, tan(x) = 64/73 --> x = tan-1( 64/73) --> x = 41°, so the direction of flight is 41°. Article posted June 11, 2012 at 09:04 PM GMT-5 • comment • Reads 241
Article posted June 11, 2012 at 09:16 PM GMT-5 • comment • Reads 229 We were required two review questions, so I think I'll choose one from the beginning of the semester just as a refresher. Number 25 on the quadrilaterals test was asking us to label the coordinates of parallelogram LAST using only three variables. I chose d, c, and b (along with 0). Point 'a' was on the y-axis so the x coordinate was 0. Then, I chose 'b' to act as the height, or y coordinate. Point 'l' is the lower, left-hand point, and it is on the x-axis, making the y coordinate 0. Then, I made the x coordinate '-c' because it is to the left of the y-axis. Point 't' is the lower right-hand point, and it also lies on the x-axis, making the y coordinate 0. Then, I chose 'd' to act as the x coordinate. Point 's' was the upper right point of the parallelogram, so the height was also 'b', making the y coordinate automatically 'b'. The x coordinate is going to be 'd+c', because 'c' is the length that the segment goes on longer past 'd', which is where you get the '+c'. Hopefully that all made sense and was (slightly) helpful! Article posted June 11, 2012 at 09:16 PM GMT-5 • comment • Reads 229
Article posted June 10, 2012 at 09:00 PM GMT-5 • comment • Reads 222 This blog is supposed to be for two weeks ago but due to problems with the website the due date was postponed. This blog I will be explaining how to do a problem from one of the previous tests that I have taken. I have chosen to explain number 4 from test 7. The problem gives you a rhombus and tells you to find the area using the diagonals of 18 and 21. To find the area you must use the formula A=.5(d1)(d2). You would use substitution of make the equation A=.5(18)(21). If you do the math the area comes out to be 189ft squared. That is how you would do a problem like that... Article posted June 10, 2012 at 09:00 PM GMT-5 • comment • Reads 222
Article posted June 10, 2012 at 09:23 PM GMT-5 • comment • Reads 311 This week in geometry I am doing the same thing as last week. I have to pick a problem from any of my previous tests and I must explain how to do it.On problem 4 from the chapter 7 retest, the problem gives you a parallelogram and tells you to find the area using a height of 24 and a base of 10. The formula you must use is A= base times height. You would set up the equation as A= 24 times 10. The answer to the problem is 240 meters squared. That is how you do an problem like that... Article posted June 10, 2012 at 09:23 PM GMT-5 • comment • Reads 311
Article posted June 12, 2012 at 05:29 PM GMT-5 • comment • Reads 234 Question: The diagonals of a square bisect all angles Answer: Always Article posted June 12, 2012 at 05:29 PM GMT-5 • comment • Reads 234
Article posted June 12, 2012 at 05:31 PM GMT-5 • comment • Reads 312 Question: A triangle has side lengths of 8 cm, 14 cm, and 11 cm. Classify the triangle. Answer:Obtuse Article posted June 12, 2012 at 05:31 PM GMT-5 • comment • Reads 312
Article posted June 9, 2012 at 10:48 AM GMT-5 • comment • Reads 303 Hey guys so the school year's almost over. Yeah, summer. Unfortunately to get to summer we have to go through probably the wort week of the school year. Finals. Yeah you all know what I'm talking about. So here is one of the review questions that I'm using to study for my geometry final. This question is on right triangle trig. Good luck! Find X and H Answer: X=13 H=8.1 Article posted June 9, 2012 at 10:48 AM GMT-5 • comment • Reads 303
Article posted June 4, 2012 at 06:43 AM GMT-5 • comment • Reads 261 Hey guys so as I said in my last blog we are learning about SOH, CAH, TOA. Well now we are learning about the law of sines. Here is a review question and its answer about the law of sines. If m Answer: Measure of angle B is 59.4. Article posted June 4, 2012 at 06:43 AM GMT-5 • comment • Reads 261
Article posted June 7, 2012 at 09:17 PM GMT-5 • comment • Reads 224 We have finals in two weeks, and I don't really have any specific emotions for the end of the year. It was an awesome freshman year and a good first highschool experience. I hope next year will be as good, but then again I won't get to have Ms. J anymore:( To study for the final we must post a review question. Mine is- If a plane takes off at 25 degrees and flies 1600 ft, what is its altitude. SPOILER BELOW To do this problem, you must first write out the formula for sine. Using SOH, it would come out to be sin25=X/1600. Then you would multiply by 1600 on both sides to isolate X. This would make it 1600*sin(25)=X. Enter 1600*sin(25)into your calculator and the resulting number is the altitude-676.18 ft Article posted June 7, 2012 at 09:17 PM GMT-5 • comment • Reads 224
Article posted June 10, 2012 at 10:45 PM GMT-5 • comment • Reads 240 Hello everyone, I'm sad to tell you that this will be my last geometry blog :( The year is finally coming to an end. Six more days of school until summer! Here is my final exam review question: A triangle has three sides with the lengths of 6:8:10 What type of triangle is it? A.) acute B.)obtuse C.) right Answer: C. Right Article posted June 10, 2012 at 10:45 PM GMT-5 • comment • Reads 240
Article posted June 10, 2012 at 11:25 AM GMT-5 • comment • Reads 208 Hello Bloggers!!! This week I will be giving you a practice problem for you to solve!! At the bottom of the blog I will show you the work to get to the answer!! Question: You know two sides and an angle (6cm, 10cm, and 56 degrees) find the missing side. Answer: X2(squared)=6(squared)+10(squared) - 2(6)(10) - cos(56) by plugging this equation: 6(squared) + 10(squared) -2(6)(10) - cos(56) into your calculator, you will come up with X2(squared)= 15.4408071 you need to find X, so you have to find the square root. Type this into your calculator and X= 3.92947924 Great Job!!!! I will be blogging soon!!!! Article posted June 10, 2012 at 11:25 AM GMT-5 • comment • Reads 208
Article posted June 9, 2012 at 02:17 PM GMT-5 • comment • Reads 217 Hey bloggers! Just as I promised on Wednesday, here is another final review question! "In parallelogram BARK, m first draw the paralleglogram: 4x = xsquared - 60 xsqaured - 4x - 60 = 0 (x + 6)(x - 10) xsquared - 10x + 6x - 60 xsquared - 4x - 60 x + 6 = 0 or x - 10 = 0 x = -6 or x = 10 7x + m 70 + m m m< ARK = 110 dgrees Wish me luck on my final! Grace Article posted June 9, 2012 at 02:17 PM GMT-5 • comment • Reads 217
Article posted June 6, 2012 at 04:34 PM GMT-5 • comment • Reads 234 Hey bloggers! I am going to show you a good review question for our Geometry final next week! "Q is the intersection of AC and BD and ABCD. Find the area of kite ABCD if AB= 10, BC= 20, AC = 30, and BQ = 5." A= 1/2 x d1 x d2 d1= DB = 2BQ = 10. d2 = AC = 30 A = 1/2 x 10 x 30 Area = 150 squared units I will be posting another question later in the week due to the website being down last weekend. Later, Grace Article posted June 6, 2012 at 04:34 PM GMT-5 • comment • Reads 234
Article posted June 10, 2012 at 08:47 PM GMT-5 • comment • Reads 219 Welcome back everyone, since we are nearing the end of the school year, I was thinking that we could all start posting review questions to prepare ourselves for the upcoming final! What is a perpendicular bisector? And, how do you find it for any triangle? A perpendicular bisector is a line that divides a side of a triangle directly in the middle, but it bisects it at a 90˚ angle. Each side of the triangle has it’s own perpendicular bisector. In a triangle, there are 3 different sides each that have a perpendicular bisector. Since there are three different types of triangles (acute, right, and obtuse), each location of the circumcenter is different. For any acute triangle, the circumcenter is inside of the triangle. The point where all three of the Perpendicular bisectors intersect is at the circumcenter. But, like always there are 2 special cases. For a right triangle, the circumcenter of a right triangle is on the hypotenuse. But for an obtuse triangle: the circumcenter of an obtuse triangle is outside of the triangle. A circumscribed circle is a circle that is drawn from the circumcenter that hits all of the vertices of the triangle. The circumcenter is always equidistant to all vertices or any other point on the circumscribed circle. If you need anymore help with perpendicular bisectors reply to this post! I hope you all begin to review now! See you all next week! -Kathleen Article posted June 10, 2012 at 08:47 PM GMT-5 • comment • Reads 219
Article posted June 10, 2012 at 10:17 AM GMT-5 • comment • Reads 204 Hello again! As we are approaching our final exam next week, we are continuing to post review questions! These are aimed to help us remember the concepts we have learned this semester! The question I'm going to ask takes us back a chapter, and deals with triangles! Be sure to blog back with any questions. Q: A triangle has side lengths of 8cm, 11cm, and 14cm. Classify this triangle as acute, obtuse, or right. A: OBTUSE To find this answer, we first begin with the Pythagorean theorem (A squared + B squared = c squared). So, when we plug in what is known, our formula becomes 64 + 121 = 196. Here's the rule to classify: IF (A squared + b squared > C squared) then the triangle is acute. IF (A squared + b squared < C squared) then the triangle is obtuse. IF (A squared + b squared = C squared) then the triangle is right. In our formula, 64 + 121 = 185. Because 185 < 196, then the triangle is obtuse! Hope that helped and you learned something knew! This is my last weekly blog... but I'll be writing occasionally until the end of the year. Talk to you then! Emma Article posted June 10, 2012 at 10:17 AM GMT-5 • comment • Reads 204
Article posted June 14, 2012 at 12:40 PM GMT-5 • comment • Reads 240 We finally made it! Finals are around the corner, and so is summer! I can’t wait to ditch my backpack and head to the beach. But first, I have to survive finals. To review for Geometry, I’ve posted one of my favorite homework problems from a past unit below. KL⎮⎮JM in isosceles trapezoid JKLM. Find the values of x and y if m⦟J=(23x-8)º, m⦟K=(12y-13)º, and m⦟M=(17x+10)º. m⦟J=m⦟M 23x-8=17x+10 6x-8=10 6x=18 x=3 m⦟J+m⦟K=180 23x-8+12y-13=180 61+12y-13=180 12y=132 y=11 I chose this problem because I was from one of the first units of the semester, and I didn’t remember that unit at all. I thought that it would be nice to review since I had forgotten about it. Article posted June 14, 2012 at 12:40 PM GMT-5 • comment • Reads 240
Article posted June 14, 2012 at 12:40 PM GMT-5 • comment • Reads 244 We finally made it! Finals are around the corner, and so is summer! I can’t wait to ditch my backpack and head to the beach. But first, I have to survive finals. To review for Geometry, I’ve posted one of my favorite homework problems from a past unit below. KL⎮⎮JM in isosceles trapezoid JKLM. Find the values of x and y if m⦟J=(23x-8)º, m⦟K=(12y-13)º, and m⦟M=(17x+10)º. m⦟J=m⦟M 23x-8=17x+10 6x-8=10 6x=18 x=3 m⦟J+m⦟K=180 23x-8+12y-13=180 61+12y-13=180 12y=132 y=11 I chose this problem because I was from one of the first units of the semester, and I didn’t remember that unit at all. I thought that it would be nice to review since I had forgotten about it. Article posted June 14, 2012 at 12:40 PM GMT-5 • comment • Reads 244
Article posted June 6, 2012 at 04:33 PM GMT-5 • comment • Reads 257 Hello fellow blog readers! Sorry that this blog is so late compared to my others. Blogmeister, as I'm sure you know, has been down for the past couple of days. I am sorry to keep you waiting for so long. Now to the important stuff. In geometry, I have a question about the latest test. My question is how to find the length of an arc. I do not remember and was wondering if you bloggers could help me out. If you comment with the correct answer, I will mention you in my next blog. Well that is all for now blog-readers! Stay tuned for this Sunday's blog and question. Stay classy and thanks for reading! Article posted June 6, 2012 at 04:33 PM GMT-5 • comment • Reads 257
Article posted June 10, 2012 at 04:52 PM GMT-5 • comment • Reads 215 The year is almost over! We have two tests this next week in geometry. One tomorrow on chapter 9 and one on June 18th on everything we've learned this semester (our final). To help review for this final I have been told to blog a review question with the answer. Do you remember the distance formula? Well for coordinate proofs we had to use the distance formula to show how one side was of equal length to the other side. To show this you use the formula d = the square root of (y2-y1)squared + (x2-x1)squared. If you plug in the coordinates from #26 on the quadrilaterals test you get AS = the square root of (b-b)squared + (a+c-0)squared. This simplifies to the square root of (a+c)squared which equals a + c. Hope this was helpful, good luck on the test and the final! Article posted June 10, 2012 at 04:52 PM GMT-5 • comment • Reads 215
Article posted June 10, 2012 at 04:11 PM GMT-5 • comment • Reads 218 AHHHH! Three more days until final exams! Before I show you the review question take a deep breath... Inhale, now exhale. Feel better? Good. Let's begin! This is from chapter 9. This chapter is on SOH CAH TOA, the law of sines, the law of cosines, and vectors. Say you had a triangle where the length of the hypotenuse (AC) is 12cm, and the short side (AB) and the long side (BC) is unknown. Angle C is given as 23 degrees and angle B is 90 degrees. To solve this problem you use CAH because you are given a right triangle, an angle (23 degrees), the hypotenuse to that angle (12cm), and the long side is present. To set it up you write; cos23 = X/12. To solve you multiple cos23 by 12, so it's 12cos23 = X. X = 11cm. I hope this was helpful! Article posted June 10, 2012 at 04:11 PM GMT-5 • comment • Reads 218
Article posted June 10, 2012 at 08:54 PM GMT-5 • comment • Reads 216 Hello everyone! Hope your week has gone nicely! Mine has been pretty interesting, because my brother had an English exchange student arrive on Thursday. He will be staying with us for the week, and I am really excited! English accents are so cool! This exchange trip has also reminded me that the end of the school year is really close! In order to prepare for the geometry final, I am going to tell you about a question we had to do for homework as a way to review. This problem was on a worksheet I did recently involving the Law of Sines: sinA = sinB a b A and B stand for angles of an oblique triangle, while a and b stand for the 2 sides opposite them. Basically, you can use this law to find one of these four measures, as long as you have an angle and the side opposite. Basically, the problem was a triangle with angles A, B, and C, and sides a, b, and c. The measure of angle A was 35º, the measure of side a was 8 cm, and the measure of side b was 12 cm. The problem asked you to find the measure of angle B. Through substitution, I came out with this equation: sin(35) = sin(B) 8 12 By multiplying both sides by 12 and then by sin to the (-1), I was able to then solve for B: sin-1(12sin(35)) = sin-1(sin(B)) 8 m This process took some time to get used to, but I've found I actually like this sorts of problems in trigonometry. Let me know if you have any questions! Have a great week! (: Article posted June 10, 2012 at 08:54 PM GMT-5 • comment • Reads 216
Article posted June 10, 2012 at 09:02 PM GMT-5 • comment • Reads 207 Hello everyone! Firstly, I would just like to explain that last week Blogmeister was experiencing some technical difficulties, so I was unable to post at that time. The post from both then and the most recent one should be uploaded now. Anyways, this week is the beginning of finals for my school! I am very nervous because I've never taken finals before, and I'm not sure what to expect. Wish me luck! In order to review once again for geometry, I am posting another review questions below. I chose this particular one because I find this area formula hard to remember sometimes, because it doesn't involve side lengths, rather, diagonal lengths. For this problem, there was a rhombus shown with diagonals of 21 ft and 18 ft. The instructions were to find the area, and the possible answers were as shown: a) 378 sq. ft b) 189 sq. ft c) 162 sq. ft d) 27(square root of)85 sq. ft. I knew that the formula for the area of a rhombus is 1/2d1*d2, so I substituted in the diameter lengths: A = 1/2 * 21 * 18 = 189 With that equation, I was able to find that the area was 189 sq. ft., or answer b. I hope this has been a helpful review question! Have a great week! (: Article posted June 10, 2012 at 09:02 PM GMT-5 • comment • Reads 207
Article posted June 11, 2012 at 08:18 AM GMT-5 • comment • Reads 256 Well, the final is coming up for geometry this. Just one more test that I need to study for. I can't say that I enjoy these types of things, but I still need to do them. To help out everybody, though, here are two problems that were on homework from this term. There are two since the blog wasn't working last week when I was supposed to post one and one from this week, so I'm just combining the two blogs into this week's. Here they are: The lengths of the diagonals of a rhombus are 2 in. and 5 in. Find the measures of the angles of the rhombus to the nearest degree. Describe each translation using an ordered pair. 2 units to the left, 1 unit down. Good luck to everyone on the final! Article posted June 11, 2012 at 08:18 AM GMT-5 • comment • Reads 256
Article posted June 10, 2012 at 07:04 PM GMT-5 • comment • Reads 190 It seems like just yesterday that we were beginning our weekly blogs, and now we are writing our last ones. This is the week of finals, and after this week, we only have one day left of school. I cannot believe that school went by this fast. I would have to say that this school year was filled with learning, and fun! On another note, we still must remain serious. We still have our final exam on Monday, and that requires a gracious amount of studying. To study for the exam, I have come up with a few review guides that will help me go over anything that I have been struggling with. One thing that I have had a bit of trouble with is vectors. Therefore, I picked out a problem about vectors from my book and worked out the answer to it. The problem was "describe each vector as an ordered pair". To do so, you had to first figure out the sine ratio as well as the cosine ratio. Then, you would order them so that they describe the vector. It took some practice to understand, but eventually, I got it! I now feel prepared for this final exam! Article posted June 10, 2012 at 07:04 PM GMT-5 • comment • Reads 190
Article posted June 4, 2012 at 09:38 AM GMT-5 • comment • Reads 251 Salutations fellow bloggers! This week in geometry class we learned about the law of sine and cosine. It isn't to difficult but I don't really understand how it saves you time. My question for the final review is: What quadrilaterals have diagonals that bisect each other? Answer: Parallelogram, rhombus, rectangle, and square Have a great week everyone! Article posted June 4, 2012 at 09:38 AM GMT-5 • comment • Reads 251
Article posted June 5, 2012 at 07:53 PM GMT-5 • comment • Reads 237 Hi everyone! Just writing because before I could add my review question, Blogmeister shut itself down! NOOOOOO!!!! Well, anyway, here’s my Trigonometry Review question: Kevin and Zane have just built a tree house. It is 20 ft off the ground, and the tree is perpendicular to the ground. When Kevin’s Mom goes out to look at the tree house, the angle of elevation from her feet to the tree house is 50*. Zane wants to know how far the tree that has the tree house in it is from the house, but he has lost the tape measure. How far away is the tree from the house? ***SPOILER ALERT*** Answer: Roughly 17 Feet --Joe Article posted June 5, 2012 at 07:53 PM GMT-5 • comment • Reads 237
Article posted June 10, 2012 at 07:42 AM GMT-5 • comment • Reads 198 HELLO! WELCOME TO THE FINAL COMPLETE WEEK OF SCHOOL! Wow, the year has gone by so fast! Did I mention that this is our final weekly blog? I can’t believe how close we are to the end of the year! It seems like just yesterday we walked into Geometry for the first time. I have had a great weekend, capped of by a RockEucharist at my church! This week in Geometry, we continued with our trigonometry unit, and we learned the law of cosines. At first I didn’t understand it, but once I got it, I had it for good. I think that on our trig. test this week, I will not have that much trouble, as this unit has been about mostly algebra and I do not have much trouble with algebra. But you never know-- it may be very hard. Our final is also coming up. Wow, that snuck up on me. Here is another review question: Ben and Griffin love modern houses. They want to make a museum about the evolution of houses. While searching for a place to put the museum, they find a modern building that is a rhombus. The sides of the house are 50 feet long. The angles formed by two intersecting lines are 120* and 60*. How much area is in the museum? ANSWER: 4330 ft. squared Have a great last week! --Joe Article posted June 10, 2012 at 07:42 AM GMT-5 • comment • Reads 198
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Article posted June 2, 2012 at 07:33 AM GMT-5 • comment • Reads 539 Today is Sunday, June 10, and I'm doing this blog a week late because the website crashed last week. It also means that there are exactly 2 school days until Finals start! I'm starting to worry it will be really hard to study without being distracted by the nice weather. To get in the spirit of studying, I'm going to post a review question from one of my past tests today on this blog. This is it: Determine which three numbers could be the sides of a right triangle. A. 64,73,98 B. 64,72,96 C. 65,72,97 The answer is C because of the Pythagorean theorem. When you square each number, the first two must add up to the third one's square. If that happens, then it is a right triangle. Choice C is the only one that happens for, so it is the correct answer. Until Next week, Article posted June 2, 2012 at 07:33 AM GMT-5 • comment • Reads 539
Article posted June 10, 2012 at 07:03 AM GMT-5 • comment • Reads 309 I'm doing the second blog for this week today, Sunday, June 10 because the website crashed last week. As I said earlier, there are exactly two school days until finals start! While this may seem terrible, I'm actually very happy because it means that we are in the home stretch before summer! I'm posting another review question this week: this will be about lines in triangles. Which line in triangles meet at the orthocenter when all three are drawn in a triangle? The answer to this question is altitudes. An altitude, also known as the height of a triangle, is a perpendicular line to the base drawn from a vertex. School may be almost done, but I can't give up now. This is the last blog of this year, and I can say that's a good thing! Article posted June 10, 2012 at 07:03 AM GMT-5 • comment • Reads 309
Article posted June 10, 2012 at 07:40 PM GMT-5 • comment • Reads 291 For this weeks blog I was asked to post two review questions for the final exam. The final exam is this week and I need to start studying, so I am going to start by making these review questions. My first question is: What does MAPA COCI stand for? A hint for this question is to remember lines in triangles and their concurrent points! My next question is: How do you find the circumference of a circle with a radius of 10 cm? Answers: Question 1: Median, Altitude, Perpendicular Bisector, Angle Bisector. Centroid, Orthocenter, Circumcenter, Incenter. Question 2: 20pi cm. Article posted June 10, 2012 at 07:40 PM GMT-5 • comment • Reads 291
Article posted June 7, 2012 at 04:31 PM GMT-5 • comment • Reads 279 On Sunday I watched a lot of TV because of all the crappy weather that we had. The best thing I watched was the golf touroment where Tiger Woods hit an unbelievable shot to clinch the victory. The problem I am working on for this blog is # 12 from the green version on test # 7. It asks for you to find the area of a parallelogram that has sides of 8 and 6. On the test, I got the problem wrong because I didn't know the area formula for parallelograms but when I went back and re-did the problem while studying for the retake I realized that I had to find the height by making a 45-45-90 triangle because the formula is a=bh. I found that the height was 3 root 2 and 8 time 3 root 2 is 24 root 2. Article posted June 7, 2012 at 04:31 PM GMT-5 • comment • Reads 279
Article posted June 7, 2012 at 04:49 PM GMT-5 • comment • Reads 356 I am blogging on a thursday for probably the first time, it seems wierd. I am getting ready to do my geometry homework. I am also going to watch the Miami Heat vs the Boston Celtics later, it is must win game for the Heat. The problem that I am going to work on in this blog is # 17 from the green version of test # 7. It asks you find the area of a deck that surrounds a hot tub if the hot tub has a diameter of 6 meters and the deck is 2 meters wide. When I took the test I knew how to do the problem but I just had a brain cramp and screwed up the final answer. The way to figure out this problem is to find the area of the hot tub and then subtract that from the area of the deck. The only formula that you need to use on this problem is a=pie(r) squared. The first thing to do is sub the number 3 into that formula and you end up with 9 pie. You then plug the number 5 into that formula because 6+ 2+2 equals 10 as a diameter and half of that equals 5. When plugging that into the equation you end up with 25 pie. The problem that I made on the test was that I forgot to subtract 9 pie from 25 pie so that is the last step in this problem. Your final answer should be 16 pie. Article posted June 7, 2012 at 04:49 PM GMT-5 • comment • Reads 356
Article posted June 10, 2012 at 10:26 PM GMT-5 • comment • Reads 224 This blog is the twenty-third of my personal weekly blog This blog was supposed to be posted last week but blogmeister was down! Oh no! We got everything squared away though, so yay! Last week was pretty gloomy, the weather hasn't been that nice until this weekend! Last week in geometry we started learning more in trigonometry and then started learning about law of sines and law of cosines! It's been pretty easy, but some things trip me up easily. Anyways the year is winding down and I can not wait until SUMMER! Midterms start this week! Here is a review question on what we have learned in Geometry! Ms. J told us that studying our tests and quizzes would be most helpful so that is exactly what I will be doing! A particular unit that I didn't fair so well on was working with the lines of triangles! Here is a review question from the Lines of Triangles Quiz that we took on February 7th. Which of the following are the slopes of two perpendicular lines? a. 3 and -3 b. 5 and 1/5 c. no slope and undefined d. -2/3 and 3/2 If you recall that the slope of any line perpendicular to another is the negative reciprocal, then this problem is easy to solve! A is not the answer because the negative reciprocal or 3 would be -1/3. B does not work because the negative reciprocal of 5 would be -1/5. c does not work because they aren't specific lines! THEREFORE D IS THE ANSWER! and to prove it.. the negative reciprocal of -2/3 would be - ( -3/2) or 3/2! TADA! Article posted June 10, 2012 at 10:26 PM GMT-5 • comment • Reads 224
Article posted June 10, 2012 at 10:34 PM GMT-5 • comment • Reads 419 This is the twenty-fourth of my personal weekly blog. Alas I have reached my last blog. Farewell classblogmeister... I won't miss you too much I promise! This week in Geometry we continued working with the law of sines and the law of cosines! We have a test this week before our final so that means a lot of studying on our parts! We pretty much did a lot of reviewing on the all the parts of trigonometry, which was helpful! This review question will be coming from the Quadrilaterals test we took on March 29th. I GOT THIS QUESTION WRONG! I honestly have no idea how I could've made the mistake, but I did, so make sure you don't jump to conclusions! Consecutive sides of a rectangle are congruent. a. sometimes b. always c. never the answer is SOMETIMES because a square (which is a rectangle) has congruent consecutive sides! YAY GEOMETRY!(: Article posted June 10, 2012 at 10:34 PM GMT-5 • comment • Reads 419
Article posted June 10, 2012 at 10:32 AM GMT-5 • comment • Reads 224 For Ch 12, the work was on transformations. The review question will be for rotations. The question as what is the point of rotation, and what is the angle of rotation. When given a pre-image and an image, to find the point of rotation, it will either be where the two images intersect or a point not connected to either image. Then to find the angle of rotation, you will need a protractor. Put your protractor on the point of rotation and pick a point you want to measure from. You always measure from the pre-image and counterclockwise unless stated to go clockwise. After you pick the point, find the point on the image that is the same, and measure. That will give you the angle of rotation Stay classy bloggers. Article posted June 10, 2012 at 10:32 AM GMT-5 • comment • Reads 224
Article posted June 10, 2012 at 11:24 AM GMT-5 • comment • Reads 217 Our final exam is Wednesday! Ahhh! And in preparation to that I am going to be posting a review question from one of the chapters we learned about this semester. In chapter 7 we learned more about areas of formulas. My review question today, will be finding the area of a rhombus (remember that the diagonals of a rhombus bisect each other). The formula for an area of a rhombus is 1/2(d1+d2), d stands for diagonal. If your given information was that one diagonal was 7cm and half of the other diagonal was 5.5cm, this is how you would solve it: We know that the other half of the diagonal is 5.5cm, because it gets bisected from the other diagonal. Now we plug in our given information to come up with 1/2(7+11) = area. Simplify to get 1/2(18) = area. Now simply multiply by 1/2 or divide by two, and we get 9cm squared = area Hopefully this helped you! I will be posting another review question soon! Goodbye! Article posted June 10, 2012 at 11:24 AM GMT-5 • comment • Reads 217
Article posted June 10, 2012 at 12:01 PM GMT-5 • comment • Reads 233 Hello again! Today, I am posting another review question. This comes from chapter 5. Question: Which of they following is the concurrent point of the altitudes? Answer: Orthocenter. I always remember which concurrent point goes with which line of a triangle because of MAPA COCI. MAPA COCI stands for: Median --> Centroid Altitude --> Orthocenter Perpendicular Bisector -->Circumcenter Angle Bisector --> Incenter This helped me remember and hopefully it will help you too! :) Now down to the sad stuff. :( I won't be saying 'see you next week' anymore..... because this is my last blog! *Gasp* I hope my blogs have helped you in some way or another, but right now I have to say goodbye for good. I know you'll miss me though! Farewell, Livy Article posted June 10, 2012 at 12:01 PM GMT-5 • comment • Reads 233
Article posted June 11, 2012 at 02:06 PM GMT-5 • comment • Reads 235 Question Given: Circle O with Diamter CD, AB is parrallel to CD, and arc AB=80 degrees Find arc CA [1] 50 [2] 60 [3] 80 [4] 100 Answer 50 Article posted June 11, 2012 at 02:06 PM GMT-5 • comment • Reads 235
Article posted June 11, 2012 at 02:10 PM GMT-5 • comment • Reads 381 Question Given: In triangle ABC, B=120, c=15, and a=15 Find C Answer 30 degrees Article posted June 11, 2012 at 02:10 PM GMT-5 • comment • Reads 381
Article posted June 11, 2012 at 09:04 PM GMT-5 • comment • Reads 241 The midterm is in two days and I'm getting a little nervous. We also have a test tomorrow that isn't helping with the nerves! Hopefully the test tomorrow will make get me feeling completely confident with the trigonometry unit, so it'll be one less thing to study for the final! My first review question that I want to go over is number 13 from practice 58. It is working with vectors which I needed a slight refresher on. The question: Homing pigeons have the ability or instinct to find their way home when released hundreds of miles away from home. Homing pigeons carried news of Olympic victories to various cities in ancient Greece. Suppose one such pigeon took off from Athens and landed in Sparta, which is 73 miles west and 64 miles south of Athens. Find the distance and its direction of flight. The first step you would need to take is plugging the coordinates into the distance formula. It comes out with 97, so Athens and Sparta are 97 miles apart. Then, you need to use the tangent ratio. This is the opposite side over the adjacent side. Therefore, tan(x) = 64/73 --> x = tan-1( 64/73) --> x = 41°, so the direction of flight is 41°. Article posted June 11, 2012 at 09:04 PM GMT-5 • comment • Reads 241
Article posted June 11, 2012 at 09:16 PM GMT-5 • comment • Reads 229 We were required two review questions, so I think I'll choose one from the beginning of the semester just as a refresher. Number 25 on the quadrilaterals test was asking us to label the coordinates of parallelogram LAST using only three variables. I chose d, c, and b (along with 0). Point 'a' was on the y-axis so the x coordinate was 0. Then, I chose 'b' to act as the height, or y coordinate. Point 'l' is the lower, left-hand point, and it is on the x-axis, making the y coordinate 0. Then, I made the x coordinate '-c' because it is to the left of the y-axis. Point 't' is the lower right-hand point, and it also lies on the x-axis, making the y coordinate 0. Then, I chose 'd' to act as the x coordinate. Point 's' was the upper right point of the parallelogram, so the height was also 'b', making the y coordinate automatically 'b'. The x coordinate is going to be 'd+c', because 'c' is the length that the segment goes on longer past 'd', which is where you get the '+c'. Hopefully that all made sense and was (slightly) helpful! Article posted June 11, 2012 at 09:16 PM GMT-5 • comment • Reads 229
Article posted June 10, 2012 at 09:00 PM GMT-5 • comment • Reads 222 This blog is supposed to be for two weeks ago but due to problems with the website the due date was postponed. This blog I will be explaining how to do a problem from one of the previous tests that I have taken. I have chosen to explain number 4 from test 7. The problem gives you a rhombus and tells you to find the area using the diagonals of 18 and 21. To find the area you must use the formula A=.5(d1)(d2). You would use substitution of make the equation A=.5(18)(21). If you do the math the area comes out to be 189ft squared. That is how you would do a problem like that... Article posted June 10, 2012 at 09:00 PM GMT-5 • comment • Reads 222
Article posted June 10, 2012 at 09:23 PM GMT-5 • comment • Reads 311 This week in geometry I am doing the same thing as last week. I have to pick a problem from any of my previous tests and I must explain how to do it.On problem 4 from the chapter 7 retest, the problem gives you a parallelogram and tells you to find the area using a height of 24 and a base of 10. The formula you must use is A= base times height. You would set up the equation as A= 24 times 10. The answer to the problem is 240 meters squared. That is how you do an problem like that... Article posted June 10, 2012 at 09:23 PM GMT-5 • comment • Reads 311
Article posted June 12, 2012 at 05:29 PM GMT-5 • comment • Reads 234 Question: The diagonals of a square bisect all angles Answer: Always Article posted June 12, 2012 at 05:29 PM GMT-5 • comment • Reads 234
Article posted June 12, 2012 at 05:31 PM GMT-5 • comment • Reads 312 Question: A triangle has side lengths of 8 cm, 14 cm, and 11 cm. Classify the triangle. Answer:Obtuse Article posted June 12, 2012 at 05:31 PM GMT-5 • comment • Reads 312
Article posted June 9, 2012 at 10:48 AM GMT-5 • comment • Reads 303 Hey guys so the school year's almost over. Yeah, summer. Unfortunately to get to summer we have to go through probably the wort week of the school year. Finals. Yeah you all know what I'm talking about. So here is one of the review questions that I'm using to study for my geometry final. This question is on right triangle trig. Good luck! Find X and H Answer: X=13 H=8.1 Article posted June 9, 2012 at 10:48 AM GMT-5 • comment • Reads 303
Article posted June 4, 2012 at 06:43 AM GMT-5 • comment • Reads 261 Hey guys so as I said in my last blog we are learning about SOH, CAH, TOA. Well now we are learning about the law of sines. Here is a review question and its answer about the law of sines. If m Answer: Measure of angle B is 59.4. Article posted June 4, 2012 at 06:43 AM GMT-5 • comment • Reads 261
Article posted June 7, 2012 at 09:17 PM GMT-5 • comment • Reads 224 We have finals in two weeks, and I don't really have any specific emotions for the end of the year. It was an awesome freshman year and a good first highschool experience. I hope next year will be as good, but then again I won't get to have Ms. J anymore:( To study for the final we must post a review question. Mine is- If a plane takes off at 25 degrees and flies 1600 ft, what is its altitude. SPOILER BELOW To do this problem, you must first write out the formula for sine. Using SOH, it would come out to be sin25=X/1600. Then you would multiply by 1600 on both sides to isolate X. This would make it 1600*sin(25)=X. Enter 1600*sin(25)into your calculator and the resulting number is the altitude-676.18 ft Article posted June 7, 2012 at 09:17 PM GMT-5 • comment • Reads 224
Article posted June 10, 2012 at 10:45 PM GMT-5 • comment • Reads 240 Hello everyone, I'm sad to tell you that this will be my last geometry blog :( The year is finally coming to an end. Six more days of school until summer! Here is my final exam review question: A triangle has three sides with the lengths of 6:8:10 What type of triangle is it? A.) acute B.)obtuse C.) right Answer: C. Right Article posted June 10, 2012 at 10:45 PM GMT-5 • comment • Reads 240
Article posted June 10, 2012 at 11:25 AM GMT-5 • comment • Reads 208 Hello Bloggers!!! This week I will be giving you a practice problem for you to solve!! At the bottom of the blog I will show you the work to get to the answer!! Question: You know two sides and an angle (6cm, 10cm, and 56 degrees) find the missing side. Answer: X2(squared)=6(squared)+10(squared) - 2(6)(10) - cos(56) by plugging this equation: 6(squared) + 10(squared) -2(6)(10) - cos(56) into your calculator, you will come up with X2(squared)= 15.4408071 you need to find X, so you have to find the square root. Type this into your calculator and X= 3.92947924 Great Job!!!! I will be blogging soon!!!! Article posted June 10, 2012 at 11:25 AM GMT-5 • comment • Reads 208
Article posted June 9, 2012 at 02:17 PM GMT-5 • comment • Reads 217 Hey bloggers! Just as I promised on Wednesday, here is another final review question! "In parallelogram BARK, m first draw the paralleglogram: 4x = xsquared - 60 xsqaured - 4x - 60 = 0 (x + 6)(x - 10) xsquared - 10x + 6x - 60 xsquared - 4x - 60 x + 6 = 0 or x - 10 = 0 x = -6 or x = 10 7x + m 70 + m m m< ARK = 110 dgrees Wish me luck on my final! Grace Article posted June 9, 2012 at 02:17 PM GMT-5 • comment • Reads 217
Article posted June 6, 2012 at 04:34 PM GMT-5 • comment • Reads 234 Hey bloggers! I am going to show you a good review question for our Geometry final next week! "Q is the intersection of AC and BD and ABCD. Find the area of kite ABCD if AB= 10, BC= 20, AC = 30, and BQ = 5." A= 1/2 x d1 x d2 d1= DB = 2BQ = 10. d2 = AC = 30 A = 1/2 x 10 x 30 Area = 150 squared units I will be posting another question later in the week due to the website being down last weekend. Later, Grace Article posted June 6, 2012 at 04:34 PM GMT-5 • comment • Reads 234
Article posted June 10, 2012 at 08:47 PM GMT-5 • comment • Reads 219 Welcome back everyone, since we are nearing the end of the school year, I was thinking that we could all start posting review questions to prepare ourselves for the upcoming final! What is a perpendicular bisector? And, how do you find it for any triangle? A perpendicular bisector is a line that divides a side of a triangle directly in the middle, but it bisects it at a 90˚ angle. Each side of the triangle has it’s own perpendicular bisector. In a triangle, there are 3 different sides each that have a perpendicular bisector. Since there are three different types of triangles (acute, right, and obtuse), each location of the circumcenter is different. For any acute triangle, the circumcenter is inside of the triangle. The point where all three of the Perpendicular bisectors intersect is at the circumcenter. But, like always there are 2 special cases. For a right triangle, the circumcenter of a right triangle is on the hypotenuse. But for an obtuse triangle: the circumcenter of an obtuse triangle is outside of the triangle. A circumscribed circle is a circle that is drawn from the circumcenter that hits all of the vertices of the triangle. The circumcenter is always equidistant to all vertices or any other point on the circumscribed circle. If you need anymore help with perpendicular bisectors reply to this post! I hope you all begin to review now! See you all next week! -Kathleen Article posted June 10, 2012 at 08:47 PM GMT-5 • comment • Reads 219
Article posted June 10, 2012 at 10:17 AM GMT-5 • comment • Reads 204 Hello again! As we are approaching our final exam next week, we are continuing to post review questions! These are aimed to help us remember the concepts we have learned this semester! The question I'm going to ask takes us back a chapter, and deals with triangles! Be sure to blog back with any questions. Q: A triangle has side lengths of 8cm, 11cm, and 14cm. Classify this triangle as acute, obtuse, or right. A: OBTUSE To find this answer, we first begin with the Pythagorean theorem (A squared + B squared = c squared). So, when we plug in what is known, our formula becomes 64 + 121 = 196. Here's the rule to classify: IF (A squared + b squared > C squared) then the triangle is acute. IF (A squared + b squared < C squared) then the triangle is obtuse. IF (A squared + b squared = C squared) then the triangle is right. In our formula, 64 + 121 = 185. Because 185 < 196, then the triangle is obtuse! Hope that helped and you learned something knew! This is my last weekly blog... but I'll be writing occasionally until the end of the year. Talk to you then! Emma Article posted June 10, 2012 at 10:17 AM GMT-5 • comment • Reads 204
Article posted June 14, 2012 at 12:40 PM GMT-5 • comment • Reads 240 We finally made it! Finals are around the corner, and so is summer! I can’t wait to ditch my backpack and head to the beach. But first, I have to survive finals. To review for Geometry, I’ve posted one of my favorite homework problems from a past unit below. KL⎮⎮JM in isosceles trapezoid JKLM. Find the values of x and y if m⦟J=(23x-8)º, m⦟K=(12y-13)º, and m⦟M=(17x+10)º. m⦟J=m⦟M 23x-8=17x+10 6x-8=10 6x=18 x=3 m⦟J+m⦟K=180 23x-8+12y-13=180 61+12y-13=180 12y=132 y=11 I chose this problem because I was from one of the first units of the semester, and I didn’t remember that unit at all. I thought that it would be nice to review since I had forgotten about it. Article posted June 14, 2012 at 12:40 PM GMT-5 • comment • Reads 240
Article posted June 14, 2012 at 12:40 PM GMT-5 • comment • Reads 244 We finally made it! Finals are around the corner, and so is summer! I can’t wait to ditch my backpack and head to the beach. But first, I have to survive finals. To review for Geometry, I’ve posted one of my favorite homework problems from a past unit below. KL⎮⎮JM in isosceles trapezoid JKLM. Find the values of x and y if m⦟J=(23x-8)º, m⦟K=(12y-13)º, and m⦟M=(17x+10)º. m⦟J=m⦟M 23x-8=17x+10 6x-8=10 6x=18 x=3 m⦟J+m⦟K=180 23x-8+12y-13=180 61+12y-13=180 12y=132 y=11 I chose this problem because I was from one of the first units of the semester, and I didn’t remember that unit at all. I thought that it would be nice to review since I had forgotten about it. Article posted June 14, 2012 at 12:40 PM GMT-5 • comment • Reads 244
Article posted June 6, 2012 at 04:33 PM GMT-5 • comment • Reads 257 Hello fellow blog readers! Sorry that this blog is so late compared to my others. Blogmeister, as I'm sure you know, has been down for the past couple of days. I am sorry to keep you waiting for so long. Now to the important stuff. In geometry, I have a question about the latest test. My question is how to find the length of an arc. I do not remember and was wondering if you bloggers could help me out. If you comment with the correct answer, I will mention you in my next blog. Well that is all for now blog-readers! Stay tuned for this Sunday's blog and question. Stay classy and thanks for reading! Article posted June 6, 2012 at 04:33 PM GMT-5 • comment • Reads 257
Article posted June 10, 2012 at 04:52 PM GMT-5 • comment • Reads 215 The year is almost over! We have two tests this next week in geometry. One tomorrow on chapter 9 and one on June 18th on everything we've learned this semester (our final). To help review for this final I have been told to blog a review question with the answer. Do you remember the distance formula? Well for coordinate proofs we had to use the distance formula to show how one side was of equal length to the other side. To show this you use the formula d = the square root of (y2-y1)squared + (x2-x1)squared. If you plug in the coordinates from #26 on the quadrilaterals test you get AS = the square root of (b-b)squared + (a+c-0)squared. This simplifies to the square root of (a+c)squared which equals a + c. Hope this was helpful, good luck on the test and the final! Article posted June 10, 2012 at 04:52 PM GMT-5 • comment • Reads 215
Article posted June 10, 2012 at 04:11 PM GMT-5 • comment • Reads 218 AHHHH! Three more days until final exams! Before I show you the review question take a deep breath... Inhale, now exhale. Feel better? Good. Let's begin! This is from chapter 9. This chapter is on SOH CAH TOA, the law of sines, the law of cosines, and vectors. Say you had a triangle where the length of the hypotenuse (AC) is 12cm, and the short side (AB) and the long side (BC) is unknown. Angle C is given as 23 degrees and angle B is 90 degrees. To solve this problem you use CAH because you are given a right triangle, an angle (23 degrees), the hypotenuse to that angle (12cm), and the long side is present. To set it up you write; cos23 = X/12. To solve you multiple cos23 by 12, so it's 12cos23 = X. X = 11cm. I hope this was helpful! Article posted June 10, 2012 at 04:11 PM GMT-5 • comment • Reads 218
Article posted June 10, 2012 at 08:54 PM GMT-5 • comment • Reads 216 Hello everyone! Hope your week has gone nicely! Mine has been pretty interesting, because my brother had an English exchange student arrive on Thursday. He will be staying with us for the week, and I am really excited! English accents are so cool! This exchange trip has also reminded me that the end of the school year is really close! In order to prepare for the geometry final, I am going to tell you about a question we had to do for homework as a way to review. This problem was on a worksheet I did recently involving the Law of Sines: sinA = sinB a b A and B stand for angles of an oblique triangle, while a and b stand for the 2 sides opposite them. Basically, you can use this law to find one of these four measures, as long as you have an angle and the side opposite. Basically, the problem was a triangle with angles A, B, and C, and sides a, b, and c. The measure of angle A was 35º, the measure of side a was 8 cm, and the measure of side b was 12 cm. The problem asked you to find the measure of angle B. Through substitution, I came out with this equation: sin(35) = sin(B) 8 12 By multiplying both sides by 12 and then by sin to the (-1), I was able to then solve for B: sin-1(12sin(35)) = sin-1(sin(B)) 8 m This process took some time to get used to, but I've found I actually like this sorts of problems in trigonometry. Let me know if you have any questions! Have a great week! (: Article posted June 10, 2012 at 08:54 PM GMT-5 • comment • Reads 216
Article posted June 10, 2012 at 09:02 PM GMT-5 • comment • Reads 207 Hello everyone! Firstly, I would just like to explain that last week Blogmeister was experiencing some technical difficulties, so I was unable to post at that time. The post from both then and the most recent one should be uploaded now. Anyways, this week is the beginning of finals for my school! I am very nervous because I've never taken finals before, and I'm not sure what to expect. Wish me luck! In order to review once again for geometry, I am posting another review questions below. I chose this particular one because I find this area formula hard to remember sometimes, because it doesn't involve side lengths, rather, diagonal lengths. For this problem, there was a rhombus shown with diagonals of 21 ft and 18 ft. The instructions were to find the area, and the possible answers were as shown: a) 378 sq. ft b) 189 sq. ft c) 162 sq. ft d) 27(square root of)85 sq. ft. I knew that the formula for the area of a rhombus is 1/2d1*d2, so I substituted in the diameter lengths: A = 1/2 * 21 * 18 = 189 With that equation, I was able to find that the area was 189 sq. ft., or answer b. I hope this has been a helpful review question! Have a great week! (: Article posted June 10, 2012 at 09:02 PM GMT-5 • comment • Reads 207
Article posted June 11, 2012 at 08:18 AM GMT-5 • comment • Reads 256 Well, the final is coming up for geometry this. Just one more test that I need to study for. I can't say that I enjoy these types of things, but I still need to do them. To help out everybody, though, here are two problems that were on homework from this term. There are two since the blog wasn't working last week when I was supposed to post one and one from this week, so I'm just combining the two blogs into this week's. Here they are: The lengths of the diagonals of a rhombus are 2 in. and 5 in. Find the measures of the angles of the rhombus to the nearest degree. Describe each translation using an ordered pair. 2 units to the left, 1 unit down. Good luck to everyone on the final! Article posted June 11, 2012 at 08:18 AM GMT-5 • comment • Reads 256
Article posted June 10, 2012 at 07:04 PM GMT-5 • comment • Reads 190 It seems like just yesterday that we were beginning our weekly blogs, and now we are writing our last ones. This is the week of finals, and after this week, we only have one day left of school. I cannot believe that school went by this fast. I would have to say that this school year was filled with learning, and fun! On another note, we still must remain serious. We still have our final exam on Monday, and that requires a gracious amount of studying. To study for the exam, I have come up with a few review guides that will help me go over anything that I have been struggling with. One thing that I have had a bit of trouble with is vectors. Therefore, I picked out a problem about vectors from my book and worked out the answer to it. The problem was "describe each vector as an ordered pair". To do so, you had to first figure out the sine ratio as well as the cosine ratio. Then, you would order them so that they describe the vector. It took some practice to understand, but eventually, I got it! I now feel prepared for this final exam! Article posted June 10, 2012 at 07:04 PM GMT-5 • comment • Reads 190
Article posted June 4, 2012 at 09:38 AM GMT-5 • comment • Reads 251 Salutations fellow bloggers! This week in geometry class we learned about the law of sine and cosine. It isn't to difficult but I don't really understand how it saves you time. My question for the final review is: What quadrilaterals have diagonals that bisect each other? Answer: Parallelogram, rhombus, rectangle, and square Have a great week everyone! Article posted June 4, 2012 at 09:38 AM GMT-5 • comment • Reads 251
Article posted June 5, 2012 at 07:53 PM GMT-5 • comment • Reads 237 Hi everyone! Just writing because before I could add my review question, Blogmeister shut itself down! NOOOOOO!!!! Well, anyway, here’s my Trigonometry Review question: Kevin and Zane have just built a tree house. It is 20 ft off the ground, and the tree is perpendicular to the ground. When Kevin’s Mom goes out to look at the tree house, the angle of elevation from her feet to the tree house is 50*. Zane wants to know how far the tree that has the tree house in it is from the house, but he has lost the tape measure. How far away is the tree from the house? ***SPOILER ALERT*** Answer: Roughly 17 Feet --Joe Article posted June 5, 2012 at 07:53 PM GMT-5 • comment • Reads 237
Article posted June 10, 2012 at 07:42 AM GMT-5 • comment • Reads 198 HELLO! WELCOME TO THE FINAL COMPLETE WEEK OF SCHOOL! Wow, the year has gone by so fast! Did I mention that this is our final weekly blog? I can’t believe how close we are to the end of the year! It seems like just yesterday we walked into Geometry for the first time. I have had a great weekend, capped of by a RockEucharist at my church! This week in Geometry, we continued with our trigonometry unit, and we learned the law of cosines. At first I didn’t understand it, but once I got it, I had it for good. I think that on our trig. test this week, I will not have that much trouble, as this unit has been about mostly algebra and I do not have much trouble with algebra. But you never know-- it may be very hard. Our final is also coming up. Wow, that snuck up on me. Here is another review question: Ben and Griffin love modern houses. They want to make a museum about the evolution of houses. While searching for a place to put the museum, they find a modern building that is a rhombus. The sides of the house are 50 feet long. The angles formed by two intersecting lines are 120* and 60*. How much area is in the museum? ANSWER: 4330 ft. squared Have a great last week! --Joe Article posted June 10, 2012 at 07:42 AM GMT-5 • comment • Reads 198
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Article posted June 2, 2012 at 07:33 AM GMT-5 • comment • Reads 539 Today is Sunday, June 10, and I'm doing this blog a week late because the website crashed last week. It also means that there are exactly 2 school days until Finals start! I'm starting to worry it will be really hard to study without being distracted by the nice weather. To get in the spirit of studying, I'm going to post a review question from one of my past tests today on this blog. This is it: Determine which three numbers could be the sides of a right triangle. A. 64,73,98 B. 64,72,96 C. 65,72,97 The answer is C because of the Pythagorean theorem. When you square each number, the first two must add up to the third one's square. If that happens, then it is a right triangle. Choice C is the only one that happens for, so it is the correct answer. Until Next week, Article posted June 2, 2012 at 07:33 AM GMT-5 • comment • Reads 539
Article posted June 10, 2012 at 07:03 AM GMT-5 • comment • Reads 309 I'm doing the second blog for this week today, Sunday, June 10 because the website crashed last week. As I said earlier, there are exactly two school days until finals start! While this may seem terrible, I'm actually very happy because it means that we are in the home stretch before summer! I'm posting another review question this week: this will be about lines in triangles. Which line in triangles meet at the orthocenter when all three are drawn in a triangle? The answer to this question is altitudes. An altitude, also known as the height of a triangle, is a perpendicular line to the base drawn from a vertex. School may be almost done, but I can't give up now. This is the last blog of this year, and I can say that's a good thing! Article posted June 10, 2012 at 07:03 AM GMT-5 • comment • Reads 309
Article posted June 10, 2012 at 07:40 PM GMT-5 • comment • Reads 291 For this weeks blog I was asked to post two review questions for the final exam. The final exam is this week and I need to start studying, so I am going to start by making these review questions. My first question is: What does MAPA COCI stand for? A hint for this question is to remember lines in triangles and their concurrent points! My next question is: How do you find the circumference of a circle with a radius of 10 cm? Answers: Question 1: Median, Altitude, Perpendicular Bisector, Angle Bisector. Centroid, Orthocenter, Circumcenter, Incenter. Question 2: 20pi cm. Article posted June 10, 2012 at 07:40 PM GMT-5 • comment • Reads 291
Article posted June 7, 2012 at 04:31 PM GMT-5 • comment • Reads 279 On Sunday I watched a lot of TV because of all the crappy weather that we had. The best thing I watched was the golf touroment where Tiger Woods hit an unbelievable shot to clinch the victory. The problem I am working on for this blog is # 12 from the green version on test # 7. It asks for you to find the area of a parallelogram that has sides of 8 and 6. On the test, I got the problem wrong because I didn't know the area formula for parallelograms but when I went back and re-did the problem while studying for the retake I realized that I had to find the height by making a 45-45-90 triangle because the formula is a=bh. I found that the height was 3 root 2 and 8 time 3 root 2 is 24 root 2. Article posted June 7, 2012 at 04:31 PM GMT-5 • comment • Reads 279
Article posted June 7, 2012 at 04:49 PM GMT-5 • comment • Reads 356 I am blogging on a thursday for probably the first time, it seems wierd. I am getting ready to do my geometry homework. I am also going to watch the Miami Heat vs the Boston Celtics later, it is must win game for the Heat. The problem that I am going to work on in this blog is # 17 from the green version of test # 7. It asks you find the area of a deck that surrounds a hot tub if the hot tub has a diameter of 6 meters and the deck is 2 meters wide. When I took the test I knew how to do the problem but I just had a brain cramp and screwed up the final answer. The way to figure out this problem is to find the area of the hot tub and then subtract that from the area of the deck. The only formula that you need to use on this problem is a=pie(r) squared. The first thing to do is sub the number 3 into that formula and you end up with 9 pie. You then plug the number 5 into that formula because 6+ 2+2 equals 10 as a diameter and half of that equals 5. When plugging that into the equation you end up with 25 pie. The problem that I made on the test was that I forgot to subtract 9 pie from 25 pie so that is the last step in this problem. Your final answer should be 16 pie. Article posted June 7, 2012 at 04:49 PM GMT-5 • comment • Reads 356
Article posted June 10, 2012 at 10:26 PM GMT-5 • comment • Reads 224 This blog is the twenty-third of my personal weekly blog This blog was supposed to be posted last week but blogmeister was down! Oh no! We got everything squared away though, so yay! Last week was pretty gloomy, the weather hasn't been that nice until this weekend! Last week in geometry we started learning more in trigonometry and then started learning about law of sines and law of cosines! It's been pretty easy, but some things trip me up easily. Anyways the year is winding down and I can not wait until SUMMER! Midterms start this week! Here is a review question on what we have learned in Geometry! Ms. J told us that studying our tests and quizzes would be most helpful so that is exactly what I will be doing! A particular unit that I didn't fair so well on was working with the lines of triangles! Here is a review question from the Lines of Triangles Quiz that we took on February 7th. Which of the following are the slopes of two perpendicular lines? a. 3 and -3 b. 5 and 1/5 c. no slope and undefined d. -2/3 and 3/2 If you recall that the slope of any line perpendicular to another is the negative reciprocal, then this problem is easy to solve! A is not the answer because the negative reciprocal or 3 would be -1/3. B does not work because the negative reciprocal of 5 would be -1/5. c does not work because they aren't specific lines! THEREFORE D IS THE ANSWER! and to prove it.. the negative reciprocal of -2/3 would be - ( -3/2) or 3/2! TADA! Article posted June 10, 2012 at 10:26 PM GMT-5 • comment • Reads 224
Article posted June 10, 2012 at 10:34 PM GMT-5 • comment • Reads 419 This is the twenty-fourth of my personal weekly blog. Alas I have reached my last blog. Farewell classblogmeister... I won't miss you too much I promise! This week in Geometry we continued working with the law of sines and the law of cosines! We have a test this week before our final so that means a lot of studying on our parts! We pretty much did a lot of reviewing on the all the parts of trigonometry, which was helpful! This review question will be coming from the Quadrilaterals test we took on March 29th. I GOT THIS QUESTION WRONG! I honestly have no idea how I could've made the mistake, but I did, so make sure you don't jump to conclusions! Consecutive sides of a rectangle are congruent. a. sometimes b. always c. never the answer is SOMETIMES because a square (which is a rectangle) has congruent consecutive sides! YAY GEOMETRY!(: Article posted June 10, 2012 at 10:34 PM GMT-5 • comment • Reads 419
Article posted June 10, 2012 at 10:32 AM GMT-5 • comment • Reads 224 For Ch 12, the work was on transformations. The review question will be for rotations. The question as what is the point of rotation, and what is the angle of rotation. When given a pre-image and an image, to find the point of rotation, it will either be where the two images intersect or a point not connected to either image. Then to find the angle of rotation, you will need a protractor. Put your protractor on the point of rotation and pick a point you want to measure from. You always measure from the pre-image and counterclockwise unless stated to go clockwise. After you pick the point, find the point on the image that is the same, and measure. That will give you the angle of rotation Stay classy bloggers. Article posted June 10, 2012 at 10:32 AM GMT-5 • comment • Reads 224
Article posted June 10, 2012 at 11:24 AM GMT-5 • comment • Reads 217 Our final exam is Wednesday! Ahhh! And in preparation to that I am going to be posting a review question from one of the chapters we learned about this semester. In chapter 7 we learned more about areas of formulas. My review question today, will be finding the area of a rhombus (remember that the diagonals of a rhombus bisect each other). The formula for an area of a rhombus is 1/2(d1+d2), d stands for diagonal. If your given information was that one diagonal was 7cm and half of the other diagonal was 5.5cm, this is how you would solve it: We know that the other half of the diagonal is 5.5cm, because it gets bisected from the other diagonal. Now we plug in our given information to come up with 1/2(7+11) = area. Simplify to get 1/2(18) = area. Now simply multiply by 1/2 or divide by two, and we get 9cm squared = area Hopefully this helped you! I will be posting another review question soon! Goodbye! Article posted June 10, 2012 at 11:24 AM GMT-5 • comment • Reads 217
Article posted June 10, 2012 at 12:01 PM GMT-5 • comment • Reads 233 Hello again! Today, I am posting another review question. This comes from chapter 5. Question: Which of they following is the concurrent point of the altitudes? Answer: Orthocenter. I always remember which concurrent point goes with which line of a triangle because of MAPA COCI. MAPA COCI stands for: Median --> Centroid Altitude --> Orthocenter Perpendicular Bisector -->Circumcenter Angle Bisector --> Incenter This helped me remember and hopefully it will help you too! :) Now down to the sad stuff. :( I won't be saying 'see you next week' anymore..... because this is my last blog! *Gasp* I hope my blogs have helped you in some way or another, but right now I have to say goodbye for good. I know you'll miss me though! Farewell, Livy Article posted June 10, 2012 at 12:01 PM GMT-5 • comment • Reads 233
Article posted June 11, 2012 at 02:06 PM GMT-5 • comment • Reads 235 Question Given: Circle O with Diamter CD, AB is parrallel to CD, and arc AB=80 degrees Find arc CA [1] 50 [2] 60 [3] 80 [4] 100 Answer 50 Article posted June 11, 2012 at 02:06 PM GMT-5 • comment • Reads 235
Article posted June 11, 2012 at 02:10 PM GMT-5 • comment • Reads 381 Question Given: In triangle ABC, B=120, c=15, and a=15 Find C Answer 30 degrees Article posted June 11, 2012 at 02:10 PM GMT-5 • comment • Reads 381
Article posted June 11, 2012 at 09:04 PM GMT-5 • comment • Reads 241 The midterm is in two days and I'm getting a little nervous. We also have a test tomorrow that isn't helping with the nerves! Hopefully the test tomorrow will make get me feeling completely confident with the trigonometry unit, so it'll be one less thing to study for the final! My first review question that I want to go over is number 13 from practice 58. It is working with vectors which I needed a slight refresher on. The question: Homing pigeons have the ability or instinct to find their way home when released hundreds of miles away from home. Homing pigeons carried news of Olympic victories to various cities in ancient Greece. Suppose one such pigeon took off from Athens and landed in Sparta, which is 73 miles west and 64 miles south of Athens. Find the distance and its direction of flight. The first step you would need to take is plugging the coordinates into the distance formula. It comes out with 97, so Athens and Sparta are 97 miles apart. Then, you need to use the tangent ratio. This is the opposite side over the adjacent side. Therefore, tan(x) = 64/73 --> x = tan-1( 64/73) --> x = 41°, so the direction of flight is 41°. Article posted June 11, 2012 at 09:04 PM GMT-5 • comment • Reads 241
Article posted June 11, 2012 at 09:16 PM GMT-5 • comment • Reads 229 We were required two review questions, so I think I'll choose one from the beginning of the semester just as a refresher. Number 25 on the quadrilaterals test was asking us to label the coordinates of parallelogram LAST using only three variables. I chose d, c, and b (along with 0). Point 'a' was on the y-axis so the x coordinate was 0. Then, I chose 'b' to act as the height, or y coordinate. Point 'l' is the lower, left-hand point, and it is on the x-axis, making the y coordinate 0. Then, I made the x coordinate '-c' because it is to the left of the y-axis. Point 't' is the lower right-hand point, and it also lies on the x-axis, making the y coordinate 0. Then, I chose 'd' to act as the x coordinate. Point 's' was the upper right point of the parallelogram, so the height was also 'b', making the y coordinate automatically 'b'. The x coordinate is going to be 'd+c', because 'c' is the length that the segment goes on longer past 'd', which is where you get the '+c'. Hopefully that all made sense and was (slightly) helpful! Article posted June 11, 2012 at 09:16 PM GMT-5 • comment • Reads 229
Article posted June 10, 2012 at 09:00 PM GMT-5 • comment • Reads 222 This blog is supposed to be for two weeks ago but due to problems with the website the due date was postponed. This blog I will be explaining how to do a problem from one of the previous tests that I have taken. I have chosen to explain number 4 from test 7. The problem gives you a rhombus and tells you to find the area using the diagonals of 18 and 21. To find the area you must use the formula A=.5(d1)(d2). You would use substitution of make the equation A=.5(18)(21). If you do the math the area comes out to be 189ft squared. That is how you would do a problem like that... Article posted June 10, 2012 at 09:00 PM GMT-5 • comment • Reads 222
Article posted June 10, 2012 at 09:23 PM GMT-5 • comment • Reads 311 This week in geometry I am doing the same thing as last week. I have to pick a problem from any of my previous tests and I must explain how to do it.On problem 4 from the chapter 7 retest, the problem gives you a parallelogram and tells you to find the area using a height of 24 and a base of 10. The formula you must use is A= base times height. You would set up the equation as A= 24 times 10. The answer to the problem is 240 meters squared. That is how you do an problem like that... Article posted June 10, 2012 at 09:23 PM GMT-5 • comment • Reads 311
Article posted June 12, 2012 at 05:29 PM GMT-5 • comment • Reads 234 Question: The diagonals of a square bisect all angles Answer: Always Article posted June 12, 2012 at 05:29 PM GMT-5 • comment • Reads 234
Article posted June 12, 2012 at 05:31 PM GMT-5 • comment • Reads 312 Question: A triangle has side lengths of 8 cm, 14 cm, and 11 cm. Classify the triangle. Answer:Obtuse Article posted June 12, 2012 at 05:31 PM GMT-5 • comment • Reads 312
Article posted June 9, 2012 at 10:48 AM GMT-5 • comment • Reads 303 Hey guys so the school year's almost over. Yeah, summer. Unfortunately to get to summer we have to go through probably the wort week of the school year. Finals. Yeah you all know what I'm talking about. So here is one of the review questions that I'm using to study for my geometry final. This question is on right triangle trig. Good luck! Find X and H Answer: X=13 H=8.1 Article posted June 9, 2012 at 10:48 AM GMT-5 • comment • Reads 303
Article posted June 4, 2012 at 06:43 AM GMT-5 • comment • Reads 261 Hey guys so as I said in my last blog we are learning about SOH, CAH, TOA. Well now we are learning about the law of sines. Here is a review question and its answer about the law of sines. If m Answer: Measure of angle B is 59.4. Article posted June 4, 2012 at 06:43 AM GMT-5 • comment • Reads 261
Article posted June 7, 2012 at 09:17 PM GMT-5 • comment • Reads 224 We have finals in two weeks, and I don't really have any specific emotions for the end of the year. It was an awesome freshman year and a good first highschool experience. I hope next year will be as good, but then again I won't get to have Ms. J anymore:( To study for the final we must post a review question. Mine is- If a plane takes off at 25 degrees and flies 1600 ft, what is its altitude. SPOILER BELOW To do this problem, you must first write out the formula for sine. Using SOH, it would come out to be sin25=X/1600. Then you would multiply by 1600 on both sides to isolate X. This would make it 1600*sin(25)=X. Enter 1600*sin(25)into your calculator and the resulting number is the altitude-676.18 ft Article posted June 7, 2012 at 09:17 PM GMT-5 • comment • Reads 224
Article posted June 10, 2012 at 10:45 PM GMT-5 • comment • Reads 240 Hello everyone, I'm sad to tell you that this will be my last geometry blog :( The year is finally coming to an end. Six more days of school until summer! Here is my final exam review question: A triangle has three sides with the lengths of 6:8:10 What type of triangle is it? A.) acute B.)obtuse C.) right Answer: C. Right Article posted June 10, 2012 at 10:45 PM GMT-5 • comment • Reads 240
Article posted June 10, 2012 at 11:25 AM GMT-5 • comment • Reads 208 Hello Bloggers!!! This week I will be giving you a practice problem for you to solve!! At the bottom of the blog I will show you the work to get to the answer!! Question: You know two sides and an angle (6cm, 10cm, and 56 degrees) find the missing side. Answer: X2(squared)=6(squared)+10(squared) - 2(6)(10) - cos(56) by plugging this equation: 6(squared) + 10(squared) -2(6)(10) - cos(56) into your calculator, you will come up with X2(squared)= 15.4408071 you need to find X, so you have to find the square root. Type this into your calculator and X= 3.92947924 Great Job!!!! I will be blogging soon!!!! Article posted June 10, 2012 at 11:25 AM GMT-5 • comment • Reads 208
Article posted June 9, 2012 at 02:17 PM GMT-5 • comment • Reads 217 Hey bloggers! Just as I promised on Wednesday, here is another final review question! "In parallelogram BARK, m first draw the paralleglogram: 4x = xsquared - 60 xsqaured - 4x - 60 = 0 (x + 6)(x - 10) xsquared - 10x + 6x - 60 xsquared - 4x - 60 x + 6 = 0 or x - 10 = 0 x = -6 or x = 10 7x + m 70 + m m m< ARK = 110 dgrees Wish me luck on my final! Grace Article posted June 9, 2012 at 02:17 PM GMT-5 • comment • Reads 217
Article posted June 6, 2012 at 04:34 PM GMT-5 • comment • Reads 234 Hey bloggers! I am going to show you a good review question for our Geometry final next week! "Q is the intersection of AC and BD and ABCD. Find the area of kite ABCD if AB= 10, BC= 20, AC = 30, and BQ = 5." A= 1/2 x d1 x d2 d1= DB = 2BQ = 10. d2 = AC = 30 A = 1/2 x 10 x 30 Area = 150 squared units I will be posting another question later in the week due to the website being down last weekend. Later, Grace Article posted June 6, 2012 at 04:34 PM GMT-5 • comment • Reads 234
Article posted June 10, 2012 at 08:47 PM GMT-5 • comment • Reads 219 Welcome back everyone, since we are nearing the end of the school year, I was thinking that we could all start posting review questions to prepare ourselves for the upcoming final! What is a perpendicular bisector? And, how do you find it for any triangle? A perpendicular bisector is a line that divides a side of a triangle directly in the middle, but it bisects it at a 90˚ angle. Each side of the triangle has it’s own perpendicular bisector. In a triangle, there are 3 different sides each that have a perpendicular bisector. Since there are three different types of triangles (acute, right, and obtuse), each location of the circumcenter is different. For any acute triangle, the circumcenter is inside of the triangle. The point where all three of the Perpendicular bisectors intersect is at the circumcenter. But, like always there are 2 special cases. For a right triangle, the circumcenter of a right triangle is on the hypotenuse. But for an obtuse triangle: the circumcenter of an obtuse triangle is outside of the triangle. A circumscribed circle is a circle that is drawn from the circumcenter that hits all of the vertices of the triangle. The circumcenter is always equidistant to all vertices or any other point on the circumscribed circle. If you need anymore help with perpendicular bisectors reply to this post! I hope you all begin to review now! See you all next week! -Kathleen Article posted June 10, 2012 at 08:47 PM GMT-5 • comment • Reads 219
Article posted June 10, 2012 at 10:17 AM GMT-5 • comment • Reads 204 Hello again! As we are approaching our final exam next week, we are continuing to post review questions! These are aimed to help us remember the concepts we have learned this semester! The question I'm going to ask takes us back a chapter, and deals with triangles! Be sure to blog back with any questions. Q: A triangle has side lengths of 8cm, 11cm, and 14cm. Classify this triangle as acute, obtuse, or right. A: OBTUSE To find this answer, we first begin with the Pythagorean theorem (A squared + B squared = c squared). So, when we plug in what is known, our formula becomes 64 + 121 = 196. Here's the rule to classify: IF (A squared + b squared > C squared) then the triangle is acute. IF (A squared + b squared < C squared) then the triangle is obtuse. IF (A squared + b squared = C squared) then the triangle is right. In our formula, 64 + 121 = 185. Because 185 < 196, then the triangle is obtuse! Hope that helped and you learned something knew! This is my last weekly blog... but I'll be writing occasionally until the end of the year. Talk to you then! Emma Article posted June 10, 2012 at 10:17 AM GMT-5 • comment • Reads 204
Article posted June 14, 2012 at 12:40 PM GMT-5 • comment • Reads 240 We finally made it! Finals are around the corner, and so is summer! I can’t wait to ditch my backpack and head to the beach. But first, I have to survive finals. To review for Geometry, I’ve posted one of my favorite homework problems from a past unit below. KL⎮⎮JM in isosceles trapezoid JKLM. Find the values of x and y if m⦟J=(23x-8)º, m⦟K=(12y-13)º, and m⦟M=(17x+10)º. m⦟J=m⦟M 23x-8=17x+10 6x-8=10 6x=18 x=3 m⦟J+m⦟K=180 23x-8+12y-13=180 61+12y-13=180 12y=132 y=11 I chose this problem because I was from one of the first units of the semester, and I didn’t remember that unit at all. I thought that it would be nice to review since I had forgotten about it. Article posted June 14, 2012 at 12:40 PM GMT-5 • comment • Reads 240
Article posted June 14, 2012 at 12:40 PM GMT-5 • comment • Reads 244 We finally made it! Finals are around the corner, and so is summer! I can’t wait to ditch my backpack and head to the beach. But first, I have to survive finals. To review for Geometry, I’ve posted one of my favorite homework problems from a past unit below. KL⎮⎮JM in isosceles trapezoid JKLM. Find the values of x and y if m⦟J=(23x-8)º, m⦟K=(12y-13)º, and m⦟M=(17x+10)º. m⦟J=m⦟M 23x-8=17x+10 6x-8=10 6x=18 x=3 m⦟J+m⦟K=180 23x-8+12y-13=180 61+12y-13=180 12y=132 y=11 I chose this problem because I was from one of the first units of the semester, and I didn’t remember that unit at all. I thought that it would be nice to review since I had forgotten about it. Article posted June 14, 2012 at 12:40 PM GMT-5 • comment • Reads 244
Article posted June 6, 2012 at 04:33 PM GMT-5 • comment • Reads 257 Hello fellow blog readers! Sorry that this blog is so late compared to my others. Blogmeister, as I'm sure you know, has been down for the past couple of days. I am sorry to keep you waiting for so long. Now to the important stuff. In geometry, I have a question about the latest test. My question is how to find the length of an arc. I do not remember and was wondering if you bloggers could help me out. If you comment with the correct answer, I will mention you in my next blog. Well that is all for now blog-readers! Stay tuned for this Sunday's blog and question. Stay classy and thanks for reading! Article posted June 6, 2012 at 04:33 PM GMT-5 • comment • Reads 257
Article posted June 10, 2012 at 04:52 PM GMT-5 • comment • Reads 215 The year is almost over! We have two tests this next week in geometry. One tomorrow on chapter 9 and one on June 18th on everything we've learned this semester (our final). To help review for this final I have been told to blog a review question with the answer. Do you remember the distance formula? Well for coordinate proofs we had to use the distance formula to show how one side was of equal length to the other side. To show this you use the formula d = the square root of (y2-y1)squared + (x2-x1)squared. If you plug in the coordinates from #26 on the quadrilaterals test you get AS = the square root of (b-b)squared + (a+c-0)squared. This simplifies to the square root of (a+c)squared which equals a + c. Hope this was helpful, good luck on the test and the final! Article posted June 10, 2012 at 04:52 PM GMT-5 • comment • Reads 215
Article posted June 10, 2012 at 04:11 PM GMT-5 • comment • Reads 218 AHHHH! Three more days until final exams! Before I show you the review question take a deep breath... Inhale, now exhale. Feel better? Good. Let's begin! This is from chapter 9. This chapter is on SOH CAH TOA, the law of sines, the law of cosines, and vectors. Say you had a triangle where the length of the hypotenuse (AC) is 12cm, and the short side (AB) and the long side (BC) is unknown. Angle C is given as 23 degrees and angle B is 90 degrees. To solve this problem you use CAH because you are given a right triangle, an angle (23 degrees), the hypotenuse to that angle (12cm), and the long side is present. To set it up you write; cos23 = X/12. To solve you multiple cos23 by 12, so it's 12cos23 = X. X = 11cm. I hope this was helpful! Article posted June 10, 2012 at 04:11 PM GMT-5 • comment • Reads 218
Article posted June 10, 2012 at 08:54 PM GMT-5 • comment • Reads 216 Hello everyone! Hope your week has gone nicely! Mine has been pretty interesting, because my brother had an English exchange student arrive on Thursday. He will be staying with us for the week, and I am really excited! English accents are so cool! This exchange trip has also reminded me that the end of the school year is really close! In order to prepare for the geometry final, I am going to tell you about a question we had to do for homework as a way to review. This problem was on a worksheet I did recently involving the Law of Sines: sinA = sinB a b A and B stand for angles of an oblique triangle, while a and b stand for the 2 sides opposite them. Basically, you can use this law to find one of these four measures, as long as you have an angle and the side opposite. Basically, the problem was a triangle with angles A, B, and C, and sides a, b, and c. The measure of angle A was 35º, the measure of side a was 8 cm, and the measure of side b was 12 cm. The problem asked you to find the measure of angle B. Through substitution, I came out with this equation: sin(35) = sin(B) 8 12 By multiplying both sides by 12 and then by sin to the (-1), I was able to then solve for B: sin-1(12sin(35)) = sin-1(sin(B)) 8 m This process took some time to get used to, but I've found I actually like this sorts of problems in trigonometry. Let me know if you have any questions! Have a great week! (: Article posted June 10, 2012 at 08:54 PM GMT-5 • comment • Reads 216
Article posted June 10, 2012 at 09:02 PM GMT-5 • comment • Reads 207 Hello everyone! Firstly, I would just like to explain that last week Blogmeister was experiencing some technical difficulties, so I was unable to post at that time. The post from both then and the most recent one should be uploaded now. Anyways, this week is the beginning of finals for my school! I am very nervous because I've never taken finals before, and I'm not sure what to expect. Wish me luck! In order to review once again for geometry, I am posting another review questions below. I chose this particular one because I find this area formula hard to remember sometimes, because it doesn't involve side lengths, rather, diagonal lengths. For this problem, there was a rhombus shown with diagonals of 21 ft and 18 ft. The instructions were to find the area, and the possible answers were as shown: a) 378 sq. ft b) 189 sq. ft c) 162 sq. ft d) 27(square root of)85 sq. ft. I knew that the formula for the area of a rhombus is 1/2d1*d2, so I substituted in the diameter lengths: A = 1/2 * 21 * 18 = 189 With that equation, I was able to find that the area was 189 sq. ft., or answer b. I hope this has been a helpful review question! Have a great week! (: Article posted June 10, 2012 at 09:02 PM GMT-5 • comment • Reads 207
Article posted June 11, 2012 at 08:18 AM GMT-5 • comment • Reads 256 Well, the final is coming up for geometry this. Just one more test that I need to study for. I can't say that I enjoy these types of things, but I still need to do them. To help out everybody, though, here are two problems that were on homework from this term. There are two since the blog wasn't working last week when I was supposed to post one and one from this week, so I'm just combining the two blogs into this week's. Here they are: The lengths of the diagonals of a rhombus are 2 in. and 5 in. Find the measures of the angles of the rhombus to the nearest degree. Describe each translation using an ordered pair. 2 units to the left, 1 unit down. Good luck to everyone on the final! Article posted June 11, 2012 at 08:18 AM GMT-5 • comment • Reads 256
Article posted June 10, 2012 at 07:04 PM GMT-5 • comment • Reads 190 It seems like just yesterday that we were beginning our weekly blogs, and now we are writing our last ones. This is the week of finals, and after this week, we only have one day left of school. I cannot believe that school went by this fast. I would have to say that this school year was filled with learning, and fun! On another note, we still must remain serious. We still have our final exam on Monday, and that requires a gracious amount of studying. To study for the exam, I have come up with a few review guides that will help me go over anything that I have been struggling with. One thing that I have had a bit of trouble with is vectors. Therefore, I picked out a problem about vectors from my book and worked out the answer to it. The problem was "describe each vector as an ordered pair". To do so, you had to first figure out the sine ratio as well as the cosine ratio. Then, you would order them so that they describe the vector. It took some practice to understand, but eventually, I got it! I now feel prepared for this final exam! Article posted June 10, 2012 at 07:04 PM GMT-5 • comment • Reads 190
Article posted June 4, 2012 at 09:38 AM GMT-5 • comment • Reads 251 Salutations fellow bloggers! This week in geometry class we learned about the law of sine and cosine. It isn't to difficult but I don't really understand how it saves you time. My question for the final review is: What quadrilaterals have diagonals that bisect each other? Answer: Parallelogram, rhombus, rectangle, and square Have a great week everyone! Article posted June 4, 2012 at 09:38 AM GMT-5 • comment • Reads 251
Article posted June 5, 2012 at 07:53 PM GMT-5 • comment • Reads 237 Hi everyone! Just writing because before I could add my review question, Blogmeister shut itself down! NOOOOOO!!!! Well, anyway, here’s my Trigonometry Review question: Kevin and Zane have just built a tree house. It is 20 ft off the ground, and the tree is perpendicular to the ground. When Kevin’s Mom goes out to look at the tree house, the angle of elevation from her feet to the tree house is 50*. Zane wants to know how far the tree that has the tree house in it is from the house, but he has lost the tape measure. How far away is the tree from the house? ***SPOILER ALERT*** Answer: Roughly 17 Feet --Joe Article posted June 5, 2012 at 07:53 PM GMT-5 • comment • Reads 237
Article posted June 10, 2012 at 07:42 AM GMT-5 • comment • Reads 198 HELLO! WELCOME TO THE FINAL COMPLETE WEEK OF SCHOOL! Wow, the year has gone by so fast! Did I mention that this is our final weekly blog? I can’t believe how close we are to the end of the year! It seems like just yesterday we walked into Geometry for the first time. I have had a great weekend, capped of by a RockEucharist at my church! This week in Geometry, we continued with our trigonometry unit, and we learned the law of cosines. At first I didn’t understand it, but once I got it, I had it for good. I think that on our trig. test this week, I will not have that much trouble, as this unit has been about mostly algebra and I do not have much trouble with algebra. But you never know-- it may be very hard. Our final is also coming up. Wow, that snuck up on me. Here is another review question: Ben and Griffin love modern houses. They want to make a museum about the evolution of houses. While searching for a place to put the museum, they find a modern building that is a rhombus. The sides of the house are 50 feet long. The angles formed by two intersecting lines are 120* and 60*. How much area is in the museum? ANSWER: 4330 ft. squared Have a great last week! --Joe Article posted June 10, 2012 at 07:42 AM GMT-5 • comment • Reads 198
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Article posted June 2, 2012 at 07:33 AM GMT-5 • comment • Reads 539 Today is Sunday, June 10, and I'm doing this blog a week late because the website crashed last week. It also means that there are exactly 2 school days until Finals start! I'm starting to worry it will be really hard to study without being distracted by the nice weather. To get in the spirit of studying, I'm going to post a review question from one of my past tests today on this blog. This is it: Determine which three numbers could be the sides of a right triangle. A. 64,73,98 B. 64,72,96 C. 65,72,97 The answer is C because of the Pythagorean theorem. When you square each number, the first two must add up to the third one's square. If that happens, then it is a right triangle. Choice C is the only one that happens for, so it is the correct answer. Until Next week, Article posted June 2, 2012 at 07:33 AM GMT-5 • comment • Reads 539
Article posted June 10, 2012 at 07:03 AM GMT-5 • comment • Reads 309 I'm doing the second blog for this week today, Sunday, June 10 because the website crashed last week. As I said earlier, there are exactly two school days until finals start! While this may seem terrible, I'm actually very happy because it means that we are in the home stretch before summer! I'm posting another review question this week: this will be about lines in triangles. Which line in triangles meet at the orthocenter when all three are drawn in a triangle? The answer to this question is altitudes. An altitude, also known as the height of a triangle, is a perpendicular line to the base drawn from a vertex. School may be almost done, but I can't give up now. This is the last blog of this year, and I can say that's a good thing! Article posted June 10, 2012 at 07:03 AM GMT-5 • comment • Reads 309
Article posted June 10, 2012 at 07:40 PM GMT-5 • comment • Reads 291 For this weeks blog I was asked to post two review questions for the final exam. The final exam is this week and I need to start studying, so I am going to start by making these review questions. My first question is: What does MAPA COCI stand for? A hint for this question is to remember lines in triangles and their concurrent points! My next question is: How do you find the circumference of a circle with a radius of 10 cm? Answers: Question 1: Median, Altitude, Perpendicular Bisector, Angle Bisector. Centroid, Orthocenter, Circumcenter, Incenter. Question 2: 20pi cm. Article posted June 10, 2012 at 07:40 PM GMT-5 • comment • Reads 291
Article posted June 7, 2012 at 04:31 PM GMT-5 • comment • Reads 279 On Sunday I watched a lot of TV because of all the crappy weather that we had. The best thing I watched was the golf touroment where Tiger Woods hit an unbelievable shot to clinch the victory. The problem I am working on for this blog is # 12 from the green version on test # 7. It asks for you to find the area of a parallelogram that has sides of 8 and 6. On the test, I got the problem wrong because I didn't know the area formula for parallelograms but when I went back and re-did the problem while studying for the retake I realized that I had to find the height by making a 45-45-90 triangle because the formula is a=bh. I found that the height was 3 root 2 and 8 time 3 root 2 is 24 root 2. Article posted June 7, 2012 at 04:31 PM GMT-5 • comment • Reads 279
Article posted June 7, 2012 at 04:49 PM GMT-5 • comment • Reads 356 I am blogging on a thursday for probably the first time, it seems wierd. I am getting ready to do my geometry homework. I am also going to watch the Miami Heat vs the Boston Celtics later, it is must win game for the Heat. The problem that I am going to work on in this blog is # 17 from the green version of test # 7. It asks you find the area of a deck that surrounds a hot tub if the hot tub has a diameter of 6 meters and the deck is 2 meters wide. When I took the test I knew how to do the problem but I just had a brain cramp and screwed up the final answer. The way to figure out this problem is to find the area of the hot tub and then subtract that from the area of the deck. The only formula that you need to use on this problem is a=pie(r) squared. The first thing to do is sub the number 3 into that formula and you end up with 9 pie. You then plug the number 5 into that formula because 6+ 2+2 equals 10 as a diameter and half of that equals 5. When plugging that into the equation you end up with 25 pie. The problem that I made on the test was that I forgot to subtract 9 pie from 25 pie so that is the last step in this problem. Your final answer should be 16 pie. Article posted June 7, 2012 at 04:49 PM GMT-5 • comment • Reads 356
Article posted June 10, 2012 at 10:26 PM GMT-5 • comment • Reads 224 This blog is the twenty-third of my personal weekly blog This blog was supposed to be posted last week but blogmeister was down! Oh no! We got everything squared away though, so yay! Last week was pretty gloomy, the weather hasn't been that nice until this weekend! Last week in geometry we started learning more in trigonometry and then started learning about law of sines and law of cosines! It's been pretty easy, but some things trip me up easily. Anyways the year is winding down and I can not wait until SUMMER! Midterms start this week! Here is a review question on what we have learned in Geometry! Ms. J told us that studying our tests and quizzes would be most helpful so that is exactly what I will be doing! A particular unit that I didn't fair so well on was working with the lines of triangles! Here is a review question from the Lines of Triangles Quiz that we took on February 7th. Which of the following are the slopes of two perpendicular lines? a. 3 and -3 b. 5 and 1/5 c. no slope and undefined d. -2/3 and 3/2 If you recall that the slope of any line perpendicular to another is the negative reciprocal, then this problem is easy to solve! A is not the answer because the negative reciprocal or 3 would be -1/3. B does not work because the negative reciprocal of 5 would be -1/5. c does not work because they aren't specific lines! THEREFORE D IS THE ANSWER! and to prove it.. the negative reciprocal of -2/3 would be - ( -3/2) or 3/2! TADA! Article posted June 10, 2012 at 10:26 PM GMT-5 • comment • Reads 224
Article posted June 10, 2012 at 10:34 PM GMT-5 • comment • Reads 419 This is the twenty-fourth of my personal weekly blog. Alas I have reached my last blog. Farewell classblogmeister... I won't miss you too much I promise! This week in Geometry we continued working with the law of sines and the law of cosines! We have a test this week before our final so that means a lot of studying on our parts! We pretty much did a lot of reviewing on the all the parts of trigonometry, which was helpful! This review question will be coming from the Quadrilaterals test we took on March 29th. I GOT THIS QUESTION WRONG! I honestly have no idea how I could've made the mistake, but I did, so make sure you don't jump to conclusions! Consecutive sides of a rectangle are congruent. a. sometimes b. always c. never the answer is SOMETIMES because a square (which is a rectangle) has congruent consecutive sides! YAY GEOMETRY!(: Article posted June 10, 2012 at 10:34 PM GMT-5 • comment • Reads 419
Article posted June 10, 2012 at 10:32 AM GMT-5 • comment • Reads 224 For Ch 12, the work was on transformations. The review question will be for rotations. The question as what is the point of rotation, and what is the angle of rotation. When given a pre-image and an image, to find the point of rotation, it will either be where the two images intersect or a point not connected to either image. Then to find the angle of rotation, you will need a protractor. Put your protractor on the point of rotation and pick a point you want to measure from. You always measure from the pre-image and counterclockwise unless stated to go clockwise. After you pick the point, find the point on the image that is the same, and measure. That will give you the angle of rotation Stay classy bloggers. Article posted June 10, 2012 at 10:32 AM GMT-5 • comment • Reads 224
Article posted June 10, 2012 at 11:24 AM GMT-5 • comment • Reads 217 Our final exam is Wednesday! Ahhh! And in preparation to that I am going to be posting a review question from one of the chapters we learned about this semester. In chapter 7 we learned more about areas of formulas. My review question today, will be finding the area of a rhombus (remember that the diagonals of a rhombus bisect each other). The formula for an area of a rhombus is 1/2(d1+d2), d stands for diagonal. If your given information was that one diagonal was 7cm and half of the other diagonal was 5.5cm, this is how you would solve it: We know that the other half of the diagonal is 5.5cm, because it gets bisected from the other diagonal. Now we plug in our given information to come up with 1/2(7+11) = area. Simplify to get 1/2(18) = area. Now simply multiply by 1/2 or divide by two, and we get 9cm squared = area Hopefully this helped you! I will be posting another review question soon! Goodbye! Article posted June 10, 2012 at 11:24 AM GMT-5 • comment • Reads 217
Article posted June 10, 2012 at 12:01 PM GMT-5 • comment • Reads 233 Hello again! Today, I am posting another review question. This comes from chapter 5. Question: Which of they following is the concurrent point of the altitudes? Answer: Orthocenter. I always remember which concurrent point goes with which line of a triangle because of MAPA COCI. MAPA COCI stands for: Median --> Centroid Altitude --> Orthocenter Perpendicular Bisector -->Circumcenter Angle Bisector --> Incenter This helped me remember and hopefully it will help you too! :) Now down to the sad stuff. :( I won't be saying 'see you next week' anymore..... because this is my last blog! *Gasp* I hope my blogs have helped you in some way or another, but right now I have to say goodbye for good. I know you'll miss me though! Farewell, Livy Article posted June 10, 2012 at 12:01 PM GMT-5 • comment • Reads 233
Article posted June 11, 2012 at 02:06 PM GMT-5 • comment • Reads 235 Question Given: Circle O with Diamter CD, AB is parrallel to CD, and arc AB=80 degrees Find arc CA [1] 50 [2] 60 [3] 80 [4] 100 Answer 50 Article posted June 11, 2012 at 02:06 PM GMT-5 • comment • Reads 235
Article posted June 11, 2012 at 02:10 PM GMT-5 • comment • Reads 381 Question Given: In triangle ABC, B=120, c=15, and a=15 Find C Answer 30 degrees Article posted June 11, 2012 at 02:10 PM GMT-5 • comment • Reads 381
Article posted June 11, 2012 at 09:04 PM GMT-5 • comment • Reads 241 The midterm is in two days and I'm getting a little nervous. We also have a test tomorrow that isn't helping with the nerves! Hopefully the test tomorrow will make get me feeling completely confident with the trigonometry unit, so it'll be one less thing to study for the final! My first review question that I want to go over is number 13 from practice 58. It is working with vectors which I needed a slight refresher on. The question: Homing pigeons have the ability or instinct to find their way home when released hundreds of miles away from home. Homing pigeons carried news of Olympic victories to various cities in ancient Greece. Suppose one such pigeon took off from Athens and landed in Sparta, which is 73 miles west and 64 miles south of Athens. Find the distance and its direction of flight. The first step you would need to take is plugging the coordinates into the distance formula. It comes out with 97, so Athens and Sparta are 97 miles apart. Then, you need to use the tangent ratio. This is the opposite side over the adjacent side. Therefore, tan(x) = 64/73 --> x = tan-1( 64/73) --> x = 41°, so the direction of flight is 41°. Article posted June 11, 2012 at 09:04 PM GMT-5 • comment • Reads 241
Article posted June 11, 2012 at 09:16 PM GMT-5 • comment • Reads 229 We were required two review questions, so I think I'll choose one from the beginning of the semester just as a refresher. Number 25 on the quadrilaterals test was asking us to label the coordinates of parallelogram LAST using only three variables. I chose d, c, and b (along with 0). Point 'a' was on the y-axis so the x coordinate was 0. Then, I chose 'b' to act as the height, or y coordinate. Point 'l' is the lower, left-hand point, and it is on the x-axis, making the y coordinate 0. Then, I made the x coordinate '-c' because it is to the left of the y-axis. Point 't' is the lower right-hand point, and it also lies on the x-axis, making the y coordinate 0. Then, I chose 'd' to act as the x coordinate. Point 's' was the upper right point of the parallelogram, so the height was also 'b', making the y coordinate automatically 'b'. The x coordinate is going to be 'd+c', because 'c' is the length that the segment goes on longer past 'd', which is where you get the '+c'. Hopefully that all made sense and was (slightly) helpful! Article posted June 11, 2012 at 09:16 PM GMT-5 • comment • Reads 229
Article posted June 10, 2012 at 09:00 PM GMT-5 • comment • Reads 222 This blog is supposed to be for two weeks ago but due to problems with the website the due date was postponed. This blog I will be explaining how to do a problem from one of the previous tests that I have taken. I have chosen to explain number 4 from test 7. The problem gives you a rhombus and tells you to find the area using the diagonals of 18 and 21. To find the area you must use the formula A=.5(d1)(d2). You would use substitution of make the equation A=.5(18)(21). If you do the math the area comes out to be 189ft squared. That is how you would do a problem like that... Article posted June 10, 2012 at 09:00 PM GMT-5 • comment • Reads 222
Article posted June 10, 2012 at 09:23 PM GMT-5 • comment • Reads 311 This week in geometry I am doing the same thing as last week. I have to pick a problem from any of my previous tests and I must explain how to do it.On problem 4 from the chapter 7 retest, the problem gives you a parallelogram and tells you to find the area using a height of 24 and a base of 10. The formula you must use is A= base times height. You would set up the equation as A= 24 times 10. The answer to the problem is 240 meters squared. That is how you do an problem like that... Article posted June 10, 2012 at 09:23 PM GMT-5 • comment • Reads 311
Article posted June 12, 2012 at 05:29 PM GMT-5 • comment • Reads 234 Question: The diagonals of a square bisect all angles Answer: Always Article posted June 12, 2012 at 05:29 PM GMT-5 • comment • Reads 234
Article posted June 12, 2012 at 05:31 PM GMT-5 • comment • Reads 312 Question: A triangle has side lengths of 8 cm, 14 cm, and 11 cm. Classify the triangle. Answer:Obtuse Article posted June 12, 2012 at 05:31 PM GMT-5 • comment • Reads 312
Article posted June 9, 2012 at 10:48 AM GMT-5 • comment • Reads 303 Hey guys so the school year's almost over. Yeah, summer. Unfortunately to get to summer we have to go through probably the wort week of the school year. Finals. Yeah you all know what I'm talking about. So here is one of the review questions that I'm using to study for my geometry final. This question is on right triangle trig. Good luck! Find X and H Answer: X=13 H=8.1 Article posted June 9, 2012 at 10:48 AM GMT-5 • comment • Reads 303
Article posted June 4, 2012 at 06:43 AM GMT-5 • comment • Reads 261 Hey guys so as I said in my last blog we are learning about SOH, CAH, TOA. Well now we are learning about the law of sines. Here is a review question and its answer about the law of sines. If m Answer: Measure of angle B is 59.4. Article posted June 4, 2012 at 06:43 AM GMT-5 • comment • Reads 261
Article posted June 7, 2012 at 09:17 PM GMT-5 • comment • Reads 224 We have finals in two weeks, and I don't really have any specific emotions for the end of the year. It was an awesome freshman year and a good first highschool experience. I hope next year will be as good, but then again I won't get to have Ms. J anymore:( To study for the final we must post a review question. Mine is- If a plane takes off at 25 degrees and flies 1600 ft, what is its altitude. SPOILER BELOW To do this problem, you must first write out the formula for sine. Using SOH, it would come out to be sin25=X/1600. Then you would multiply by 1600 on both sides to isolate X. This would make it 1600*sin(25)=X. Enter 1600*sin(25)into your calculator and the resulting number is the altitude-676.18 ft Article posted June 7, 2012 at 09:17 PM GMT-5 • comment • Reads 224
Article posted June 10, 2012 at 10:45 PM GMT-5 • comment • Reads 240 Hello everyone, I'm sad to tell you that this will be my last geometry blog :( The year is finally coming to an end. Six more days of school until summer! Here is my final exam review question: A triangle has three sides with the lengths of 6:8:10 What type of triangle is it? A.) acute B.)obtuse C.) right Answer: C. Right Article posted June 10, 2012 at 10:45 PM GMT-5 • comment • Reads 240
Article posted June 10, 2012 at 11:25 AM GMT-5 • comment • Reads 208 Hello Bloggers!!! This week I will be giving you a practice problem for you to solve!! At the bottom of the blog I will show you the work to get to the answer!! Question: You know two sides and an angle (6cm, 10cm, and 56 degrees) find the missing side. Answer: X2(squared)=6(squared)+10(squared) - 2(6)(10) - cos(56) by plugging this equation: 6(squared) + 10(squared) -2(6)(10) - cos(56) into your calculator, you will come up with X2(squared)= 15.4408071 you need to find X, so you have to find the square root. Type this into your calculator and X= 3.92947924 Great Job!!!! I will be blogging soon!!!! Article posted June 10, 2012 at 11:25 AM GMT-5 • comment • Reads 208
Article posted June 9, 2012 at 02:17 PM GMT-5 • comment • Reads 217 Hey bloggers! Just as I promised on Wednesday, here is another final review question! "In parallelogram BARK, m first draw the paralleglogram: 4x = xsquared - 60 xsqaured - 4x - 60 = 0 (x + 6)(x - 10) xsquared - 10x + 6x - 60 xsquared - 4x - 60 x + 6 = 0 or x - 10 = 0 x = -6 or x = 10 7x + m 70 + m m m< ARK = 110 dgrees Wish me luck on my final! Grace Article posted June 9, 2012 at 02:17 PM GMT-5 • comment • Reads 217
Article posted June 6, 2012 at 04:34 PM GMT-5 • comment • Reads 234 Hey bloggers! I am going to show you a good review question for our Geometry final next week! "Q is the intersection of AC and BD and ABCD. Find the area of kite ABCD if AB= 10, BC= 20, AC = 30, and BQ = 5." A= 1/2 x d1 x d2 d1= DB = 2BQ = 10. d2 = AC = 30 A = 1/2 x 10 x 30 Area = 150 squared units I will be posting another question later in the week due to the website being down last weekend. Later, Grace Article posted June 6, 2012 at 04:34 PM GMT-5 • comment • Reads 234
Article posted June 10, 2012 at 08:47 PM GMT-5 • comment • Reads 219 Welcome back everyone, since we are nearing the end of the school year, I was thinking that we could all start posting review questions to prepare ourselves for the upcoming final! What is a perpendicular bisector? And, how do you find it for any triangle? A perpendicular bisector is a line that divides a side of a triangle directly in the middle, but it bisects it at a 90˚ angle. Each side of the triangle has it’s own perpendicular bisector. In a triangle, there are 3 different sides each that have a perpendicular bisector. Since there are three different types of triangles (acute, right, and obtuse), each location of the circumcenter is different. For any acute triangle, the circumcenter is inside of the triangle. The point where all three of the Perpendicular bisectors intersect is at the circumcenter. But, like always there are 2 special cases. For a right triangle, the circumcenter of a right triangle is on the hypotenuse. But for an obtuse triangle: the circumcenter of an obtuse triangle is outside of the triangle. A circumscribed circle is a circle that is drawn from the circumcenter that hits all of the vertices of the triangle. The circumcenter is always equidistant to all vertices or any other point on the circumscribed circle. If you need anymore help with perpendicular bisectors reply to this post! I hope you all begin to review now! See you all next week! -Kathleen Article posted June 10, 2012 at 08:47 PM GMT-5 • comment • Reads 219
Article posted June 10, 2012 at 10:17 AM GMT-5 • comment • Reads 204 Hello again! As we are approaching our final exam next week, we are continuing to post review questions! These are aimed to help us remember the concepts we have learned this semester! The question I'm going to ask takes us back a chapter, and deals with triangles! Be sure to blog back with any questions. Q: A triangle has side lengths of 8cm, 11cm, and 14cm. Classify this triangle as acute, obtuse, or right. A: OBTUSE To find this answer, we first begin with the Pythagorean theorem (A squared + B squared = c squared). So, when we plug in what is known, our formula becomes 64 + 121 = 196. Here's the rule to classify: IF (A squared + b squared > C squared) then the triangle is acute. IF (A squared + b squared < C squared) then the triangle is obtuse. IF (A squared + b squared = C squared) then the triangle is right. In our formula, 64 + 121 = 185. Because 185 < 196, then the triangle is obtuse! Hope that helped and you learned something knew! This is my last weekly blog... but I'll be writing occasionally until the end of the year. Talk to you then! Emma Article posted June 10, 2012 at 10:17 AM GMT-5 • comment • Reads 204
Article posted June 14, 2012 at 12:40 PM GMT-5 • comment • Reads 240 We finally made it! Finals are around the corner, and so is summer! I can’t wait to ditch my backpack and head to the beach. But first, I have to survive finals. To review for Geometry, I’ve posted one of my favorite homework problems from a past unit below. KL⎮⎮JM in isosceles trapezoid JKLM. Find the values of x and y if m⦟J=(23x-8)º, m⦟K=(12y-13)º, and m⦟M=(17x+10)º. m⦟J=m⦟M 23x-8=17x+10 6x-8=10 6x=18 x=3 m⦟J+m⦟K=180 23x-8+12y-13=180 61+12y-13=180 12y=132 y=11 I chose this problem because I was from one of the first units of the semester, and I didn’t remember that unit at all. I thought that it would be nice to review since I had forgotten about it. Article posted June 14, 2012 at 12:40 PM GMT-5 • comment • Reads 240
Article posted June 14, 2012 at 12:40 PM GMT-5 • comment • Reads 244 We finally made it! Finals are around the corner, and so is summer! I can’t wait to ditch my backpack and head to the beach. But first, I have to survive finals. To review for Geometry, I’ve posted one of my favorite homework problems from a past unit below. KL⎮⎮JM in isosceles trapezoid JKLM. Find the values of x and y if m⦟J=(23x-8)º, m⦟K=(12y-13)º, and m⦟M=(17x+10)º. m⦟J=m⦟M 23x-8=17x+10 6x-8=10 6x=18 x=3 m⦟J+m⦟K=180 23x-8+12y-13=180 61+12y-13=180 12y=132 y=11 I chose this problem because I was from one of the first units of the semester, and I didn’t remember that unit at all. I thought that it would be nice to review since I had forgotten about it. Article posted June 14, 2012 at 12:40 PM GMT-5 • comment • Reads 244
Article posted June 6, 2012 at 04:33 PM GMT-5 • comment • Reads 257 Hello fellow blog readers! Sorry that this blog is so late compared to my others. Blogmeister, as I'm sure you know, has been down for the past couple of days. I am sorry to keep you waiting for so long. Now to the important stuff. In geometry, I have a question about the latest test. My question is how to find the length of an arc. I do not remember and was wondering if you bloggers could help me out. If you comment with the correct answer, I will mention you in my next blog. Well that is all for now blog-readers! Stay tuned for this Sunday's blog and question. Stay classy and thanks for reading! Article posted June 6, 2012 at 04:33 PM GMT-5 • comment • Reads 257
Article posted June 10, 2012 at 04:52 PM GMT-5 • comment • Reads 215 The year is almost over! We have two tests this next week in geometry. One tomorrow on chapter 9 and one on June 18th on everything we've learned this semester (our final). To help review for this final I have been told to blog a review question with the answer. Do you remember the distance formula? Well for coordinate proofs we had to use the distance formula to show how one side was of equal length to the other side. To show this you use the formula d = the square root of (y2-y1)squared + (x2-x1)squared. If you plug in the coordinates from #26 on the quadrilaterals test you get AS = the square root of (b-b)squared + (a+c-0)squared. This simplifies to the square root of (a+c)squared which equals a + c. Hope this was helpful, good luck on the test and the final! Article posted June 10, 2012 at 04:52 PM GMT-5 • comment • Reads 215
Article posted June 10, 2012 at 04:11 PM GMT-5 • comment • Reads 218 AHHHH! Three more days until final exams! Before I show you the review question take a deep breath... Inhale, now exhale. Feel better? Good. Let's begin! This is from chapter 9. This chapter is on SOH CAH TOA, the law of sines, the law of cosines, and vectors. Say you had a triangle where the length of the hypotenuse (AC) is 12cm, and the short side (AB) and the long side (BC) is unknown. Angle C is given as 23 degrees and angle B is 90 degrees. To solve this problem you use CAH because you are given a right triangle, an angle (23 degrees), the hypotenuse to that angle (12cm), and the long side is present. To set it up you write; cos23 = X/12. To solve you multiple cos23 by 12, so it's 12cos23 = X. X = 11cm. I hope this was helpful! Article posted June 10, 2012 at 04:11 PM GMT-5 • comment • Reads 218
Article posted June 10, 2012 at 08:54 PM GMT-5 • comment • Reads 216 Hello everyone! Hope your week has gone nicely! Mine has been pretty interesting, because my brother had an English exchange student arrive on Thursday. He will be staying with us for the week, and I am really excited! English accents are so cool! This exchange trip has also reminded me that the end of the school year is really close! In order to prepare for the geometry final, I am going to tell you about a question we had to do for homework as a way to review. This problem was on a worksheet I did recently involving the Law of Sines: sinA = sinB a b A and B stand for angles of an oblique triangle, while a and b stand for the 2 sides opposite them. Basically, you can use this law to find one of these four measures, as long as you have an angle and the side opposite. Basically, the problem was a triangle with angles A, B, and C, and sides a, b, and c. The measure of angle A was 35º, the measure of side a was 8 cm, and the measure of side b was 12 cm. The problem asked you to find the measure of angle B. Through substitution, I came out with this equation: sin(35) = sin(B) 8 12 By multiplying both sides by 12 and then by sin to the (-1), I was able to then solve for B: sin-1(12sin(35)) = sin-1(sin(B)) 8 m This process took some time to get used to, but I've found I actually like this sorts of problems in trigonometry. Let me know if you have any questions! Have a great week! (: Article posted June 10, 2012 at 08:54 PM GMT-5 • comment • Reads 216
Article posted June 10, 2012 at 09:02 PM GMT-5 • comment • Reads 207 Hello everyone! Firstly, I would just like to explain that last week Blogmeister was experiencing some technical difficulties, so I was unable to post at that time. The post from both then and the most recent one should be uploaded now. Anyways, this week is the beginning of finals for my school! I am very nervous because I've never taken finals before, and I'm not sure what to expect. Wish me luck! In order to review once again for geometry, I am posting another review questions below. I chose this particular one because I find this area formula hard to remember sometimes, because it doesn't involve side lengths, rather, diagonal lengths. For this problem, there was a rhombus shown with diagonals of 21 ft and 18 ft. The instructions were to find the area, and the possible answers were as shown: a) 378 sq. ft b) 189 sq. ft c) 162 sq. ft d) 27(square root of)85 sq. ft. I knew that the formula for the area of a rhombus is 1/2d1*d2, so I substituted in the diameter lengths: A = 1/2 * 21 * 18 = 189 With that equation, I was able to find that the area was 189 sq. ft., or answer b. I hope this has been a helpful review question! Have a great week! (: Article posted June 10, 2012 at 09:02 PM GMT-5 • comment • Reads 207
Article posted June 11, 2012 at 08:18 AM GMT-5 • comment • Reads 256 Well, the final is coming up for geometry this. Just one more test that I need to study for. I can't say that I enjoy these types of things, but I still need to do them. To help out everybody, though, here are two problems that were on homework from this term. There are two since the blog wasn't working last week when I was supposed to post one and one from this week, so I'm just combining the two blogs into this week's. Here they are: The lengths of the diagonals of a rhombus are 2 in. and 5 in. Find the measures of the angles of the rhombus to the nearest degree. Describe each translation using an ordered pair. 2 units to the left, 1 unit down. Good luck to everyone on the final! Article posted June 11, 2012 at 08:18 AM GMT-5 • comment • Reads 256
Article posted June 10, 2012 at 07:04 PM GMT-5 • comment • Reads 190 It seems like just yesterday that we were beginning our weekly blogs, and now we are writing our last ones. This is the week of finals, and after this week, we only have one day left of school. I cannot believe that school went by this fast. I would have to say that this school year was filled with learning, and fun! On another note, we still must remain serious. We still have our final exam on Monday, and that requires a gracious amount of studying. To study for the exam, I have come up with a few review guides that will help me go over anything that I have been struggling with. One thing that I have had a bit of trouble with is vectors. Therefore, I picked out a problem about vectors from my book and worked out the answer to it. The problem was "describe each vector as an ordered pair". To do so, you had to first figure out the sine ratio as well as the cosine ratio. Then, you would order them so that they describe the vector. It took some practice to understand, but eventually, I got it! I now feel prepared for this final exam! Article posted June 10, 2012 at 07:04 PM GMT-5 • comment • Reads 190
Article posted June 4, 2012 at 09:38 AM GMT-5 • comment • Reads 251 Salutations fellow bloggers! This week in geometry class we learned about the law of sine and cosine. It isn't to difficult but I don't really understand how it saves you time. My question for the final review is: What quadrilaterals have diagonals that bisect each other? Answer: Parallelogram, rhombus, rectangle, and square Have a great week everyone! Article posted June 4, 2012 at 09:38 AM GMT-5 • comment • Reads 251
Article posted June 5, 2012 at 07:53 PM GMT-5 • comment • Reads 237 Hi everyone! Just writing because before I could add my review question, Blogmeister shut itself down! NOOOOOO!!!! Well, anyway, here’s my Trigonometry Review question: Kevin and Zane have just built a tree house. It is 20 ft off the ground, and the tree is perpendicular to the ground. When Kevin’s Mom goes out to look at the tree house, the angle of elevation from her feet to the tree house is 50*. Zane wants to know how far the tree that has the tree house in it is from the house, but he has lost the tape measure. How far away is the tree from the house? ***SPOILER ALERT*** Answer: Roughly 17 Feet --Joe Article posted June 5, 2012 at 07:53 PM GMT-5 • comment • Reads 237
Article posted June 10, 2012 at 07:42 AM GMT-5 • comment • Reads 198 HELLO! WELCOME TO THE FINAL COMPLETE WEEK OF SCHOOL! Wow, the year has gone by so fast! Did I mention that this is our final weekly blog? I can’t believe how close we are to the end of the year! It seems like just yesterday we walked into Geometry for the first time. I have had a great weekend, capped of by a RockEucharist at my church! This week in Geometry, we continued with our trigonometry unit, and we learned the law of cosines. At first I didn’t understand it, but once I got it, I had it for good. I think that on our trig. test this week, I will not have that much trouble, as this unit has been about mostly algebra and I do not have much trouble with algebra. But you never know-- it may be very hard. Our final is also coming up. Wow, that snuck up on me. Here is another review question: Ben and Griffin love modern houses. They want to make a museum about the evolution of houses. While searching for a place to put the museum, they find a modern building that is a rhombus. The sides of the house are 50 feet long. The angles formed by two intersecting lines are 120* and 60*. How much area is in the museum? ANSWER: 4330 ft. squared Have a great last week! --Joe Article posted June 10, 2012 at 07:42 AM GMT-5 • comment • Reads 198
My Classes & Students
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<MASK>
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<MASK>
(3) 如果测量只是让集合 $f_{1} \leq f \leq f_{2}$ 中的特征值通过，则基简化为
$$\开始{对齐} \Psi(x, t) &=\sum_{f}^{\prime} c_{f}(t) \psi_{f}(x) & & ; t \geq t_{0} \ \sum_{f}^{\prime}\left|c_{f}(t)\right|^{2} &=1 & & \end{对齐}$$
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<MASK>
(3) 如果测量只是让集合 $f_{1} \leq f \leq f_{2}$ 中的特征值通过，则基简化为
$$\开始{对齐} \Psi(x, t) &=\sum_{f}^{\prime} c_{f}(t) \psi_{f}(x) & & ; t \geq t_{0} \ \sum_{f}^{\prime}\left|c_{f}(t)\right|^{2} &=1 & & \end{对齐}$$
<MASK>
<UNMASK>
19th Ave New York, NY 95822, USA
# 物理代考| Reduction of the Basis量子力学代写
<MASK>
8.2 Reduction of the Basis
Let us try to formalize this measurement theory. Suppose we are looking at a single particle, and we have a complete set of the eigenfunctions of some hermitian operator with real eigenvalues at our disposal
$$F \psi_{f}(x)=f \psi_{f}(x) \quad ; \text { eigenfunctions }$$
Order the eigenvalues $f_{0} \leq f_{1} \leq f_{2} \cdots$, and expand the wave function $\Psi(x, t)$ in this complete set of eigenfuctions
$$\Psi(x, t)=\sum_{f} c_{f}(t) \psi_{f}(x) \quad ; \text { complete set }$$
The state is normalized, so that
$$\sum_{f}\left|c_{f}(t)\right|^{2}=1$$
Measurement theory then assumes the following:
(1) If we make a precise measurement of the quantity $F$, we will observe one of the eigenvalues $f$;
(2) If we perform a pure pass measurement at a time $t_{0}$ that lets the eigenvalue $f$ through, then the wave function is reduced to $^{4}$
\begin{aligned} \Psi(x, t) &=c_{f}(t) \psi_{f}(x) \quad ; t \geq t_{0} \ \left|c_{f}(t)\right|^{2} &=1 \end{aligned}
The measurement is reproducible and the basis is reduced.
(3) If the measurement simply lets the eigenvalues in the set $f_{1} \leq f \leq f_{2}$ through, then the basis is reduced to
\begin{aligned} \Psi(x, t) &=\sum_{f}^{\prime} c_{f}(t) \psi_{f}(x) & & ; t \geq t_{0} \ \sum_{f}^{\prime}\left|c_{f}(t)\right|^{2} &=1 & & \end{aligned}
where the sum $\sum_{f}^{\prime}$ goes over $f_{1} \leq f \leq f_{2}$.
${ }^{4}$ Note that the coefficient $c_{f}(t)$ must be rescaled to achieve this norm (see Prob. 10.2).
## 物理代考
<MASK>
$$F \psi_{f}(x)=f \psi_{f}(x) \quad ; \text { 特征函数 }$$
<MASK>
(3) 如果测量只是让集合 $f_{1} \leq f \leq f_{2}$ 中的特征值通过，则基简化为
$$\开始{对齐} \Psi(x, t) &=\sum_{f}^{\prime} c_{f}(t) \psi_{f}(x) & & ; t \geq t_{0} \ \sum_{f}^{\prime}\left|c_{f}(t)\right|^{2} &=1 & & \end{对齐}$$
<MASK>
Matlab代写
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19th Ave New York, NY 95822, USA
# 物理代考| Reduction of the Basis量子力学代写
<MASK>
8.2 Reduction of the Basis
Let us try to formalize this measurement theory. Suppose we are looking at a single particle, and we have a complete set of the eigenfunctions of some hermitian operator with real eigenvalues at our disposal
$$F \psi_{f}(x)=f \psi_{f}(x) \quad ; \text { eigenfunctions }$$
Order the eigenvalues $f_{0} \leq f_{1} \leq f_{2} \cdots$, and expand the wave function $\Psi(x, t)$ in this complete set of eigenfuctions
$$\Psi(x, t)=\sum_{f} c_{f}(t) \psi_{f}(x) \quad ; \text { complete set }$$
The state is normalized, so that
$$\sum_{f}\left|c_{f}(t)\right|^{2}=1$$
Measurement theory then assumes the following:
(1) If we make a precise measurement of the quantity $F$, we will observe one of the eigenvalues $f$;
(2) If we perform a pure pass measurement at a time $t_{0}$ that lets the eigenvalue $f$ through, then the wave function is reduced to $^{4}$
\begin{aligned} \Psi(x, t) &=c_{f}(t) \psi_{f}(x) \quad ; t \geq t_{0} \ \left|c_{f}(t)\right|^{2} &=1 \end{aligned}
The measurement is reproducible and the basis is reduced.
(3) If the measurement simply lets the eigenvalues in the set $f_{1} \leq f \leq f_{2}$ through, then the basis is reduced to
\begin{aligned} \Psi(x, t) &=\sum_{f}^{\prime} c_{f}(t) \psi_{f}(x) & & ; t \geq t_{0} \ \sum_{f}^{\prime}\left|c_{f}(t)\right|^{2} &=1 & & \end{aligned}
where the sum $\sum_{f}^{\prime}$ goes over $f_{1} \leq f \leq f_{2}$.
${ }^{4}$ Note that the coefficient $c_{f}(t)$ must be rescaled to achieve this norm (see Prob. 10.2).
## 物理代考
<MASK>
$$F \psi_{f}(x)=f \psi_{f}(x) \quad ; \text { 特征函数 }$$
<MASK>
(3) 如果测量只是让集合 $f_{1} \leq f \leq f_{2}$ 中的特征值通过，则基简化为
$$\开始{对齐} \Psi(x, t) &=\sum_{f}^{\prime} c_{f}(t) \psi_{f}(x) & & ; t \geq t_{0} \ \sum_{f}^{\prime}\left|c_{f}(t)\right|^{2} &=1 & & \end{对齐}$$
<MASK>
Matlab代写
<UNMASK>
19th Ave New York, NY 95822, USA
# 物理代考| Reduction of the Basis量子力学代写
<MASK>
8.2 Reduction of the Basis
Let us try to formalize this measurement theory. Suppose we are looking at a single particle, and we have a complete set of the eigenfunctions of some hermitian operator with real eigenvalues at our disposal
$$F \psi_{f}(x)=f \psi_{f}(x) \quad ; \text { eigenfunctions }$$
Order the eigenvalues $f_{0} \leq f_{1} \leq f_{2} \cdots$, and expand the wave function $\Psi(x, t)$ in this complete set of eigenfuctions
$$\Psi(x, t)=\sum_{f} c_{f}(t) \psi_{f}(x) \quad ; \text { complete set }$$
The state is normalized, so that
$$\sum_{f}\left|c_{f}(t)\right|^{2}=1$$
Measurement theory then assumes the following:
(1) If we make a precise measurement of the quantity $F$, we will observe one of the eigenvalues $f$;
(2) If we perform a pure pass measurement at a time $t_{0}$ that lets the eigenvalue $f$ through, then the wave function is reduced to $^{4}$
\begin{aligned} \Psi(x, t) &=c_{f}(t) \psi_{f}(x) \quad ; t \geq t_{0} \ \left|c_{f}(t)\right|^{2} &=1 \end{aligned}
The measurement is reproducible and the basis is reduced.
(3) If the measurement simply lets the eigenvalues in the set $f_{1} \leq f \leq f_{2}$ through, then the basis is reduced to
\begin{aligned} \Psi(x, t) &=\sum_{f}^{\prime} c_{f}(t) \psi_{f}(x) & & ; t \geq t_{0} \ \sum_{f}^{\prime}\left|c_{f}(t)\right|^{2} &=1 & & \end{aligned}
where the sum $\sum_{f}^{\prime}$ goes over $f_{1} \leq f \leq f_{2}$.
${ }^{4}$ Note that the coefficient $c_{f}(t)$ must be rescaled to achieve this norm (see Prob. 10.2).
## 物理代考
8.2 降低基数
$$F \psi_{f}(x)=f \psi_{f}(x) \quad ; \text { 特征函数 }$$
$$\Psi(x, t)=\sum_{f} c_{f}(t) \psi_{f}(x) \quad ; \text { 完整集 }$$
$$\sum_{f}\left|c_{f}(t)\right|^{2}=1$$
(1) 如果我们对数量 $F$ 进行精确测量，我们将观察到一个特征值 $f$；
(2) 如果我们在时间 $t_{0}$ 执行纯通过测量，让特征值 $f$ 通过，那么波函数将简化为 $^{4}$
$$\开始{对齐} \Psi(x, t) &=c_{f}(t) \psi_{f}(x) \quad ; t \geq t_{0} \ \left|c_{f}(t)\right|^{2} &=1 \end{对齐}$$
(3) 如果测量只是让集合 $f_{1} \leq f \leq f_{2}$ 中的特征值通过，则基简化为
$$\开始{对齐} \Psi(x, t) &=\sum_{f}^{\prime} c_{f}(t) \psi_{f}(x) & & ; t \geq t_{0} \ \sum_{f}^{\prime}\left|c_{f}(t)\right|^{2} &=1 & & \end{对齐}$$
${ }^{4}$ 注意，必须重新调整系数 $c_{f}(t)$ 才能达到这个标准（见 Prob. 10.2）。
Matlab代写
|
19th Ave New York, NY 95822, USA
# 物理代考| Reduction of the Basis量子力学代写
<MASK>
8.2 Reduction of the Basis
Let us try to formalize this measurement theory. Suppose we are looking at a single particle, and we have a complete set of the eigenfunctions of some hermitian operator with real eigenvalues at our disposal
$$F \psi_{f}(x)=f \psi_{f}(x) \quad ; \text { eigenfunctions }$$
Order the eigenvalues $f_{0} \leq f_{1} \leq f_{2} \cdots$, and expand the wave function $\Psi(x, t)$ in this complete set of eigenfuctions
$$\Psi(x, t)=\sum_{f} c_{f}(t) \psi_{f}(x) \quad ; \text { complete set }$$
The state is normalized, so that
$$\sum_{f}\left|c_{f}(t)\right|^{2}=1$$
Measurement theory then assumes the following:
(1) If we make a precise measurement of the quantity $F$, we will observe one of the eigenvalues $f$;
(2) If we perform a pure pass measurement at a time $t_{0}$ that lets the eigenvalue $f$ through, then the wave function is reduced to $^{4}$
\begin{aligned} \Psi(x, t) &=c_{f}(t) \psi_{f}(x) \quad ; t \geq t_{0} \ \left|c_{f}(t)\right|^{2} &=1 \end{aligned}
The measurement is reproducible and the basis is reduced.
(3) If the measurement simply lets the eigenvalues in the set $f_{1} \leq f \leq f_{2}$ through, then the basis is reduced to
\begin{aligned} \Psi(x, t) &=\sum_{f}^{\prime} c_{f}(t) \psi_{f}(x) & & ; t \geq t_{0} \ \sum_{f}^{\prime}\left|c_{f}(t)\right|^{2} &=1 & & \end{aligned}
where the sum $\sum_{f}^{\prime}$ goes over $f_{1} \leq f \leq f_{2}$.
${ }^{4}$ Note that the coefficient $c_{f}(t)$ must be rescaled to achieve this norm (see Prob. 10.2).
## 物理代考
8.2 降低基数
$$F \psi_{f}(x)=f \psi_{f}(x) \quad ; \text { 特征函数 }$$
$$\Psi(x, t)=\sum_{f} c_{f}(t) \psi_{f}(x) \quad ; \text { 完整集 }$$
$$\sum_{f}\left|c_{f}(t)\right|^{2}=1$$
(1) 如果我们对数量 $F$ 进行精确测量，我们将观察到一个特征值 $f$；
(2) 如果我们在时间 $t_{0}$ 执行纯通过测量，让特征值 $f$ 通过，那么波函数将简化为 $^{4}$
$$\开始{对齐} \Psi(x, t) &=c_{f}(t) \psi_{f}(x) \quad ; t \geq t_{0} \ \left|c_{f}(t)\right|^{2} &=1 \end{对齐}$$
(3) 如果测量只是让集合 $f_{1} \leq f \leq f_{2}$ 中的特征值通过，则基简化为
$$\开始{对齐} \Psi(x, t) &=\sum_{f}^{\prime} c_{f}(t) \psi_{f}(x) & & ; t \geq t_{0} \ \sum_{f}^{\prime}\left|c_{f}(t)\right|^{2} &=1 & & \end{对齐}$$
${ }^{4}$ 注意，必须重新调整系数 $c_{f}(t)$ 才能达到这个标准（见 Prob. 10.2）。
Matlab代写
<UNMASK>
19th Ave New York, NY 95822, USA
# 物理代考| Reduction of the Basis量子力学代写
## 物理代写
8.2 Reduction of the Basis
Let us try to formalize this measurement theory. Suppose we are looking at a single particle, and we have a complete set of the eigenfunctions of some hermitian operator with real eigenvalues at our disposal
$$F \psi_{f}(x)=f \psi_{f}(x) \quad ; \text { eigenfunctions }$$
Order the eigenvalues $f_{0} \leq f_{1} \leq f_{2} \cdots$, and expand the wave function $\Psi(x, t)$ in this complete set of eigenfuctions
$$\Psi(x, t)=\sum_{f} c_{f}(t) \psi_{f}(x) \quad ; \text { complete set }$$
The state is normalized, so that
$$\sum_{f}\left|c_{f}(t)\right|^{2}=1$$
Measurement theory then assumes the following:
(1) If we make a precise measurement of the quantity $F$, we will observe one of the eigenvalues $f$;
(2) If we perform a pure pass measurement at a time $t_{0}$ that lets the eigenvalue $f$ through, then the wave function is reduced to $^{4}$
\begin{aligned} \Psi(x, t) &=c_{f}(t) \psi_{f}(x) \quad ; t \geq t_{0} \ \left|c_{f}(t)\right|^{2} &=1 \end{aligned}
The measurement is reproducible and the basis is reduced.
(3) If the measurement simply lets the eigenvalues in the set $f_{1} \leq f \leq f_{2}$ through, then the basis is reduced to
\begin{aligned} \Psi(x, t) &=\sum_{f}^{\prime} c_{f}(t) \psi_{f}(x) & & ; t \geq t_{0} \ \sum_{f}^{\prime}\left|c_{f}(t)\right|^{2} &=1 & & \end{aligned}
where the sum $\sum_{f}^{\prime}$ goes over $f_{1} \leq f \leq f_{2}$.
${ }^{4}$ Note that the coefficient $c_{f}(t)$ must be rescaled to achieve this norm (see Prob. 10.2).
## 物理代考
8.2 降低基数
$$F \psi_{f}(x)=f \psi_{f}(x) \quad ; \text { 特征函数 }$$
$$\Psi(x, t)=\sum_{f} c_{f}(t) \psi_{f}(x) \quad ; \text { 完整集 }$$
$$\sum_{f}\left|c_{f}(t)\right|^{2}=1$$
(1) 如果我们对数量 $F$ 进行精确测量，我们将观察到一个特征值 $f$；
(2) 如果我们在时间 $t_{0}$ 执行纯通过测量，让特征值 $f$ 通过，那么波函数将简化为 $^{4}$
$$\开始{对齐} \Psi(x, t) &=c_{f}(t) \psi_{f}(x) \quad ; t \geq t_{0} \ \left|c_{f}(t)\right|^{2} &=1 \end{对齐}$$
(3) 如果测量只是让集合 $f_{1} \leq f \leq f_{2}$ 中的特征值通过，则基简化为
$$\开始{对齐} \Psi(x, t) &=\sum_{f}^{\prime} c_{f}(t) \psi_{f}(x) & & ; t \geq t_{0} \ \sum_{f}^{\prime}\left|c_{f}(t)\right|^{2} &=1 & & \end{对齐}$$
${ }^{4}$ 注意，必须重新调整系数 $c_{f}(t)$ 才能达到这个标准（见 Prob. 10.2）。
Matlab代写
|
<MASK>
### mahmoud2011
<MASK>
3. Aug 29, 2011
<MASK>
### mahmoud2011
<MASK>
<UNMASK>
<MASK>
$sech^{-1}(x) = ln(\frac{1\pm \sqrt{1-x^{2}}}{x}) : 0<x\leq 1$
<MASK>
thanks
<MASK>
### mahmoud2011
<MASK>
3. Aug 29, 2011
<MASK>
For the first question , I tried to check my solution I tried to check the solution directly from the definition of inverse function,
<MASK>
$f(x) = sech(x) : x\geq 0$
<MASK>
### mahmoud2011
<MASK>
|
<MASK>
$sech^{-1}(x) = ln(\frac{1\pm \sqrt{1-x^{2}}}{x}) : 0<x\leq 1$
<MASK>
thanks
<MASK>
### mahmoud2011
<MASK>
3. Aug 29, 2011
<MASK>
For the first question , I tried to check my solution I tried to check the solution directly from the definition of inverse function,
<MASK>
$f(x) = sech(x) : x\geq 0$
<MASK>
### mahmoud2011
<MASK>
<UNMASK>
<MASK>
$sech^{-1}(x) = ln(\frac{1\pm \sqrt{1-x^{2}}}{x}) : 0<x\leq 1$
And this not function which of them I will choose
<MASK>
thanks
<MASK>
### mahmoud2011
<MASK>
3. Aug 29, 2011
<MASK>
For the first question , I tried to check my solution I tried to check the solution directly from the definition of inverse function,
<MASK>
$f(x) = sech(x) : x\geq 0$
<MASK>
$(f o g)(x) = x : 0<x\leq 1$
Whereas $(g o f)(x) = x : x\leq0$
<MASK>
Exactly what does the question say? The reason you are getting "$\pm$" is because sech is NOT a one-to-one function and so does NOT HAVE an inverse over all real numbers. You can, of course, restrict the domain so that it does have an inverse. How the domain is restricted will determine the sign.
<MASK>
### mahmoud2011
<MASK>
|
<MASK>
$sech^{-1}(x) = ln(\frac{1\pm \sqrt{1-x^{2}}}{x}) : 0<x\leq 1$
And this not function which of them I will choose
<MASK>
thanks
<MASK>
### mahmoud2011
<MASK>
3. Aug 29, 2011
<MASK>
For the first question , I tried to check my solution I tried to check the solution directly from the definition of inverse function,
<MASK>
$f(x) = sech(x) : x\geq 0$
<MASK>
$(f o g)(x) = x : 0<x\leq 1$
Whereas $(g o f)(x) = x : x\leq0$
<MASK>
Exactly what does the question say? The reason you are getting "$\pm$" is because sech is NOT a one-to-one function and so does NOT HAVE an inverse over all real numbers. You can, of course, restrict the domain so that it does have an inverse. How the domain is restricted will determine the sign.
<MASK>
### mahmoud2011
<MASK>
<UNMASK>
<MASK>
1. Aug 29, 2011
### mahmoud2011
<MASK>
$sech^{-1}(x) = ln(\frac{1\pm \sqrt{1-x^{2}}}{x}) : 0<x\leq 1$
And this not function which of them I will choose
Another Question is how can I prove without the graph that csch (x) is one - to -one
thanks
<MASK>
### mahmoud2011
<MASK>
3. Aug 29, 2011
### mahmoud2011
For the first question , I tried to check my solution I tried to check the solution directly from the definition of inverse function,
<MASK>
$f(x) = sech(x) : x\geq 0$
<MASK>
$(f o g)(x) = x : 0<x\leq 1$
Whereas $(g o f)(x) = x : x\leq0$
<MASK>
Exactly what does the question say? The reason you are getting "$\pm$" is because sech is NOT a one-to-one function and so does NOT HAVE an inverse over all real numbers. You can, of course, restrict the domain so that it does have an inverse. How the domain is restricted will determine the sign.
<MASK>
### mahmoud2011
I have restricted the domain such that x is greater than or equal zero and I also had Both signs So I Checked the solutions from definition of Inverse function And Choosed the positive sign As I have written above.
|
<MASK>
1. Aug 29, 2011
### mahmoud2011
<MASK>
$sech^{-1}(x) = ln(\frac{1\pm \sqrt{1-x^{2}}}{x}) : 0<x\leq 1$
And this not function which of them I will choose
Another Question is how can I prove without the graph that csch (x) is one - to -one
thanks
<MASK>
### mahmoud2011
<MASK>
3. Aug 29, 2011
### mahmoud2011
For the first question , I tried to check my solution I tried to check the solution directly from the definition of inverse function,
<MASK>
$f(x) = sech(x) : x\geq 0$
<MASK>
$(f o g)(x) = x : 0<x\leq 1$
Whereas $(g o f)(x) = x : x\leq0$
<MASK>
Exactly what does the question say? The reason you are getting "$\pm$" is because sech is NOT a one-to-one function and so does NOT HAVE an inverse over all real numbers. You can, of course, restrict the domain so that it does have an inverse. How the domain is restricted will determine the sign.
<MASK>
### mahmoud2011
I have restricted the domain such that x is greater than or equal zero and I also had Both signs So I Checked the solutions from definition of Inverse function And Choosed the positive sign As I have written above.
<UNMASK>
# Homework Help: Finding Inverse Hyperbolic secant in terms of logarithms ?
1. Aug 29, 2011
### mahmoud2011
The Problem is when I Compute the Inverse I have to solutions
$sech^{-1}(x) = ln(\frac{1\pm \sqrt{1-x^{2}}}{x}) : 0<x\leq 1$
And this not function which of them I will choose
Another Question is how can I prove without the graph that csch (x) is one - to -one
thanks
Last edited: Aug 29, 2011
2. Aug 29, 2011
### mahmoud2011
For the second question I proved it depending on the definition of one to one function and csch(x) ,where since sinh(x) is one to one ,the whenever sinh(c)=sinh(d) , c=d
Then it follows that whenever 1/sinh(c) = 1/sinh(d) , c=d
And thus the result follows and that's what I did .
3. Aug 29, 2011
### mahmoud2011
For the first question , I tried to check my solution I tried to check the solution directly from the definition of inverse function,
Let $h(x) = ln(\frac{1-\sqrt{1-x^2}}{x}) : 0<x\leq 1$
$g(x)=ln (\frac{1+ \sqrt{1-x^2}}{x}) : 0<x\leq 1$
$f(x) = sech(x) : x\geq 0$
then we have after some Algebra ,
$(f o h)(x) = x : 0<x\leq 1$
Whereas $(h o f)(x) = -x : x\leq0$
And Thus h is not the inverse of f , Checking g we will see that it do its job it is the inverse where
$(f o g)(x) = x : 0<x\leq 1$
Whereas $(g o f)(x) = x : x\leq0$
4. Aug 30, 2011
### HallsofIvy
Exactly what does the question say? The reason you are getting "$\pm$" is because sech is NOT a one-to-one function and so does NOT HAVE an inverse over all real numbers. You can, of course, restrict the domain so that it does have an inverse. How the domain is restricted will determine the sign.
5. Aug 30, 2011
### mahmoud2011
I have restricted the domain such that x is greater than or equal zero and I also had Both signs So I Checked the solutions from definition of Inverse function And Choosed the positive sign As I have written above.
|
<MASK>
We can see a ray of light is incident on this surface and another ray which is parallel to this ray is also incident on this surface. Plane AB is incident at an angle ‘ i ‘ on the reflecting surface MN. As these rays are incident from the surface, so we call it incident ray. If we draw a perpendicular from point ‘A’ to this ray of light, Point A, and point B will have a line joining them and this is called as wavefront and this wavefront is incident on the surface.
<MASK>
### Refraction using Huygen’s principle
<MASK>
## FAQs on Refraction & Reflection of Plane Waves using Huygens Principle - Physics Class 12 - NEET
<MASK>
### Top Courses for NEET
<MASK>
<UNMASK>
Refraction & Reflection of Plane Waves using Huygens Principle
# Refraction & Reflection of Plane Waves using Huygens Principle | Physics Class 12 - NEET PDF Download
<MASK>
We can see a ray of light is incident on this surface and another ray which is parallel to this ray is also incident on this surface. Plane AB is incident at an angle ‘ i ‘ on the reflecting surface MN. As these rays are incident from the surface, so we call it incident ray. If we draw a perpendicular from point ‘A’ to this ray of light, Point A, and point B will have a line joining them and this is called as wavefront and this wavefront is incident on the surface.
<MASK>
BC = vr
<MASK>
AE = BC = vr
<MASK>
### Refraction using Huygen’s principle
<MASK>
A plane wave AB is incident at an angle i on the surface PP' separating medium 1 and medium 2. The plane wave undergoes refraction and CE represents the refracted
<MASK>
BC = v1r
So to determine the shape of the refracted wavefront, we draw a sphere of radius v2r from the point A in the second medium. Let CE represent a tangent plane drawn from the point C on to the sphere. Then, AE = v2r, and CE would represent the refracted wavefront. If we now consider the triangles ABC and AEC, we readily obtain
<MASK>
n1 sin i = n2 sin r
The document Refraction & Reflection of Plane Waves using Huygens Principle | Physics Class 12 - NEET is a part of the NEET Course Physics Class 12.
All you need of NEET at this link: NEET
<MASK>
105 videos|425 docs|114 tests
## FAQs on Refraction & Reflection of Plane Waves using Huygens Principle - Physics Class 12 - NEET
1. What is refraction and reflection of plane waves?
Ans. Refraction is the bending of a wave as it passes from one medium to another, while reflection is the bouncing back of a wave after striking a surface. In the context of plane waves, refraction and reflection refer to the behavior of the wavefronts as they encounter different mediums or surfaces.
2. How does Huygens Principle explain refraction and reflection of plane waves?
Ans. According to Huygens Principle, every point on a wavefront acts as a source of secondary spherical wavelets. These secondary wavelets combine to form the new wavefront. In the case of refraction, the change in speed of the wavefront as it enters a new medium causes the wavefront to change direction. In reflection, the wavefront encounters a surface and the secondary wavelets bounce back, resulting in the reflection of the wave.
3. Can you explain how refraction and reflection affect the direction of plane waves?
Ans. Refraction causes a change in the direction of plane waves when they pass from one medium to another. This change occurs due to the change in speed of the wavefront in different mediums. The angle at which the wavefront enters the new medium, known as the angle of incidence, determines the angle at which it is refracted, known as the angle of refraction. Reflection, on the other hand, causes the wavefront to bounce back in the opposite direction when it encounters a surface.
4. What factors determine the extent of refraction and reflection in plane waves?
Ans. The extent of refraction depends on the change in speed and the angle of incidence of the wavefront. The greater the difference in speed between the two mediums, the greater the change in direction of the wavefront. Additionally, the angle of incidence plays a role in determining the angle of refraction. The extent of reflection depends on the nature of the surface encountered by the wavefront, as well as the angle of incidence. Smooth surfaces tend to have more reflection compared to rough surfaces.
5. How are refraction and reflection of plane waves useful in practical applications?
Ans. Refraction and reflection of plane waves have numerous practical applications. For example, in optics, lenses are designed based on the principles of refraction to focus or diverge light. In telecommunications, the reflection of radio waves off the ionosphere allows for long-distance communication. Refraction is also used in the field of medicine for techniques such as ultrasound imaging and laser eye surgery.
## Physics Class 12
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Refraction & Reflection of Plane Waves using Huygens Principle
# Refraction & Reflection of Plane Waves using Huygens Principle | Physics Class 12 - NEET PDF Download
<MASK>
We can see a ray of light is incident on this surface and another ray which is parallel to this ray is also incident on this surface. Plane AB is incident at an angle ‘ i ‘ on the reflecting surface MN. As these rays are incident from the surface, so we call it incident ray. If we draw a perpendicular from point ‘A’ to this ray of light, Point A, and point B will have a line joining them and this is called as wavefront and this wavefront is incident on the surface.
<MASK>
BC = vr
<MASK>
AE = BC = vr
<MASK>
### Refraction using Huygen’s principle
<MASK>
A plane wave AB is incident at an angle i on the surface PP' separating medium 1 and medium 2. The plane wave undergoes refraction and CE represents the refracted
<MASK>
BC = v1r
So to determine the shape of the refracted wavefront, we draw a sphere of radius v2r from the point A in the second medium. Let CE represent a tangent plane drawn from the point C on to the sphere. Then, AE = v2r, and CE would represent the refracted wavefront. If we now consider the triangles ABC and AEC, we readily obtain
<MASK>
n1 sin i = n2 sin r
The document Refraction & Reflection of Plane Waves using Huygens Principle | Physics Class 12 - NEET is a part of the NEET Course Physics Class 12.
All you need of NEET at this link: NEET
<MASK>
105 videos|425 docs|114 tests
## FAQs on Refraction & Reflection of Plane Waves using Huygens Principle - Physics Class 12 - NEET
1. What is refraction and reflection of plane waves?
Ans. Refraction is the bending of a wave as it passes from one medium to another, while reflection is the bouncing back of a wave after striking a surface. In the context of plane waves, refraction and reflection refer to the behavior of the wavefronts as they encounter different mediums or surfaces.
2. How does Huygens Principle explain refraction and reflection of plane waves?
Ans. According to Huygens Principle, every point on a wavefront acts as a source of secondary spherical wavelets. These secondary wavelets combine to form the new wavefront. In the case of refraction, the change in speed of the wavefront as it enters a new medium causes the wavefront to change direction. In reflection, the wavefront encounters a surface and the secondary wavelets bounce back, resulting in the reflection of the wave.
3. Can you explain how refraction and reflection affect the direction of plane waves?
Ans. Refraction causes a change in the direction of plane waves when they pass from one medium to another. This change occurs due to the change in speed of the wavefront in different mediums. The angle at which the wavefront enters the new medium, known as the angle of incidence, determines the angle at which it is refracted, known as the angle of refraction. Reflection, on the other hand, causes the wavefront to bounce back in the opposite direction when it encounters a surface.
4. What factors determine the extent of refraction and reflection in plane waves?
Ans. The extent of refraction depends on the change in speed and the angle of incidence of the wavefront. The greater the difference in speed between the two mediums, the greater the change in direction of the wavefront. Additionally, the angle of incidence plays a role in determining the angle of refraction. The extent of reflection depends on the nature of the surface encountered by the wavefront, as well as the angle of incidence. Smooth surfaces tend to have more reflection compared to rough surfaces.
5. How are refraction and reflection of plane waves useful in practical applications?
Ans. Refraction and reflection of plane waves have numerous practical applications. For example, in optics, lenses are designed based on the principles of refraction to focus or diverge light. In telecommunications, the reflection of radio waves off the ionosphere allows for long-distance communication. Refraction is also used in the field of medicine for techniques such as ultrasound imaging and laser eye surgery.
## Physics Class 12
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Refraction & Reflection of Plane Waves using Huygens Principle
# Refraction & Reflection of Plane Waves using Huygens Principle | Physics Class 12 - NEET PDF Download
<MASK>
We can see a ray of light is incident on this surface and another ray which is parallel to this ray is also incident on this surface. Plane AB is incident at an angle ‘ i ‘ on the reflecting surface MN. As these rays are incident from the surface, so we call it incident ray. If we draw a perpendicular from point ‘A’ to this ray of light, Point A, and point B will have a line joining them and this is called as wavefront and this wavefront is incident on the surface.
<MASK>
BC = vr
<MASK>
AE = BC = vr
If we now consider the triangles EAC and BAC we will find that they are congruent and therefore, the angles ‘ i ‘ and ‘r ‘ would be equal. This is the law of reflection
### Refraction using Huygen’s principle
<MASK>
A plane wave AB is incident at an angle i on the surface PP' separating medium 1 and medium 2. The plane wave undergoes refraction and CE represents the refracted
<MASK>
BC = v1r
So to determine the shape of the refracted wavefront, we draw a sphere of radius v2r from the point A in the second medium. Let CE represent a tangent plane drawn from the point C on to the sphere. Then, AE = v2r, and CE would represent the refracted wavefront. If we now consider the triangles ABC and AEC, we readily obtain
<MASK>
n1 sin i = n2 sin r
The document Refraction & Reflection of Plane Waves using Huygens Principle | Physics Class 12 - NEET is a part of the NEET Course Physics Class 12.
All you need of NEET at this link: NEET
<MASK>
105 videos|425 docs|114 tests
## FAQs on Refraction & Reflection of Plane Waves using Huygens Principle - Physics Class 12 - NEET
1. What is refraction and reflection of plane waves?
Ans. Refraction is the bending of a wave as it passes from one medium to another, while reflection is the bouncing back of a wave after striking a surface. In the context of plane waves, refraction and reflection refer to the behavior of the wavefronts as they encounter different mediums or surfaces.
2. How does Huygens Principle explain refraction and reflection of plane waves?
Ans. According to Huygens Principle, every point on a wavefront acts as a source of secondary spherical wavelets. These secondary wavelets combine to form the new wavefront. In the case of refraction, the change in speed of the wavefront as it enters a new medium causes the wavefront to change direction. In reflection, the wavefront encounters a surface and the secondary wavelets bounce back, resulting in the reflection of the wave.
3. Can you explain how refraction and reflection affect the direction of plane waves?
Ans. Refraction causes a change in the direction of plane waves when they pass from one medium to another. This change occurs due to the change in speed of the wavefront in different mediums. The angle at which the wavefront enters the new medium, known as the angle of incidence, determines the angle at which it is refracted, known as the angle of refraction. Reflection, on the other hand, causes the wavefront to bounce back in the opposite direction when it encounters a surface.
4. What factors determine the extent of refraction and reflection in plane waves?
Ans. The extent of refraction depends on the change in speed and the angle of incidence of the wavefront. The greater the difference in speed between the two mediums, the greater the change in direction of the wavefront. Additionally, the angle of incidence plays a role in determining the angle of refraction. The extent of reflection depends on the nature of the surface encountered by the wavefront, as well as the angle of incidence. Smooth surfaces tend to have more reflection compared to rough surfaces.
5. How are refraction and reflection of plane waves useful in practical applications?
Ans. Refraction and reflection of plane waves have numerous practical applications. For example, in optics, lenses are designed based on the principles of refraction to focus or diverge light. In telecommunications, the reflection of radio waves off the ionosphere allows for long-distance communication. Refraction is also used in the field of medicine for techniques such as ultrasound imaging and laser eye surgery.
## Physics Class 12
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Refraction & Reflection of Plane Waves using Huygens Principle
# Refraction & Reflection of Plane Waves using Huygens Principle | Physics Class 12 - NEET PDF Download
<MASK>
We can see a ray of light is incident on this surface and another ray which is parallel to this ray is also incident on this surface. Plane AB is incident at an angle ‘ i ‘ on the reflecting surface MN. As these rays are incident from the surface, so we call it incident ray. If we draw a perpendicular from point ‘A’ to this ray of light, Point A, and point B will have a line joining them and this is called as wavefront and this wavefront is incident on the surface.
<MASK>
BC = vr
<MASK>
AE = BC = vr
If we now consider the triangles EAC and BAC we will find that they are congruent and therefore, the angles ‘ i ‘ and ‘r ‘ would be equal. This is the law of reflection
### Refraction using Huygen’s principle
<MASK>
A plane wave AB is incident at an angle i on the surface PP' separating medium 1 and medium 2. The plane wave undergoes refraction and CE represents the refracted
<MASK>
BC = v1r
So to determine the shape of the refracted wavefront, we draw a sphere of radius v2r from the point A in the second medium. Let CE represent a tangent plane drawn from the point C on to the sphere. Then, AE = v2r, and CE would represent the refracted wavefront. If we now consider the triangles ABC and AEC, we readily obtain
<MASK>
n1 sin i = n2 sin r
The document Refraction & Reflection of Plane Waves using Huygens Principle | Physics Class 12 - NEET is a part of the NEET Course Physics Class 12.
All you need of NEET at this link: NEET
<MASK>
105 videos|425 docs|114 tests
## FAQs on Refraction & Reflection of Plane Waves using Huygens Principle - Physics Class 12 - NEET
1. What is refraction and reflection of plane waves?
Ans. Refraction is the bending of a wave as it passes from one medium to another, while reflection is the bouncing back of a wave after striking a surface. In the context of plane waves, refraction and reflection refer to the behavior of the wavefronts as they encounter different mediums or surfaces.
2. How does Huygens Principle explain refraction and reflection of plane waves?
Ans. According to Huygens Principle, every point on a wavefront acts as a source of secondary spherical wavelets. These secondary wavelets combine to form the new wavefront. In the case of refraction, the change in speed of the wavefront as it enters a new medium causes the wavefront to change direction. In reflection, the wavefront encounters a surface and the secondary wavelets bounce back, resulting in the reflection of the wave.
3. Can you explain how refraction and reflection affect the direction of plane waves?
Ans. Refraction causes a change in the direction of plane waves when they pass from one medium to another. This change occurs due to the change in speed of the wavefront in different mediums. The angle at which the wavefront enters the new medium, known as the angle of incidence, determines the angle at which it is refracted, known as the angle of refraction. Reflection, on the other hand, causes the wavefront to bounce back in the opposite direction when it encounters a surface.
4. What factors determine the extent of refraction and reflection in plane waves?
Ans. The extent of refraction depends on the change in speed and the angle of incidence of the wavefront. The greater the difference in speed between the two mediums, the greater the change in direction of the wavefront. Additionally, the angle of incidence plays a role in determining the angle of refraction. The extent of reflection depends on the nature of the surface encountered by the wavefront, as well as the angle of incidence. Smooth surfaces tend to have more reflection compared to rough surfaces.
5. How are refraction and reflection of plane waves useful in practical applications?
Ans. Refraction and reflection of plane waves have numerous practical applications. For example, in optics, lenses are designed based on the principles of refraction to focus or diverge light. In telecommunications, the reflection of radio waves off the ionosphere allows for long-distance communication. Refraction is also used in the field of medicine for techniques such as ultrasound imaging and laser eye surgery.
## Physics Class 12
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Refraction & Reflection of Plane Waves using Huygens Principle
# Refraction & Reflection of Plane Waves using Huygens Principle | Physics Class 12 - NEET PDF Download
As we know that when light falls on an object, it bends and move through the material, this is what refraction is. Also when the light bounces off the medium it is called a reflection. Let us know study reflection and refraction of waves by Huygen’s principle.
<MASK>
We can see a ray of light is incident on this surface and another ray which is parallel to this ray is also incident on this surface. Plane AB is incident at an angle ‘ i ‘ on the reflecting surface MN. As these rays are incident from the surface, so we call it incident ray. If we draw a perpendicular from point ‘A’ to this ray of light, Point A, and point B will have a line joining them and this is called as wavefront and this wavefront is incident on the surface.
<MASK>
BC = vr
In order the construct the reflected wavefront we draw a sphere of radius vr from the point A. Let CE represent the tangent plane drawn from the point C to this sphere. So,
AE = BC = vr
If we now consider the triangles EAC and BAC we will find that they are congruent and therefore, the angles ‘ i ‘ and ‘r ‘ would be equal. This is the law of reflection
### Refraction using Huygen’s principle
We know that when a light travels from one transparent medium to another transparent medium its path changes. So the laws of refraction state that the angle of incidence is the angle between the incident ray and the normal and the angle of refraction is the angle between the refracted ray and the normal.
The incident ray, reflected ray and the normal, to the interface of any two given mediums all lie in the same plane. We also know that the ratio of the sine of the angle of incidence and sine of the angle of refraction is constant.
A plane wave AB is incident at an angle i on the surface PP' separating medium 1 and medium 2. The plane wave undergoes refraction and CE represents the refracted
<MASK>
We can see a ray of light is incident on this surface and another ray which is parallel to this ray is also incident on this surface. As these rays are incident from the surface, so we call it incident ray.
Let PP’ represent the medium 1 and medium 2. The speed of the light in this medium is represented by v1 and v2. If we draw a perpendicular from point ‘A’ to this ray of light, Point A, and point B will have a line joining them and this is called as wavefront and this wavefront is incident on the surface.
If ‘ r ‘ represents the time taken by the wavefront from the point B to C then the distance,
BC = v1r
So to determine the shape of the refracted wavefront, we draw a sphere of radius v2r from the point A in the second medium. Let CE represent a tangent plane drawn from the point C on to the sphere. Then, AE = v2r, and CE would represent the refracted wavefront. If we now consider the triangles ABC and AEC, we readily obtain
where’ i ‘ and ‘ r ‘ are the angles of incidence and refraction, respectively. Substituting the values of v1 and v2 in terms of we get the Snell’s Law,
n1 sin i = n2 sin r
The document Refraction & Reflection of Plane Waves using Huygens Principle | Physics Class 12 - NEET is a part of the NEET Course Physics Class 12.
All you need of NEET at this link: NEET
<MASK>
105 videos|425 docs|114 tests
## FAQs on Refraction & Reflection of Plane Waves using Huygens Principle - Physics Class 12 - NEET
1. What is refraction and reflection of plane waves?
Ans. Refraction is the bending of a wave as it passes from one medium to another, while reflection is the bouncing back of a wave after striking a surface. In the context of plane waves, refraction and reflection refer to the behavior of the wavefronts as they encounter different mediums or surfaces.
2. How does Huygens Principle explain refraction and reflection of plane waves?
Ans. According to Huygens Principle, every point on a wavefront acts as a source of secondary spherical wavelets. These secondary wavelets combine to form the new wavefront. In the case of refraction, the change in speed of the wavefront as it enters a new medium causes the wavefront to change direction. In reflection, the wavefront encounters a surface and the secondary wavelets bounce back, resulting in the reflection of the wave.
3. Can you explain how refraction and reflection affect the direction of plane waves?
Ans. Refraction causes a change in the direction of plane waves when they pass from one medium to another. This change occurs due to the change in speed of the wavefront in different mediums. The angle at which the wavefront enters the new medium, known as the angle of incidence, determines the angle at which it is refracted, known as the angle of refraction. Reflection, on the other hand, causes the wavefront to bounce back in the opposite direction when it encounters a surface.
4. What factors determine the extent of refraction and reflection in plane waves?
Ans. The extent of refraction depends on the change in speed and the angle of incidence of the wavefront. The greater the difference in speed between the two mediums, the greater the change in direction of the wavefront. Additionally, the angle of incidence plays a role in determining the angle of refraction. The extent of reflection depends on the nature of the surface encountered by the wavefront, as well as the angle of incidence. Smooth surfaces tend to have more reflection compared to rough surfaces.
5. How are refraction and reflection of plane waves useful in practical applications?
Ans. Refraction and reflection of plane waves have numerous practical applications. For example, in optics, lenses are designed based on the principles of refraction to focus or diverge light. In telecommunications, the reflection of radio waves off the ionosphere allows for long-distance communication. Refraction is also used in the field of medicine for techniques such as ultrasound imaging and laser eye surgery.
## Physics Class 12
105 videos|425 docs|114 tests
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<MASK>
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;
|
Refraction & Reflection of Plane Waves using Huygens Principle
# Refraction & Reflection of Plane Waves using Huygens Principle | Physics Class 12 - NEET PDF Download
As we know that when light falls on an object, it bends and move through the material, this is what refraction is. Also when the light bounces off the medium it is called a reflection. Let us know study reflection and refraction of waves by Huygen’s principle.
<MASK>
We can see a ray of light is incident on this surface and another ray which is parallel to this ray is also incident on this surface. Plane AB is incident at an angle ‘ i ‘ on the reflecting surface MN. As these rays are incident from the surface, so we call it incident ray. If we draw a perpendicular from point ‘A’ to this ray of light, Point A, and point B will have a line joining them and this is called as wavefront and this wavefront is incident on the surface.
<MASK>
BC = vr
In order the construct the reflected wavefront we draw a sphere of radius vr from the point A. Let CE represent the tangent plane drawn from the point C to this sphere. So,
AE = BC = vr
If we now consider the triangles EAC and BAC we will find that they are congruent and therefore, the angles ‘ i ‘ and ‘r ‘ would be equal. This is the law of reflection
### Refraction using Huygen’s principle
We know that when a light travels from one transparent medium to another transparent medium its path changes. So the laws of refraction state that the angle of incidence is the angle between the incident ray and the normal and the angle of refraction is the angle between the refracted ray and the normal.
The incident ray, reflected ray and the normal, to the interface of any two given mediums all lie in the same plane. We also know that the ratio of the sine of the angle of incidence and sine of the angle of refraction is constant.
A plane wave AB is incident at an angle i on the surface PP' separating medium 1 and medium 2. The plane wave undergoes refraction and CE represents the refracted
<MASK>
We can see a ray of light is incident on this surface and another ray which is parallel to this ray is also incident on this surface. As these rays are incident from the surface, so we call it incident ray.
Let PP’ represent the medium 1 and medium 2. The speed of the light in this medium is represented by v1 and v2. If we draw a perpendicular from point ‘A’ to this ray of light, Point A, and point B will have a line joining them and this is called as wavefront and this wavefront is incident on the surface.
If ‘ r ‘ represents the time taken by the wavefront from the point B to C then the distance,
BC = v1r
So to determine the shape of the refracted wavefront, we draw a sphere of radius v2r from the point A in the second medium. Let CE represent a tangent plane drawn from the point C on to the sphere. Then, AE = v2r, and CE would represent the refracted wavefront. If we now consider the triangles ABC and AEC, we readily obtain
where’ i ‘ and ‘ r ‘ are the angles of incidence and refraction, respectively. Substituting the values of v1 and v2 in terms of we get the Snell’s Law,
n1 sin i = n2 sin r
The document Refraction & Reflection of Plane Waves using Huygens Principle | Physics Class 12 - NEET is a part of the NEET Course Physics Class 12.
All you need of NEET at this link: NEET
<MASK>
105 videos|425 docs|114 tests
## FAQs on Refraction & Reflection of Plane Waves using Huygens Principle - Physics Class 12 - NEET
1. What is refraction and reflection of plane waves?
Ans. Refraction is the bending of a wave as it passes from one medium to another, while reflection is the bouncing back of a wave after striking a surface. In the context of plane waves, refraction and reflection refer to the behavior of the wavefronts as they encounter different mediums or surfaces.
2. How does Huygens Principle explain refraction and reflection of plane waves?
Ans. According to Huygens Principle, every point on a wavefront acts as a source of secondary spherical wavelets. These secondary wavelets combine to form the new wavefront. In the case of refraction, the change in speed of the wavefront as it enters a new medium causes the wavefront to change direction. In reflection, the wavefront encounters a surface and the secondary wavelets bounce back, resulting in the reflection of the wave.
3. Can you explain how refraction and reflection affect the direction of plane waves?
Ans. Refraction causes a change in the direction of plane waves when they pass from one medium to another. This change occurs due to the change in speed of the wavefront in different mediums. The angle at which the wavefront enters the new medium, known as the angle of incidence, determines the angle at which it is refracted, known as the angle of refraction. Reflection, on the other hand, causes the wavefront to bounce back in the opposite direction when it encounters a surface.
4. What factors determine the extent of refraction and reflection in plane waves?
Ans. The extent of refraction depends on the change in speed and the angle of incidence of the wavefront. The greater the difference in speed between the two mediums, the greater the change in direction of the wavefront. Additionally, the angle of incidence plays a role in determining the angle of refraction. The extent of reflection depends on the nature of the surface encountered by the wavefront, as well as the angle of incidence. Smooth surfaces tend to have more reflection compared to rough surfaces.
5. How are refraction and reflection of plane waves useful in practical applications?
Ans. Refraction and reflection of plane waves have numerous practical applications. For example, in optics, lenses are designed based on the principles of refraction to focus or diverge light. In telecommunications, the reflection of radio waves off the ionosphere allows for long-distance communication. Refraction is also used in the field of medicine for techniques such as ultrasound imaging and laser eye surgery.
## Physics Class 12
105 videos|425 docs|114 tests
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<UNMASK>
Refraction & Reflection of Plane Waves using Huygens Principle
# Refraction & Reflection of Plane Waves using Huygens Principle | Physics Class 12 - NEET PDF Download
As we know that when light falls on an object, it bends and move through the material, this is what refraction is. Also when the light bounces off the medium it is called a reflection. Let us know study reflection and refraction of waves by Huygen’s principle.
Reflection using Huygens Principle
We can see a ray of light is incident on this surface and another ray which is parallel to this ray is also incident on this surface. Plane AB is incident at an angle ‘ i ‘ on the reflecting surface MN. As these rays are incident from the surface, so we call it incident ray. If we draw a perpendicular from point ‘A’ to this ray of light, Point A, and point B will have a line joining them and this is called as wavefront and this wavefront is incident on the surface.
These incident wavefront is carrying two points, point A and point B, so we can say that from point B to point C light is travelling a distance. If ‘ v ‘ represents the speed of the wave in the medium and if ‘ r ‘ represents the time taken by the wavefront from the point B to C then the distance
BC = vr
In order the construct the reflected wavefront we draw a sphere of radius vr from the point A. Let CE represent the tangent plane drawn from the point C to this sphere. So,
AE = BC = vr
If we now consider the triangles EAC and BAC we will find that they are congruent and therefore, the angles ‘ i ‘ and ‘r ‘ would be equal. This is the law of reflection
### Refraction using Huygen’s principle
We know that when a light travels from one transparent medium to another transparent medium its path changes. So the laws of refraction state that the angle of incidence is the angle between the incident ray and the normal and the angle of refraction is the angle between the refracted ray and the normal.
The incident ray, reflected ray and the normal, to the interface of any two given mediums all lie in the same plane. We also know that the ratio of the sine of the angle of incidence and sine of the angle of refraction is constant.
A plane wave AB is incident at an angle i on the surface PP' separating medium 1 and medium 2. The plane wave undergoes refraction and CE represents the refracted
wavefront. the figure corresponds to v2 < v1 so that the refracted waves bends towards the normal.
We can see a ray of light is incident on this surface and another ray which is parallel to this ray is also incident on this surface. As these rays are incident from the surface, so we call it incident ray.
Let PP’ represent the medium 1 and medium 2. The speed of the light in this medium is represented by v1 and v2. If we draw a perpendicular from point ‘A’ to this ray of light, Point A, and point B will have a line joining them and this is called as wavefront and this wavefront is incident on the surface.
If ‘ r ‘ represents the time taken by the wavefront from the point B to C then the distance,
BC = v1r
So to determine the shape of the refracted wavefront, we draw a sphere of radius v2r from the point A in the second medium. Let CE represent a tangent plane drawn from the point C on to the sphere. Then, AE = v2r, and CE would represent the refracted wavefront. If we now consider the triangles ABC and AEC, we readily obtain
where’ i ‘ and ‘ r ‘ are the angles of incidence and refraction, respectively. Substituting the values of v1 and v2 in terms of we get the Snell’s Law,
n1 sin i = n2 sin r
The document Refraction & Reflection of Plane Waves using Huygens Principle | Physics Class 12 - NEET is a part of the NEET Course Physics Class 12.
All you need of NEET at this link: NEET
## Physics Class 12
105 videos|425 docs|114 tests
## FAQs on Refraction & Reflection of Plane Waves using Huygens Principle - Physics Class 12 - NEET
1. What is refraction and reflection of plane waves?
Ans. Refraction is the bending of a wave as it passes from one medium to another, while reflection is the bouncing back of a wave after striking a surface. In the context of plane waves, refraction and reflection refer to the behavior of the wavefronts as they encounter different mediums or surfaces.
2. How does Huygens Principle explain refraction and reflection of plane waves?
Ans. According to Huygens Principle, every point on a wavefront acts as a source of secondary spherical wavelets. These secondary wavelets combine to form the new wavefront. In the case of refraction, the change in speed of the wavefront as it enters a new medium causes the wavefront to change direction. In reflection, the wavefront encounters a surface and the secondary wavelets bounce back, resulting in the reflection of the wave.
3. Can you explain how refraction and reflection affect the direction of plane waves?
Ans. Refraction causes a change in the direction of plane waves when they pass from one medium to another. This change occurs due to the change in speed of the wavefront in different mediums. The angle at which the wavefront enters the new medium, known as the angle of incidence, determines the angle at which it is refracted, known as the angle of refraction. Reflection, on the other hand, causes the wavefront to bounce back in the opposite direction when it encounters a surface.
4. What factors determine the extent of refraction and reflection in plane waves?
Ans. The extent of refraction depends on the change in speed and the angle of incidence of the wavefront. The greater the difference in speed between the two mediums, the greater the change in direction of the wavefront. Additionally, the angle of incidence plays a role in determining the angle of refraction. The extent of reflection depends on the nature of the surface encountered by the wavefront, as well as the angle of incidence. Smooth surfaces tend to have more reflection compared to rough surfaces.
5. How are refraction and reflection of plane waves useful in practical applications?
Ans. Refraction and reflection of plane waves have numerous practical applications. For example, in optics, lenses are designed based on the principles of refraction to focus or diverge light. In telecommunications, the reflection of radio waves off the ionosphere allows for long-distance communication. Refraction is also used in the field of medicine for techniques such as ultrasound imaging and laser eye surgery.
## Physics Class 12
105 videos|425 docs|114 tests
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<MASK>
## Recommended: Please try your approach on {IDE} first, before moving on to the solution.
<MASK>
## Python3
<MASK>
<UNMASK>
<MASK>
Input : N = 2, M =2
m[][] = { { 1, 2 },
{ 3, 4 } };
Output :3
Longest path is either 1 2 4 or
1 3 4.
```
## Recommended: Please try your approach on {IDE} first, before moving on to the solution.
The idea is to use dynamic programming. Maintain the 2D matrix, dp[][], where dp[i][j] store the value of length of longest increasing sequence for sub matrix starting from ith row and j-th column.
Let the longest increasing sub sequence values for m[i+1][j] and m[i][j+1] be known already as v1 and v2 respectively. Then the value for m[i][j] will be max(v1, v2) + 1.
We can start from m[n-1][m-1] as base case with length of longest increasing sub sequence be 1, moving upwards and leftwards updating the value of cells. Then the LIP value for cell m[0][0] will be the answer.
<MASK>
## C++
<MASK>
## Java
<MASK>
## Python3
`# Python3 program to find longest ` `# increasing path in a matrix. ` `MAX` `=` `20` ` ` `# Return the length of ` `# LIP in 2D matrix ` `def` `LIP(dp, mat, n, m, x, y): ` ` ` ` ``# If value not calculated yet. ` ` ``if` `(dp[x][y] < ``0``): ` ` ``result ``=` `0` ` ` ` ``# If reach bottom left cell, ` ` ``# return 1. ` ` ``if` `(x ``=``=` `n ``-` `1` `and` `y ``=``=` `m ``-` `1``): ` ` ``dp[x][y] ``=` `1` ` ``return` `dp[x][y] ` ` ` ` ``# If reach the corner ` ` ``# of the matrix. ` ` ``if` `(x ``=``=` `n ``-` `1` `or` `y ``=``=` `m ``-` `1``): ` ` ``result ``=` `1` ` ` ` ``# If value greater than below cell. ` ` ``elif` `(mat[x][y] < mat[x ``+` `1``][y]): ` ` ``result ``=` `1` `+` `LIP(dp, mat, n, ` ` ``m, x ``+` `1``, y) ` ` ` ` ``# If value greater than left cell. ` ` ``elif` `(mat[x][y] < mat[x][y ``+` `1``]): ` ` ``result ``=` `max``(result, ``1` `+` `LIP(dp, mat, n, ` ` ``m, x, y ``+` `1``)) ` ` ``dp[x][y] ``=` `result ` ` ``return` `dp[x][y] ` ` ` `# Wrapper function ` `def` `wrapper(mat, n, m): ` ` ``dp ``=` `[[``7` `for` `i ``in` `range``(``MAX``)] ` ` ``for` `i ``in` `range``(``MAX``)] ` ` ``return` `LIP(dp, mat, n, m, ``0``, ``0``) ` ` ` `# Driver Code ` `mat ``=` `[[``1``, ``2``, ``3``, ``4` `], ` ` ``[``2``, ``2``, ``3``, ``4` `], ` ` ``[``3``, ``2``, ``3``, ``4` `], ` ` ``[``4``, ``5``, ``6``, ``7` `]] ` `n ``=` `4` `m ``=` `4` `print``(wrapper(mat, n, m)) ` ` ` `# This code is contributed ` `# by Sahil Shelangia `
<MASK>
|
<MASK>
Input : N = 2, M =2
m[][] = { { 1, 2 },
{ 3, 4 } };
Output :3
Longest path is either 1 2 4 or
1 3 4.
```
## Recommended: Please try your approach on {IDE} first, before moving on to the solution.
The idea is to use dynamic programming. Maintain the 2D matrix, dp[][], where dp[i][j] store the value of length of longest increasing sequence for sub matrix starting from ith row and j-th column.
Let the longest increasing sub sequence values for m[i+1][j] and m[i][j+1] be known already as v1 and v2 respectively. Then the value for m[i][j] will be max(v1, v2) + 1.
We can start from m[n-1][m-1] as base case with length of longest increasing sub sequence be 1, moving upwards and leftwards updating the value of cells. Then the LIP value for cell m[0][0] will be the answer.
<MASK>
## C++
<MASK>
## Java
<MASK>
## Python3
`# Python3 program to find longest ` `# increasing path in a matrix. ` `MAX` `=` `20` ` ` `# Return the length of ` `# LIP in 2D matrix ` `def` `LIP(dp, mat, n, m, x, y): ` ` ` ` ``# If value not calculated yet. ` ` ``if` `(dp[x][y] < ``0``): ` ` ``result ``=` `0` ` ` ` ``# If reach bottom left cell, ` ` ``# return 1. ` ` ``if` `(x ``=``=` `n ``-` `1` `and` `y ``=``=` `m ``-` `1``): ` ` ``dp[x][y] ``=` `1` ` ``return` `dp[x][y] ` ` ` ` ``# If reach the corner ` ` ``# of the matrix. ` ` ``if` `(x ``=``=` `n ``-` `1` `or` `y ``=``=` `m ``-` `1``): ` ` ``result ``=` `1` ` ` ` ``# If value greater than below cell. ` ` ``elif` `(mat[x][y] < mat[x ``+` `1``][y]): ` ` ``result ``=` `1` `+` `LIP(dp, mat, n, ` ` ``m, x ``+` `1``, y) ` ` ` ` ``# If value greater than left cell. ` ` ``elif` `(mat[x][y] < mat[x][y ``+` `1``]): ` ` ``result ``=` `max``(result, ``1` `+` `LIP(dp, mat, n, ` ` ``m, x, y ``+` `1``)) ` ` ``dp[x][y] ``=` `result ` ` ``return` `dp[x][y] ` ` ` `# Wrapper function ` `def` `wrapper(mat, n, m): ` ` ``dp ``=` `[[``7` `for` `i ``in` `range``(``MAX``)] ` ` ``for` `i ``in` `range``(``MAX``)] ` ` ``return` `LIP(dp, mat, n, m, ``0``, ``0``) ` ` ` `# Driver Code ` `mat ``=` `[[``1``, ``2``, ``3``, ``4` `], ` ` ``[``2``, ``2``, ``3``, ``4` `], ` ` ``[``3``, ``2``, ``3``, ``4` `], ` ` ``[``4``, ``5``, ``6``, ``7` `]] ` `n ``=` `4` `m ``=` `4` `print``(wrapper(mat, n, m)) ` ` ` `# This code is contributed ` `# by Sahil Shelangia `
<MASK>
<UNMASK>
<MASK>
Input : N = 2, M =2
m[][] = { { 1, 2 },
{ 3, 4 } };
Output :3
Longest path is either 1 2 4 or
1 3 4.
```
## Recommended: Please try your approach on {IDE} first, before moving on to the solution.
The idea is to use dynamic programming. Maintain the 2D matrix, dp[][], where dp[i][j] store the value of length of longest increasing sequence for sub matrix starting from ith row and j-th column.
Let the longest increasing sub sequence values for m[i+1][j] and m[i][j+1] be known already as v1 and v2 respectively. Then the value for m[i][j] will be max(v1, v2) + 1.
We can start from m[n-1][m-1] as base case with length of longest increasing sub sequence be 1, moving upwards and leftwards updating the value of cells. Then the LIP value for cell m[0][0] will be the answer.
<MASK>
## C++
<MASK>
## Java
<MASK>
## Python3
`# Python3 program to find longest ` `# increasing path in a matrix. ` `MAX` `=` `20` ` ` `# Return the length of ` `# LIP in 2D matrix ` `def` `LIP(dp, mat, n, m, x, y): ` ` ` ` ``# If value not calculated yet. ` ` ``if` `(dp[x][y] < ``0``): ` ` ``result ``=` `0` ` ` ` ``# If reach bottom left cell, ` ` ``# return 1. ` ` ``if` `(x ``=``=` `n ``-` `1` `and` `y ``=``=` `m ``-` `1``): ` ` ``dp[x][y] ``=` `1` ` ``return` `dp[x][y] ` ` ` ` ``# If reach the corner ` ` ``# of the matrix. ` ` ``if` `(x ``=``=` `n ``-` `1` `or` `y ``=``=` `m ``-` `1``): ` ` ``result ``=` `1` ` ` ` ``# If value greater than below cell. ` ` ``elif` `(mat[x][y] < mat[x ``+` `1``][y]): ` ` ``result ``=` `1` `+` `LIP(dp, mat, n, ` ` ``m, x ``+` `1``, y) ` ` ` ` ``# If value greater than left cell. ` ` ``elif` `(mat[x][y] < mat[x][y ``+` `1``]): ` ` ``result ``=` `max``(result, ``1` `+` `LIP(dp, mat, n, ` ` ``m, x, y ``+` `1``)) ` ` ``dp[x][y] ``=` `result ` ` ``return` `dp[x][y] ` ` ` `# Wrapper function ` `def` `wrapper(mat, n, m): ` ` ``dp ``=` `[[``7` `for` `i ``in` `range``(``MAX``)] ` ` ``for` `i ``in` `range``(``MAX``)] ` ` ``return` `LIP(dp, mat, n, m, ``0``, ``0``) ` ` ` `# Driver Code ` `mat ``=` `[[``1``, ``2``, ``3``, ``4` `], ` ` ``[``2``, ``2``, ``3``, ``4` `], ` ` ``[``3``, ``2``, ``3``, ``4` `], ` ` ``[``4``, ``5``, ``6``, ``7` `]] ` `n ``=` `4` `m ``=` `4` `print``(wrapper(mat, n, m)) ` ` ` `# This code is contributed ` `# by Sahil Shelangia `
<MASK>
`// C# program to find longest increasing ` `// path in a matrix. ` `using` `System; ` ` ` `public` `class` `GFG ` `{ ` ` ` ` ``// Return the length of LIP in 2D matrix ` ` ``static` `int` `LIP(``int` `[,]dp, ``int` `[,]mat, ``int` `n, ` ` ``int` `m, ``int` `x, ``int` `y) ` ` ``{ ` ` ``// If value not calculated yet. ` ` ``if` `(dp[x,y] < 0) ` ` ``{ ` ` ``int` `result = 0; ` ` ` ` ``// If reach bottom left cell, return 1. ` ` ``if` `(x == n - 1 && y == m - 1) ` ` ``return` `dp[x, y] = 1; ` ` ` ` ``// If reach the corner of the matrix. ` ` ``if` `(x == n - 1 || y == m - 1) ` ` ``result = 1; ` ` ` ` ``// If value greater than below cell. ` ` ``if` `(x + 1 < n && mat[x, y] < mat[x + 1, y]) ` ` ``result = 1 + LIP(dp, mat, n, m, x + 1, y); ` ` ` ` ``// If value greater than left cell. ` ` ``if` `(y + 1 < m && mat[x, y] < mat[x, y + 1]) ` ` ``result = Math.Max(result, 1 + ` ` ``LIP(dp, mat, n, m, x, y + 1)); ` ` ` ` ``dp[x, y] = result; ` ` ``} ` ` ` ` ``return` `dp[x,y]; ` ` ``} ` ` ` ` ``// Wrapper function ` ` ``static` `int` `wrapper(``int` `[,]mat, ``int` `n, ``int` `m) ` ` ``{ ` ` ``int` `[,]dp = ``new` `int``[10, 10]; ` ` ``for``(``int` `i = 0; i < 10; i++) ` ` ``{ ` ` ``for``(``int` `j = 0; j < 10; j++) ` ` ``{ ` ` ``dp[i, j] = -1; ` ` ``} ` ` ``} ` ` ` ` ``return` `LIP(dp, mat, n, m, 0, 0); ` ` ``} ` ` ` ` ``/* Driver code */` ` ``public` `static` `void` `Main() ` ` ``{ ` ` ``int` `[,]mat= { { 1, 2, 3, 4 }, ` ` ``{ 2, 2, 3, 4 }, ` ` ``{ 3, 2, 3, 4 }, ` ` ``{ 4, 5, 6, 7 }, }; ` ` ``int` `n = 4, m = 4; ` ` ``Console.WriteLine(wrapper(mat, n, m)); ` ` ``} ` `} ` ` ` `/* This code contributed by PrinciRaj1992 */`
<MASK>
|
<MASK>
Input : N = 2, M =2
m[][] = { { 1, 2 },
{ 3, 4 } };
Output :3
Longest path is either 1 2 4 or
1 3 4.
```
## Recommended: Please try your approach on {IDE} first, before moving on to the solution.
The idea is to use dynamic programming. Maintain the 2D matrix, dp[][], where dp[i][j] store the value of length of longest increasing sequence for sub matrix starting from ith row and j-th column.
Let the longest increasing sub sequence values for m[i+1][j] and m[i][j+1] be known already as v1 and v2 respectively. Then the value for m[i][j] will be max(v1, v2) + 1.
We can start from m[n-1][m-1] as base case with length of longest increasing sub sequence be 1, moving upwards and leftwards updating the value of cells. Then the LIP value for cell m[0][0] will be the answer.
<MASK>
## C++
<MASK>
## Java
<MASK>
## Python3
`# Python3 program to find longest ` `# increasing path in a matrix. ` `MAX` `=` `20` ` ` `# Return the length of ` `# LIP in 2D matrix ` `def` `LIP(dp, mat, n, m, x, y): ` ` ` ` ``# If value not calculated yet. ` ` ``if` `(dp[x][y] < ``0``): ` ` ``result ``=` `0` ` ` ` ``# If reach bottom left cell, ` ` ``# return 1. ` ` ``if` `(x ``=``=` `n ``-` `1` `and` `y ``=``=` `m ``-` `1``): ` ` ``dp[x][y] ``=` `1` ` ``return` `dp[x][y] ` ` ` ` ``# If reach the corner ` ` ``# of the matrix. ` ` ``if` `(x ``=``=` `n ``-` `1` `or` `y ``=``=` `m ``-` `1``): ` ` ``result ``=` `1` ` ` ` ``# If value greater than below cell. ` ` ``elif` `(mat[x][y] < mat[x ``+` `1``][y]): ` ` ``result ``=` `1` `+` `LIP(dp, mat, n, ` ` ``m, x ``+` `1``, y) ` ` ` ` ``# If value greater than left cell. ` ` ``elif` `(mat[x][y] < mat[x][y ``+` `1``]): ` ` ``result ``=` `max``(result, ``1` `+` `LIP(dp, mat, n, ` ` ``m, x, y ``+` `1``)) ` ` ``dp[x][y] ``=` `result ` ` ``return` `dp[x][y] ` ` ` `# Wrapper function ` `def` `wrapper(mat, n, m): ` ` ``dp ``=` `[[``7` `for` `i ``in` `range``(``MAX``)] ` ` ``for` `i ``in` `range``(``MAX``)] ` ` ``return` `LIP(dp, mat, n, m, ``0``, ``0``) ` ` ` `# Driver Code ` `mat ``=` `[[``1``, ``2``, ``3``, ``4` `], ` ` ``[``2``, ``2``, ``3``, ``4` `], ` ` ``[``3``, ``2``, ``3``, ``4` `], ` ` ``[``4``, ``5``, ``6``, ``7` `]] ` `n ``=` `4` `m ``=` `4` `print``(wrapper(mat, n, m)) ` ` ` `# This code is contributed ` `# by Sahil Shelangia `
<MASK>
`// C# program to find longest increasing ` `// path in a matrix. ` `using` `System; ` ` ` `public` `class` `GFG ` `{ ` ` ` ` ``// Return the length of LIP in 2D matrix ` ` ``static` `int` `LIP(``int` `[,]dp, ``int` `[,]mat, ``int` `n, ` ` ``int` `m, ``int` `x, ``int` `y) ` ` ``{ ` ` ``// If value not calculated yet. ` ` ``if` `(dp[x,y] < 0) ` ` ``{ ` ` ``int` `result = 0; ` ` ` ` ``// If reach bottom left cell, return 1. ` ` ``if` `(x == n - 1 && y == m - 1) ` ` ``return` `dp[x, y] = 1; ` ` ` ` ``// If reach the corner of the matrix. ` ` ``if` `(x == n - 1 || y == m - 1) ` ` ``result = 1; ` ` ` ` ``// If value greater than below cell. ` ` ``if` `(x + 1 < n && mat[x, y] < mat[x + 1, y]) ` ` ``result = 1 + LIP(dp, mat, n, m, x + 1, y); ` ` ` ` ``// If value greater than left cell. ` ` ``if` `(y + 1 < m && mat[x, y] < mat[x, y + 1]) ` ` ``result = Math.Max(result, 1 + ` ` ``LIP(dp, mat, n, m, x, y + 1)); ` ` ` ` ``dp[x, y] = result; ` ` ``} ` ` ` ` ``return` `dp[x,y]; ` ` ``} ` ` ` ` ``// Wrapper function ` ` ``static` `int` `wrapper(``int` `[,]mat, ``int` `n, ``int` `m) ` ` ``{ ` ` ``int` `[,]dp = ``new` `int``[10, 10]; ` ` ``for``(``int` `i = 0; i < 10; i++) ` ` ``{ ` ` ``for``(``int` `j = 0; j < 10; j++) ` ` ``{ ` ` ``dp[i, j] = -1; ` ` ``} ` ` ``} ` ` ` ` ``return` `LIP(dp, mat, n, m, 0, 0); ` ` ``} ` ` ` ` ``/* Driver code */` ` ``public` `static` `void` `Main() ` ` ``{ ` ` ``int` `[,]mat= { { 1, 2, 3, 4 }, ` ` ``{ 2, 2, 3, 4 }, ` ` ``{ 3, 2, 3, 4 }, ` ` ``{ 4, 5, 6, 7 }, }; ` ` ``int` `n = 4, m = 4; ` ` ``Console.WriteLine(wrapper(mat, n, m)); ` ` ``} ` `} ` ` ` `/* This code contributed by PrinciRaj1992 */`
<MASK>
<UNMASK>
# Longest Increasing Path in Matrix
<MASK>
Examples:
```Input : N = 4, M = 4
m[][] = { { 1, 2, 3, 4 },
{ 2, 2, 3, 4 },
{ 3, 2, 3, 4 },
{ 4, 5, 6, 7 } };
Output : 7
Longest path is 1 2 3 4 5 6 7.
Input : N = 2, M =2
m[][] = { { 1, 2 },
{ 3, 4 } };
Output :3
Longest path is either 1 2 4 or
1 3 4.
```
## Recommended: Please try your approach on {IDE} first, before moving on to the solution.
The idea is to use dynamic programming. Maintain the 2D matrix, dp[][], where dp[i][j] store the value of length of longest increasing sequence for sub matrix starting from ith row and j-th column.
Let the longest increasing sub sequence values for m[i+1][j] and m[i][j+1] be known already as v1 and v2 respectively. Then the value for m[i][j] will be max(v1, v2) + 1.
We can start from m[n-1][m-1] as base case with length of longest increasing sub sequence be 1, moving upwards and leftwards updating the value of cells. Then the LIP value for cell m[0][0] will be the answer.
Below is the implementation of this approach:
## C++
`// CPP program to find longest increasing ` `// path in a matrix. ` `#include ` `#define MAX 10 ` `using` `namespace` `std; ` ` ` `// Return the length of LIP in 2D matrix ` `int` `LIP(``int` `dp[][MAX], ``int` `mat[][MAX], ``int` `n, ``int` `m, ``int` `x, ``int` `y) ` `{ ` ` ``// If value not calculated yet. ` ` ``if` `(dp[x][y] < 0) ` ` ``{ ` ` ``int` `result = 0; ` ` ` ` ``// If reach bottom left cell, return 1. ` ` ``if` `(x == n-1 && y == m-1) ` ` ``return` `dp[x][y] = 1; ` ` ` ` ``// If reach the corner of the matrix. ` ` ``if` `(x == n-1 || y == m-1) ` ` ``result = 1; ` ` ` ` ``// If value greater than below cell. ` ` ``if` `(mat[x][y] < mat[x+1][y]) ` ` ``result = 1 + LIP(dp, mat, n, m, x+1, y); ` ` ` ` ``// If value greater than left cell. ` ` ``if` `(mat[x][y] < mat[x][y+1]) ` ` ``result = max(result, 1 + LIP(dp, mat, n, m, x, y+1)); ` ` ` ` ``dp[x][y] = result; ` ` ``} ` ` ` ` ``return` `dp[x][y]; ` `} ` ` ` `// Wrapper function ` `int` `wrapper(``int` `mat[][MAX], ``int` `n, ``int` `m) ` `{ ` ` ``int` `dp[MAX][MAX]; ` ` ``memset``(dp, -1, ``sizeof` `dp); ` ` ` ` ``return` `LIP(dp, mat, n, m, 0, 0); ` `} ` ` ` `// Driven Program ` `int` `main() ` `{ ` ` ``int` `mat[][MAX] = { ` ` ``{ 1, 2, 3, 4 }, ` ` ``{ 2, 2, 3, 4 }, ` ` ``{ 3, 2, 3, 4 }, ` ` ``{ 4, 5, 6, 7 }, ` ` ``}; ` ` ``int` `n = 4, m = 4; ` ` ``cout << wrapper(mat, n, m) << endl; ` ` ` ` ``return` `0; ` `} `
/div>
## Java
`// Java program to find longest increasing ` `// path in a matrix. ` `import` `java.util.*; ` ` ` `class` `GFG { ` ` ` ` ``// Return the length of LIP in 2D matrix ` ` ``static` `int` `LIP(``int` `dp[][], ``int` `mat[][], ``int` `n, ` ` ``int` `m, ``int` `x, ``int` `y) ` ` ``{ ` ` ``// If value not calculated yet. ` ` ``if` `(dp[x][y] < ``0``) ` ` ``{ ` ` ``int` `result = ``0``; ` ` ` ` ``// If reach bottom left cell, return 1. ` ` ``if` `(x == n-``1` `&& y == m-``1``) ` ` ``return` `dp[x][y] = ``1``; ` ` ` ` ``// If reach the corner of the matrix. ` ` ``if` `(x == n-``1` `|| y == m-``1``) ` ` ``result = ``1``; ` ` ` ` ``// If value greater than below cell. ` ` ``if` `(x + ``1` `< n && mat[x][y] < mat[x+``1``][y]) ` ` ``result = ``1` `+ LIP(dp, mat, n, m, x+``1``, y); ` ` ` ` ``// If value greater than left cell. ` ` ``if` `(y + ``1` `< m && mat[x][y] < mat[x][y+``1``]) ` ` ``result = Math.max(result, ``1` `+ ` ` ``LIP(dp, mat, n, m, x, y+``1``)); ` ` ` ` ``dp[x][y] = result; ` ` ``} ` ` ` ` ``return` `dp[x][y]; ` ` ``} ` ` ` ` ``// Wrapper function ` ` ``static` `int` `wrapper(``int` `mat[][], ``int` `n, ``int` `m) ` ` ``{ ` ` ``int` `dp[][] = ``new` `int``[``10``][``10``]; ` ` ``for``(``int` `i = ``0``; i < ``10``; i++) ` ` ``Arrays.fill(dp[i],-``1``); ` ` ` ` ``return` `LIP(dp, mat, n, m, ``0``, ``0``); ` ` ``} ` ` ` ` ``/* Driver program to test above function */` ` ``public` `static` `void` `main(String[] args) ` ` ``{ ` ` ``int` `mat[][] = { ` ` ``{ ``1``, ``2``, ``3``, ``4` `}, ` ` ``{ ``2``, ``2``, ``3``, ``4` `}, ` ` ``{ ``3``, ``2``, ``3``, ``4` `}, ` ` ``{ ``4``, ``5``, ``6``, ``7` `}, ` ` ``}; ` ` ``int` `n = ``4``, m = ``4``; ` ` ``System.out.println(wrapper(mat, n, m)); ` ` ` ` ``} ` `} ` ` ` `// This code is contributed by Arnav Kr. Mandal. `
## Python3
`# Python3 program to find longest ` `# increasing path in a matrix. ` `MAX` `=` `20` ` ` `# Return the length of ` `# LIP in 2D matrix ` `def` `LIP(dp, mat, n, m, x, y): ` ` ` ` ``# If value not calculated yet. ` ` ``if` `(dp[x][y] < ``0``): ` ` ``result ``=` `0` ` ` ` ``# If reach bottom left cell, ` ` ``# return 1. ` ` ``if` `(x ``=``=` `n ``-` `1` `and` `y ``=``=` `m ``-` `1``): ` ` ``dp[x][y] ``=` `1` ` ``return` `dp[x][y] ` ` ` ` ``# If reach the corner ` ` ``# of the matrix. ` ` ``if` `(x ``=``=` `n ``-` `1` `or` `y ``=``=` `m ``-` `1``): ` ` ``result ``=` `1` ` ` ` ``# If value greater than below cell. ` ` ``elif` `(mat[x][y] < mat[x ``+` `1``][y]): ` ` ``result ``=` `1` `+` `LIP(dp, mat, n, ` ` ``m, x ``+` `1``, y) ` ` ` ` ``# If value greater than left cell. ` ` ``elif` `(mat[x][y] < mat[x][y ``+` `1``]): ` ` ``result ``=` `max``(result, ``1` `+` `LIP(dp, mat, n, ` ` ``m, x, y ``+` `1``)) ` ` ``dp[x][y] ``=` `result ` ` ``return` `dp[x][y] ` ` ` `# Wrapper function ` `def` `wrapper(mat, n, m): ` ` ``dp ``=` `[[``7` `for` `i ``in` `range``(``MAX``)] ` ` ``for` `i ``in` `range``(``MAX``)] ` ` ``return` `LIP(dp, mat, n, m, ``0``, ``0``) ` ` ` `# Driver Code ` `mat ``=` `[[``1``, ``2``, ``3``, ``4` `], ` ` ``[``2``, ``2``, ``3``, ``4` `], ` ` ``[``3``, ``2``, ``3``, ``4` `], ` ` ``[``4``, ``5``, ``6``, ``7` `]] ` `n ``=` `4` `m ``=` `4` `print``(wrapper(mat, n, m)) ` ` ` `# This code is contributed ` `# by Sahil Shelangia `
<MASK>
`// C# program to find longest increasing ` `// path in a matrix. ` `using` `System; ` ` ` `public` `class` `GFG ` `{ ` ` ` ` ``// Return the length of LIP in 2D matrix ` ` ``static` `int` `LIP(``int` `[,]dp, ``int` `[,]mat, ``int` `n, ` ` ``int` `m, ``int` `x, ``int` `y) ` ` ``{ ` ` ``// If value not calculated yet. ` ` ``if` `(dp[x,y] < 0) ` ` ``{ ` ` ``int` `result = 0; ` ` ` ` ``// If reach bottom left cell, return 1. ` ` ``if` `(x == n - 1 && y == m - 1) ` ` ``return` `dp[x, y] = 1; ` ` ` ` ``// If reach the corner of the matrix. ` ` ``if` `(x == n - 1 || y == m - 1) ` ` ``result = 1; ` ` ` ` ``// If value greater than below cell. ` ` ``if` `(x + 1 < n && mat[x, y] < mat[x + 1, y]) ` ` ``result = 1 + LIP(dp, mat, n, m, x + 1, y); ` ` ` ` ``// If value greater than left cell. ` ` ``if` `(y + 1 < m && mat[x, y] < mat[x, y + 1]) ` ` ``result = Math.Max(result, 1 + ` ` ``LIP(dp, mat, n, m, x, y + 1)); ` ` ` ` ``dp[x, y] = result; ` ` ``} ` ` ` ` ``return` `dp[x,y]; ` ` ``} ` ` ` ` ``// Wrapper function ` ` ``static` `int` `wrapper(``int` `[,]mat, ``int` `n, ``int` `m) ` ` ``{ ` ` ``int` `[,]dp = ``new` `int``[10, 10]; ` ` ``for``(``int` `i = 0; i < 10; i++) ` ` ``{ ` ` ``for``(``int` `j = 0; j < 10; j++) ` ` ``{ ` ` ``dp[i, j] = -1; ` ` ``} ` ` ``} ` ` ` ` ``return` `LIP(dp, mat, n, m, 0, 0); ` ` ``} ` ` ` ` ``/* Driver code */` ` ``public` `static` `void` `Main() ` ` ``{ ` ` ``int` `[,]mat= { { 1, 2, 3, 4 }, ` ` ``{ 2, 2, 3, 4 }, ` ` ``{ 3, 2, 3, 4 }, ` ` ``{ 4, 5, 6, 7 }, }; ` ` ``int` `n = 4, m = 4; ` ` ``Console.WriteLine(wrapper(mat, n, m)); ` ` ``} ` `} ` ` ` `/* This code contributed by PrinciRaj1992 */`
Output:
```7
```
<MASK>
|
# Longest Increasing Path in Matrix
<MASK>
Examples:
```Input : N = 4, M = 4
m[][] = { { 1, 2, 3, 4 },
{ 2, 2, 3, 4 },
{ 3, 2, 3, 4 },
{ 4, 5, 6, 7 } };
Output : 7
Longest path is 1 2 3 4 5 6 7.
Input : N = 2, M =2
m[][] = { { 1, 2 },
{ 3, 4 } };
Output :3
Longest path is either 1 2 4 or
1 3 4.
```
## Recommended: Please try your approach on {IDE} first, before moving on to the solution.
The idea is to use dynamic programming. Maintain the 2D matrix, dp[][], where dp[i][j] store the value of length of longest increasing sequence for sub matrix starting from ith row and j-th column.
Let the longest increasing sub sequence values for m[i+1][j] and m[i][j+1] be known already as v1 and v2 respectively. Then the value for m[i][j] will be max(v1, v2) + 1.
We can start from m[n-1][m-1] as base case with length of longest increasing sub sequence be 1, moving upwards and leftwards updating the value of cells. Then the LIP value for cell m[0][0] will be the answer.
Below is the implementation of this approach:
## C++
`// CPP program to find longest increasing ` `// path in a matrix. ` `#include ` `#define MAX 10 ` `using` `namespace` `std; ` ` ` `// Return the length of LIP in 2D matrix ` `int` `LIP(``int` `dp[][MAX], ``int` `mat[][MAX], ``int` `n, ``int` `m, ``int` `x, ``int` `y) ` `{ ` ` ``// If value not calculated yet. ` ` ``if` `(dp[x][y] < 0) ` ` ``{ ` ` ``int` `result = 0; ` ` ` ` ``// If reach bottom left cell, return 1. ` ` ``if` `(x == n-1 && y == m-1) ` ` ``return` `dp[x][y] = 1; ` ` ` ` ``// If reach the corner of the matrix. ` ` ``if` `(x == n-1 || y == m-1) ` ` ``result = 1; ` ` ` ` ``// If value greater than below cell. ` ` ``if` `(mat[x][y] < mat[x+1][y]) ` ` ``result = 1 + LIP(dp, mat, n, m, x+1, y); ` ` ` ` ``// If value greater than left cell. ` ` ``if` `(mat[x][y] < mat[x][y+1]) ` ` ``result = max(result, 1 + LIP(dp, mat, n, m, x, y+1)); ` ` ` ` ``dp[x][y] = result; ` ` ``} ` ` ` ` ``return` `dp[x][y]; ` `} ` ` ` `// Wrapper function ` `int` `wrapper(``int` `mat[][MAX], ``int` `n, ``int` `m) ` `{ ` ` ``int` `dp[MAX][MAX]; ` ` ``memset``(dp, -1, ``sizeof` `dp); ` ` ` ` ``return` `LIP(dp, mat, n, m, 0, 0); ` `} ` ` ` `// Driven Program ` `int` `main() ` `{ ` ` ``int` `mat[][MAX] = { ` ` ``{ 1, 2, 3, 4 }, ` ` ``{ 2, 2, 3, 4 }, ` ` ``{ 3, 2, 3, 4 }, ` ` ``{ 4, 5, 6, 7 }, ` ` ``}; ` ` ``int` `n = 4, m = 4; ` ` ``cout << wrapper(mat, n, m) << endl; ` ` ` ` ``return` `0; ` `} `
/div>
## Java
`// Java program to find longest increasing ` `// path in a matrix. ` `import` `java.util.*; ` ` ` `class` `GFG { ` ` ` ` ``// Return the length of LIP in 2D matrix ` ` ``static` `int` `LIP(``int` `dp[][], ``int` `mat[][], ``int` `n, ` ` ``int` `m, ``int` `x, ``int` `y) ` ` ``{ ` ` ``// If value not calculated yet. ` ` ``if` `(dp[x][y] < ``0``) ` ` ``{ ` ` ``int` `result = ``0``; ` ` ` ` ``// If reach bottom left cell, return 1. ` ` ``if` `(x == n-``1` `&& y == m-``1``) ` ` ``return` `dp[x][y] = ``1``; ` ` ` ` ``// If reach the corner of the matrix. ` ` ``if` `(x == n-``1` `|| y == m-``1``) ` ` ``result = ``1``; ` ` ` ` ``// If value greater than below cell. ` ` ``if` `(x + ``1` `< n && mat[x][y] < mat[x+``1``][y]) ` ` ``result = ``1` `+ LIP(dp, mat, n, m, x+``1``, y); ` ` ` ` ``// If value greater than left cell. ` ` ``if` `(y + ``1` `< m && mat[x][y] < mat[x][y+``1``]) ` ` ``result = Math.max(result, ``1` `+ ` ` ``LIP(dp, mat, n, m, x, y+``1``)); ` ` ` ` ``dp[x][y] = result; ` ` ``} ` ` ` ` ``return` `dp[x][y]; ` ` ``} ` ` ` ` ``// Wrapper function ` ` ``static` `int` `wrapper(``int` `mat[][], ``int` `n, ``int` `m) ` ` ``{ ` ` ``int` `dp[][] = ``new` `int``[``10``][``10``]; ` ` ``for``(``int` `i = ``0``; i < ``10``; i++) ` ` ``Arrays.fill(dp[i],-``1``); ` ` ` ` ``return` `LIP(dp, mat, n, m, ``0``, ``0``); ` ` ``} ` ` ` ` ``/* Driver program to test above function */` ` ``public` `static` `void` `main(String[] args) ` ` ``{ ` ` ``int` `mat[][] = { ` ` ``{ ``1``, ``2``, ``3``, ``4` `}, ` ` ``{ ``2``, ``2``, ``3``, ``4` `}, ` ` ``{ ``3``, ``2``, ``3``, ``4` `}, ` ` ``{ ``4``, ``5``, ``6``, ``7` `}, ` ` ``}; ` ` ``int` `n = ``4``, m = ``4``; ` ` ``System.out.println(wrapper(mat, n, m)); ` ` ` ` ``} ` `} ` ` ` `// This code is contributed by Arnav Kr. Mandal. `
## Python3
`# Python3 program to find longest ` `# increasing path in a matrix. ` `MAX` `=` `20` ` ` `# Return the length of ` `# LIP in 2D matrix ` `def` `LIP(dp, mat, n, m, x, y): ` ` ` ` ``# If value not calculated yet. ` ` ``if` `(dp[x][y] < ``0``): ` ` ``result ``=` `0` ` ` ` ``# If reach bottom left cell, ` ` ``# return 1. ` ` ``if` `(x ``=``=` `n ``-` `1` `and` `y ``=``=` `m ``-` `1``): ` ` ``dp[x][y] ``=` `1` ` ``return` `dp[x][y] ` ` ` ` ``# If reach the corner ` ` ``# of the matrix. ` ` ``if` `(x ``=``=` `n ``-` `1` `or` `y ``=``=` `m ``-` `1``): ` ` ``result ``=` `1` ` ` ` ``# If value greater than below cell. ` ` ``elif` `(mat[x][y] < mat[x ``+` `1``][y]): ` ` ``result ``=` `1` `+` `LIP(dp, mat, n, ` ` ``m, x ``+` `1``, y) ` ` ` ` ``# If value greater than left cell. ` ` ``elif` `(mat[x][y] < mat[x][y ``+` `1``]): ` ` ``result ``=` `max``(result, ``1` `+` `LIP(dp, mat, n, ` ` ``m, x, y ``+` `1``)) ` ` ``dp[x][y] ``=` `result ` ` ``return` `dp[x][y] ` ` ` `# Wrapper function ` `def` `wrapper(mat, n, m): ` ` ``dp ``=` `[[``7` `for` `i ``in` `range``(``MAX``)] ` ` ``for` `i ``in` `range``(``MAX``)] ` ` ``return` `LIP(dp, mat, n, m, ``0``, ``0``) ` ` ` `# Driver Code ` `mat ``=` `[[``1``, ``2``, ``3``, ``4` `], ` ` ``[``2``, ``2``, ``3``, ``4` `], ` ` ``[``3``, ``2``, ``3``, ``4` `], ` ` ``[``4``, ``5``, ``6``, ``7` `]] ` `n ``=` `4` `m ``=` `4` `print``(wrapper(mat, n, m)) ` ` ` `# This code is contributed ` `# by Sahil Shelangia `
<MASK>
`// C# program to find longest increasing ` `// path in a matrix. ` `using` `System; ` ` ` `public` `class` `GFG ` `{ ` ` ` ` ``// Return the length of LIP in 2D matrix ` ` ``static` `int` `LIP(``int` `[,]dp, ``int` `[,]mat, ``int` `n, ` ` ``int` `m, ``int` `x, ``int` `y) ` ` ``{ ` ` ``// If value not calculated yet. ` ` ``if` `(dp[x,y] < 0) ` ` ``{ ` ` ``int` `result = 0; ` ` ` ` ``// If reach bottom left cell, return 1. ` ` ``if` `(x == n - 1 && y == m - 1) ` ` ``return` `dp[x, y] = 1; ` ` ` ` ``// If reach the corner of the matrix. ` ` ``if` `(x == n - 1 || y == m - 1) ` ` ``result = 1; ` ` ` ` ``// If value greater than below cell. ` ` ``if` `(x + 1 < n && mat[x, y] < mat[x + 1, y]) ` ` ``result = 1 + LIP(dp, mat, n, m, x + 1, y); ` ` ` ` ``// If value greater than left cell. ` ` ``if` `(y + 1 < m && mat[x, y] < mat[x, y + 1]) ` ` ``result = Math.Max(result, 1 + ` ` ``LIP(dp, mat, n, m, x, y + 1)); ` ` ` ` ``dp[x, y] = result; ` ` ``} ` ` ` ` ``return` `dp[x,y]; ` ` ``} ` ` ` ` ``// Wrapper function ` ` ``static` `int` `wrapper(``int` `[,]mat, ``int` `n, ``int` `m) ` ` ``{ ` ` ``int` `[,]dp = ``new` `int``[10, 10]; ` ` ``for``(``int` `i = 0; i < 10; i++) ` ` ``{ ` ` ``for``(``int` `j = 0; j < 10; j++) ` ` ``{ ` ` ``dp[i, j] = -1; ` ` ``} ` ` ``} ` ` ` ` ``return` `LIP(dp, mat, n, m, 0, 0); ` ` ``} ` ` ` ` ``/* Driver code */` ` ``public` `static` `void` `Main() ` ` ``{ ` ` ``int` `[,]mat= { { 1, 2, 3, 4 }, ` ` ``{ 2, 2, 3, 4 }, ` ` ``{ 3, 2, 3, 4 }, ` ` ``{ 4, 5, 6, 7 }, }; ` ` ``int` `n = 4, m = 4; ` ` ``Console.WriteLine(wrapper(mat, n, m)); ` ` ``} ` `} ` ` ` `/* This code contributed by PrinciRaj1992 */`
Output:
```7
```
<MASK>
<UNMASK>
# Longest Increasing Path in Matrix
Given a matrix of N rows and M columns. From m[i][j], we can move to m[i+1][j], if m[i+1][j] > m[i][j], or can move to m[i][j+1] if m[i][j+1] > m[i][j]. The task is print longest path length if we start from (0, 0).
Examples:
```Input : N = 4, M = 4
m[][] = { { 1, 2, 3, 4 },
{ 2, 2, 3, 4 },
{ 3, 2, 3, 4 },
{ 4, 5, 6, 7 } };
Output : 7
Longest path is 1 2 3 4 5 6 7.
Input : N = 2, M =2
m[][] = { { 1, 2 },
{ 3, 4 } };
Output :3
Longest path is either 1 2 4 or
1 3 4.
```
## Recommended: Please try your approach on {IDE} first, before moving on to the solution.
The idea is to use dynamic programming. Maintain the 2D matrix, dp[][], where dp[i][j] store the value of length of longest increasing sequence for sub matrix starting from ith row and j-th column.
Let the longest increasing sub sequence values for m[i+1][j] and m[i][j+1] be known already as v1 and v2 respectively. Then the value for m[i][j] will be max(v1, v2) + 1.
We can start from m[n-1][m-1] as base case with length of longest increasing sub sequence be 1, moving upwards and leftwards updating the value of cells. Then the LIP value for cell m[0][0] will be the answer.
Below is the implementation of this approach:
## C++
`// CPP program to find longest increasing ` `// path in a matrix. ` `#include ` `#define MAX 10 ` `using` `namespace` `std; ` ` ` `// Return the length of LIP in 2D matrix ` `int` `LIP(``int` `dp[][MAX], ``int` `mat[][MAX], ``int` `n, ``int` `m, ``int` `x, ``int` `y) ` `{ ` ` ``// If value not calculated yet. ` ` ``if` `(dp[x][y] < 0) ` ` ``{ ` ` ``int` `result = 0; ` ` ` ` ``// If reach bottom left cell, return 1. ` ` ``if` `(x == n-1 && y == m-1) ` ` ``return` `dp[x][y] = 1; ` ` ` ` ``// If reach the corner of the matrix. ` ` ``if` `(x == n-1 || y == m-1) ` ` ``result = 1; ` ` ` ` ``// If value greater than below cell. ` ` ``if` `(mat[x][y] < mat[x+1][y]) ` ` ``result = 1 + LIP(dp, mat, n, m, x+1, y); ` ` ` ` ``// If value greater than left cell. ` ` ``if` `(mat[x][y] < mat[x][y+1]) ` ` ``result = max(result, 1 + LIP(dp, mat, n, m, x, y+1)); ` ` ` ` ``dp[x][y] = result; ` ` ``} ` ` ` ` ``return` `dp[x][y]; ` `} ` ` ` `// Wrapper function ` `int` `wrapper(``int` `mat[][MAX], ``int` `n, ``int` `m) ` `{ ` ` ``int` `dp[MAX][MAX]; ` ` ``memset``(dp, -1, ``sizeof` `dp); ` ` ` ` ``return` `LIP(dp, mat, n, m, 0, 0); ` `} ` ` ` `// Driven Program ` `int` `main() ` `{ ` ` ``int` `mat[][MAX] = { ` ` ``{ 1, 2, 3, 4 }, ` ` ``{ 2, 2, 3, 4 }, ` ` ``{ 3, 2, 3, 4 }, ` ` ``{ 4, 5, 6, 7 }, ` ` ``}; ` ` ``int` `n = 4, m = 4; ` ` ``cout << wrapper(mat, n, m) << endl; ` ` ` ` ``return` `0; ` `} `
/div>
## Java
`// Java program to find longest increasing ` `// path in a matrix. ` `import` `java.util.*; ` ` ` `class` `GFG { ` ` ` ` ``// Return the length of LIP in 2D matrix ` ` ``static` `int` `LIP(``int` `dp[][], ``int` `mat[][], ``int` `n, ` ` ``int` `m, ``int` `x, ``int` `y) ` ` ``{ ` ` ``// If value not calculated yet. ` ` ``if` `(dp[x][y] < ``0``) ` ` ``{ ` ` ``int` `result = ``0``; ` ` ` ` ``// If reach bottom left cell, return 1. ` ` ``if` `(x == n-``1` `&& y == m-``1``) ` ` ``return` `dp[x][y] = ``1``; ` ` ` ` ``// If reach the corner of the matrix. ` ` ``if` `(x == n-``1` `|| y == m-``1``) ` ` ``result = ``1``; ` ` ` ` ``// If value greater than below cell. ` ` ``if` `(x + ``1` `< n && mat[x][y] < mat[x+``1``][y]) ` ` ``result = ``1` `+ LIP(dp, mat, n, m, x+``1``, y); ` ` ` ` ``// If value greater than left cell. ` ` ``if` `(y + ``1` `< m && mat[x][y] < mat[x][y+``1``]) ` ` ``result = Math.max(result, ``1` `+ ` ` ``LIP(dp, mat, n, m, x, y+``1``)); ` ` ` ` ``dp[x][y] = result; ` ` ``} ` ` ` ` ``return` `dp[x][y]; ` ` ``} ` ` ` ` ``// Wrapper function ` ` ``static` `int` `wrapper(``int` `mat[][], ``int` `n, ``int` `m) ` ` ``{ ` ` ``int` `dp[][] = ``new` `int``[``10``][``10``]; ` ` ``for``(``int` `i = ``0``; i < ``10``; i++) ` ` ``Arrays.fill(dp[i],-``1``); ` ` ` ` ``return` `LIP(dp, mat, n, m, ``0``, ``0``); ` ` ``} ` ` ` ` ``/* Driver program to test above function */` ` ``public` `static` `void` `main(String[] args) ` ` ``{ ` ` ``int` `mat[][] = { ` ` ``{ ``1``, ``2``, ``3``, ``4` `}, ` ` ``{ ``2``, ``2``, ``3``, ``4` `}, ` ` ``{ ``3``, ``2``, ``3``, ``4` `}, ` ` ``{ ``4``, ``5``, ``6``, ``7` `}, ` ` ``}; ` ` ``int` `n = ``4``, m = ``4``; ` ` ``System.out.println(wrapper(mat, n, m)); ` ` ` ` ``} ` `} ` ` ` `// This code is contributed by Arnav Kr. Mandal. `
## Python3
`# Python3 program to find longest ` `# increasing path in a matrix. ` `MAX` `=` `20` ` ` `# Return the length of ` `# LIP in 2D matrix ` `def` `LIP(dp, mat, n, m, x, y): ` ` ` ` ``# If value not calculated yet. ` ` ``if` `(dp[x][y] < ``0``): ` ` ``result ``=` `0` ` ` ` ``# If reach bottom left cell, ` ` ``# return 1. ` ` ``if` `(x ``=``=` `n ``-` `1` `and` `y ``=``=` `m ``-` `1``): ` ` ``dp[x][y] ``=` `1` ` ``return` `dp[x][y] ` ` ` ` ``# If reach the corner ` ` ``# of the matrix. ` ` ``if` `(x ``=``=` `n ``-` `1` `or` `y ``=``=` `m ``-` `1``): ` ` ``result ``=` `1` ` ` ` ``# If value greater than below cell. ` ` ``elif` `(mat[x][y] < mat[x ``+` `1``][y]): ` ` ``result ``=` `1` `+` `LIP(dp, mat, n, ` ` ``m, x ``+` `1``, y) ` ` ` ` ``# If value greater than left cell. ` ` ``elif` `(mat[x][y] < mat[x][y ``+` `1``]): ` ` ``result ``=` `max``(result, ``1` `+` `LIP(dp, mat, n, ` ` ``m, x, y ``+` `1``)) ` ` ``dp[x][y] ``=` `result ` ` ``return` `dp[x][y] ` ` ` `# Wrapper function ` `def` `wrapper(mat, n, m): ` ` ``dp ``=` `[[``7` `for` `i ``in` `range``(``MAX``)] ` ` ``for` `i ``in` `range``(``MAX``)] ` ` ``return` `LIP(dp, mat, n, m, ``0``, ``0``) ` ` ` `# Driver Code ` `mat ``=` `[[``1``, ``2``, ``3``, ``4` `], ` ` ``[``2``, ``2``, ``3``, ``4` `], ` ` ``[``3``, ``2``, ``3``, ``4` `], ` ` ``[``4``, ``5``, ``6``, ``7` `]] ` `n ``=` `4` `m ``=` `4` `print``(wrapper(mat, n, m)) ` ` ` `# This code is contributed ` `# by Sahil Shelangia `
## C#
`// C# program to find longest increasing ` `// path in a matrix. ` `using` `System; ` ` ` `public` `class` `GFG ` `{ ` ` ` ` ``// Return the length of LIP in 2D matrix ` ` ``static` `int` `LIP(``int` `[,]dp, ``int` `[,]mat, ``int` `n, ` ` ``int` `m, ``int` `x, ``int` `y) ` ` ``{ ` ` ``// If value not calculated yet. ` ` ``if` `(dp[x,y] < 0) ` ` ``{ ` ` ``int` `result = 0; ` ` ` ` ``// If reach bottom left cell, return 1. ` ` ``if` `(x == n - 1 && y == m - 1) ` ` ``return` `dp[x, y] = 1; ` ` ` ` ``// If reach the corner of the matrix. ` ` ``if` `(x == n - 1 || y == m - 1) ` ` ``result = 1; ` ` ` ` ``// If value greater than below cell. ` ` ``if` `(x + 1 < n && mat[x, y] < mat[x + 1, y]) ` ` ``result = 1 + LIP(dp, mat, n, m, x + 1, y); ` ` ` ` ``// If value greater than left cell. ` ` ``if` `(y + 1 < m && mat[x, y] < mat[x, y + 1]) ` ` ``result = Math.Max(result, 1 + ` ` ``LIP(dp, mat, n, m, x, y + 1)); ` ` ` ` ``dp[x, y] = result; ` ` ``} ` ` ` ` ``return` `dp[x,y]; ` ` ``} ` ` ` ` ``// Wrapper function ` ` ``static` `int` `wrapper(``int` `[,]mat, ``int` `n, ``int` `m) ` ` ``{ ` ` ``int` `[,]dp = ``new` `int``[10, 10]; ` ` ``for``(``int` `i = 0; i < 10; i++) ` ` ``{ ` ` ``for``(``int` `j = 0; j < 10; j++) ` ` ``{ ` ` ``dp[i, j] = -1; ` ` ``} ` ` ``} ` ` ` ` ``return` `LIP(dp, mat, n, m, 0, 0); ` ` ``} ` ` ` ` ``/* Driver code */` ` ``public` `static` `void` `Main() ` ` ``{ ` ` ``int` `[,]mat= { { 1, 2, 3, 4 }, ` ` ``{ 2, 2, 3, 4 }, ` ` ``{ 3, 2, 3, 4 }, ` ` ``{ 4, 5, 6, 7 }, }; ` ` ``int` `n = 4, m = 4; ` ` ``Console.WriteLine(wrapper(mat, n, m)); ` ` ``} ` `} ` ` ` `/* This code contributed by PrinciRaj1992 */`
Output:
```7
```
Time Complexity: O(N*M).
|
<MASK>
### Deepak K Kapur
<MASK>
Now take m=1kg and d=1m
<MASK>
2. May 28, 2015
<MASK>
3. May 28, 2015
<MASK>
4. May 28, 2015
### Deepak K Kapur
But,
1. d=1m and one meter is always one meter. How does 1m depend on time?
<MASK>
Thanks.
<MASK>
### Staff: Mentor
You are confusing mass with matter. You can touch and feel matter. Mass is a property that systems of matter have. It also does not have an independent existence.
<MASK>
So, it means that in this equation no kind of relation between energy and time can be ever found out....is it so??
<MASK>
### Staff: Mentor
<MASK>
### Fredrik
<MASK>
13. May 28, 2015
### Deepak K Kapur
<MASK>
14. May 28, 2015
<MASK>
Cant we say in simple language that when a mass of 1kg moves a distance of 1m then as per E=mc2, we have E=1/t2.
<MASK>
You're just substituting different representations of the same number c in the formula $E=mc^2$! Its like getting the formula $E_k=\frac 1 2 mv^2$ and writing it as $E_k=\frac{16}{32} mv^{\sqrt{4}}$!
EDIT:
More clearly, if you take d=1 m, then your t is 1/c. So the representation you're using is $E=m(\frac{1}{\frac{1}{c}})^2=m(\frac{1}{\frac{1}{299792458}})^2 \approx m(\frac{1}{3.33 \times 10^{-9}})^2$!!!
Last edited: May 28, 2015
16. May 29, 2015
<MASK>
18. May 30, 2015
<MASK>
19. May 30, 2015
<MASK>
You really need to stop trying to break the equation. If you get to a point where you think you have, all it means is that you've confused yourself.
<MASK>
Thanks, u hav been very helpful....
<MASK>
### Staff: Mentor
<MASK>
Staff Emeritus
In non-relativistic classical mechanics, the total energy of a free particle of mass m moving at speed v is $\frac 1 2 mv^2$. In special relativity, it's $\frac{mc^2}{\sqrt{1-\frac{v^2}{c^2}}}$. Note that this is equal to $mc^2$ if and only if $v=0$.
<MASK>
Also plz explain the meaning of 'square' of c?
Last edited: May 30, 2015
24. May 30, 2015
<MASK>
25. May 30, 2015
<MASK>
<UNMASK>
<MASK>
### Deepak K Kapur
If i put c=d/t in
E=mc2, then E=m×d2/t2
Now take m=1kg and d=1m
Does this mean that E is inversely proportional to time?
2. May 28, 2015
### Fredrik
<MASK>
3. May 28, 2015
### PeroK
<MASK>
Look up "dimensional analysis".
4. May 28, 2015
### Deepak K Kapur
But,
1. d=1m and one meter is always one meter. How does 1m depend on time?
<MASK>
Thanks.
<MASK>
6. May 28, 2015
### Fredrik
<MASK>
When d=1 m, then t is (1 m)/(299792458 m/s), so what we get is
$$E=(1~\mathrm{kg}) \frac{(1~\mathrm{m})^2}{\left(\frac{1~\mathrm{m}}{(299792458~\mathrm{m/s}}\right)^2} = (1~\mathrm{kg}) (299792458~\mathrm{m/s})^2 =(299792458)^2~\mathrm{kgm^2/s^2} =(299792458)^2~\mathrm{J}.$$
<MASK>
### Staff: Mentor
You are confusing mass with matter. You can touch and feel matter. Mass is a property that systems of matter have. It also does not have an independent existence.
Besides, there is no reason that the operation of multiplication requires "abstract" or "concrete" entities to be multiplied by themselves. The concepts of "abstract" and "concrete" don't even enter in to the operation.
8. May 28, 2015
<MASK>
So, it means that in this equation no kind of relation between energy and time can be ever found out....is it so??
9. May 28, 2015
<MASK>
10. May 28, 2015
### Staff: Mentor
Mathematical operations are certainly abstract, whether the quantities represented by the math are abstract or not. However, "enters the 'abstract' realm" sounds more like something that you will find in a fantasy novel than in a scientific textbook, so I don't have a direct answer to your question.
11. May 28, 2015
### Fredrik
Staff Emeritus
Not necessarily. But it means that most candidates for such relations can be immediately discarded because they contradict experiments that don't contradict relativity.
<MASK>
12. May 28, 2015
<MASK>
No we don't. The units do not disappear.
13. May 28, 2015
### Deepak K Kapur
<MASK>
14. May 28, 2015
<MASK>
Cant we say in simple language that when a mass of 1kg moves a distance of 1m then as per E=mc2, we have E=1/t2.
<MASK>
### ShayanJ
<MASK>
You're just substituting different representations of the same number c in the formula $E=mc^2$! Its like getting the formula $E_k=\frac 1 2 mv^2$ and writing it as $E_k=\frac{16}{32} mv^{\sqrt{4}}$!
EDIT:
More clearly, if you take d=1 m, then your t is 1/c. So the representation you're using is $E=m(\frac{1}{\frac{1}{c}})^2=m(\frac{1}{\frac{1}{299792458}})^2 \approx m(\frac{1}{3.33 \times 10^{-9}})^2$!!!
Last edited: May 28, 2015
16. May 29, 2015
<MASK>
Last edited: May 29, 2015
17. May 30, 2015
<MASK>
2. How is time constant here? If i put t= 2 sec, the distance gets doubled to keep the speed of light constant. So, how is time constant?
18. May 30, 2015
<MASK>
19. May 30, 2015
### Staff: Mentor
<MASK>
You really need to stop trying to break the equation. If you get to a point where you think you have, all it means is that you've confused yourself.
20. May 30, 2015
<MASK>
I will not argue further on the time issue...
But,
1. if v in KE=1/2mv2 implies the motion of mass, why doesnt c in E=mc2 imply the motion of mass.
2. What does c in this equation mean?
Thanks, u hav been very helpful....
Last edited: May 30, 2015
21. May 30, 2015
### Staff: Mentor
There is no "implies" here: V is the speed of the object, C is the speed of light. Though the two equations bear some similarities, they are not the same equation.
Again, c is the speed of light. Or do you really mean what does the equation mean? The equation provides a conversion that demonstrates the equivalence of matter and energy. So, for example, if you weigh the fuel of a nuclear reactor after it is spent and find it to be less than its starting weight, this equation tells you how much energy was released by converting the lost matter into energy.
http://en.m.wikipedia.org/wiki/Mass–energy_equivalence
22. May 30, 2015
### Fredrik
Staff Emeritus
In non-relativistic classical mechanics, the total energy of a free particle of mass m moving at speed v is $\frac 1 2 mv^2$. In special relativity, it's $\frac{mc^2}{\sqrt{1-\frac{v^2}{c^2}}}$. Note that this is equal to $mc^2$ if and only if $v=0$.
23. May 30, 2015
### Deepak K Kapur
<MASK>
Also plz explain the meaning of 'square' of c?
Last edited: May 30, 2015
24. May 30, 2015
### Deepak K Kapur
Also, i read on the net that one can have c=1 also. If this is the case, what meaning is left in 'conversion' then. If c=1, does it mean that energy and mass are same?
<MASK>
25. May 30, 2015
<MASK>
No, that isn't what it means. I can't really elaborate on something that isn't true.
C^2 is the conversion factor to equate matter and energy.
|
<MASK>
### Deepak K Kapur
If i put c=d/t in
E=mc2, then E=m×d2/t2
Now take m=1kg and d=1m
Does this mean that E is inversely proportional to time?
2. May 28, 2015
### Fredrik
<MASK>
3. May 28, 2015
### PeroK
<MASK>
Look up "dimensional analysis".
4. May 28, 2015
### Deepak K Kapur
But,
1. d=1m and one meter is always one meter. How does 1m depend on time?
<MASK>
Thanks.
<MASK>
6. May 28, 2015
### Fredrik
<MASK>
When d=1 m, then t is (1 m)/(299792458 m/s), so what we get is
$$E=(1~\mathrm{kg}) \frac{(1~\mathrm{m})^2}{\left(\frac{1~\mathrm{m}}{(299792458~\mathrm{m/s}}\right)^2} = (1~\mathrm{kg}) (299792458~\mathrm{m/s})^2 =(299792458)^2~\mathrm{kgm^2/s^2} =(299792458)^2~\mathrm{J}.$$
<MASK>
### Staff: Mentor
You are confusing mass with matter. You can touch and feel matter. Mass is a property that systems of matter have. It also does not have an independent existence.
Besides, there is no reason that the operation of multiplication requires "abstract" or "concrete" entities to be multiplied by themselves. The concepts of "abstract" and "concrete" don't even enter in to the operation.
8. May 28, 2015
<MASK>
So, it means that in this equation no kind of relation between energy and time can be ever found out....is it so??
9. May 28, 2015
<MASK>
10. May 28, 2015
### Staff: Mentor
Mathematical operations are certainly abstract, whether the quantities represented by the math are abstract or not. However, "enters the 'abstract' realm" sounds more like something that you will find in a fantasy novel than in a scientific textbook, so I don't have a direct answer to your question.
11. May 28, 2015
### Fredrik
Staff Emeritus
Not necessarily. But it means that most candidates for such relations can be immediately discarded because they contradict experiments that don't contradict relativity.
<MASK>
12. May 28, 2015
<MASK>
No we don't. The units do not disappear.
13. May 28, 2015
### Deepak K Kapur
<MASK>
14. May 28, 2015
<MASK>
Cant we say in simple language that when a mass of 1kg moves a distance of 1m then as per E=mc2, we have E=1/t2.
<MASK>
### ShayanJ
<MASK>
You're just substituting different representations of the same number c in the formula $E=mc^2$! Its like getting the formula $E_k=\frac 1 2 mv^2$ and writing it as $E_k=\frac{16}{32} mv^{\sqrt{4}}$!
EDIT:
More clearly, if you take d=1 m, then your t is 1/c. So the representation you're using is $E=m(\frac{1}{\frac{1}{c}})^2=m(\frac{1}{\frac{1}{299792458}})^2 \approx m(\frac{1}{3.33 \times 10^{-9}})^2$!!!
Last edited: May 28, 2015
16. May 29, 2015
<MASK>
Last edited: May 29, 2015
17. May 30, 2015
<MASK>
2. How is time constant here? If i put t= 2 sec, the distance gets doubled to keep the speed of light constant. So, how is time constant?
18. May 30, 2015
<MASK>
19. May 30, 2015
### Staff: Mentor
<MASK>
You really need to stop trying to break the equation. If you get to a point where you think you have, all it means is that you've confused yourself.
20. May 30, 2015
<MASK>
I will not argue further on the time issue...
But,
1. if v in KE=1/2mv2 implies the motion of mass, why doesnt c in E=mc2 imply the motion of mass.
2. What does c in this equation mean?
Thanks, u hav been very helpful....
Last edited: May 30, 2015
21. May 30, 2015
### Staff: Mentor
There is no "implies" here: V is the speed of the object, C is the speed of light. Though the two equations bear some similarities, they are not the same equation.
Again, c is the speed of light. Or do you really mean what does the equation mean? The equation provides a conversion that demonstrates the equivalence of matter and energy. So, for example, if you weigh the fuel of a nuclear reactor after it is spent and find it to be less than its starting weight, this equation tells you how much energy was released by converting the lost matter into energy.
http://en.m.wikipedia.org/wiki/Mass–energy_equivalence
22. May 30, 2015
### Fredrik
Staff Emeritus
In non-relativistic classical mechanics, the total energy of a free particle of mass m moving at speed v is $\frac 1 2 mv^2$. In special relativity, it's $\frac{mc^2}{\sqrt{1-\frac{v^2}{c^2}}}$. Note that this is equal to $mc^2$ if and only if $v=0$.
23. May 30, 2015
### Deepak K Kapur
<MASK>
Also plz explain the meaning of 'square' of c?
Last edited: May 30, 2015
24. May 30, 2015
### Deepak K Kapur
Also, i read on the net that one can have c=1 also. If this is the case, what meaning is left in 'conversion' then. If c=1, does it mean that energy and mass are same?
<MASK>
25. May 30, 2015
<MASK>
No, that isn't what it means. I can't really elaborate on something that isn't true.
C^2 is the conversion factor to equate matter and energy.
<UNMASK>
<MASK>
### Deepak K Kapur
If i put c=d/t in
E=mc2, then E=m×d2/t2
Now take m=1kg and d=1m
Does this mean that E is inversely proportional to time?
2. May 28, 2015
### Fredrik
<MASK>
3. May 28, 2015
### PeroK
<MASK>
Look up "dimensional analysis".
4. May 28, 2015
### Deepak K Kapur
But,
1. d=1m and one meter is always one meter. How does 1m depend on time?
<MASK>
Thanks.
<MASK>
### Deepak K Kapur
One more question (silly one).
Mass is something concrete, we can see it, touch it, feel it etc.
Speed is something abstract, we can't see it, touch it etc. apart from mass. It doesnt seem to have independent existence.
Then, how on earth can we multiply a concrete entity with an abstract one??
6. May 28, 2015
### Fredrik
Staff Emeritus
1 m doesn't depend on time, but this d is defined by d=ct. If it's not, then we're no longer talking about velocity c. So d "depends on" t in the sense that the number represented by the variable d can be calculated if you know the number represented by the variable t.
When d=1 m, then t is (1 m)/(299792458 m/s), so what we get is
$$E=(1~\mathrm{kg}) \frac{(1~\mathrm{m})^2}{\left(\frac{1~\mathrm{m}}{(299792458~\mathrm{m/s}}\right)^2} = (1~\mathrm{kg}) (299792458~\mathrm{m/s})^2 =(299792458)^2~\mathrm{kgm^2/s^2} =(299792458)^2~\mathrm{J}.$$
7. May 28, 2015
### Staff: Mentor
You are confusing mass with matter. You can touch and feel matter. Mass is a property that systems of matter have. It also does not have an independent existence.
Besides, there is no reason that the operation of multiplication requires "abstract" or "concrete" entities to be multiplied by themselves. The concepts of "abstract" and "concrete" don't even enter in to the operation.
8. May 28, 2015
### Deepak K Kapur
So, it means that in this equation no kind of relation between energy and time can be ever found out....is it so??
9. May 28, 2015
### Deepak K Kapur
Does it then mean that when we do mathematics ( which is something abstract), then every kind of entity, whether concrete or abstract enters the 'abstract' realm?
10. May 28, 2015
### Staff: Mentor
Mathematical operations are certainly abstract, whether the quantities represented by the math are abstract or not. However, "enters the 'abstract' realm" sounds more like something that you will find in a fantasy novel than in a scientific textbook, so I don't have a direct answer to your question.
11. May 28, 2015
### Fredrik
Staff Emeritus
Not necessarily. But it means that most candidates for such relations can be immediately discarded because they contradict experiments that don't contradict relativity.
I would say that aspects of the real world are represented by abstract mathematical things in the theory.
12. May 28, 2015
### Staff: Mentor
No we don't. The units do not disappear.
13. May 28, 2015
### Deepak K Kapur
<MASK>
14. May 28, 2015
<MASK>
Cant we say in simple language that when a mass of 1kg moves a distance of 1m then as per E=mc2, we have E=1/t2.
<MASK>
15. May 28, 2015
### ShayanJ
t can not be a variable. The point is, $c=299792458 \ \frac m s=\frac{299792458 \ m}{1 \ s}=\frac{149896229 \ m}{.5 \ s}=\frac{59958491.6 \ m}{.2 \ s}=\dots$
<MASK>
You're just substituting different representations of the same number c in the formula $E=mc^2$! Its like getting the formula $E_k=\frac 1 2 mv^2$ and writing it as $E_k=\frac{16}{32} mv^{\sqrt{4}}$!
EDIT:
More clearly, if you take d=1 m, then your t is 1/c. So the representation you're using is $E=m(\frac{1}{\frac{1}{c}})^2=m(\frac{1}{\frac{1}{299792458}})^2 \approx m(\frac{1}{3.33 \times 10^{-9}})^2$!!!
Last edited: May 28, 2015
16. May 29, 2015
### Staff: Mentor
<MASK>
Last edited: May 29, 2015
17. May 30, 2015
<MASK>
1. 'c' in this equation implies that motion of mass is involved. So, why not describe the motion of a kg of mass?
2. How is time constant here? If i put t= 2 sec, the distance gets doubled to keep the speed of light constant. So, how is time constant?
18. May 30, 2015
<MASK>
19. May 30, 2015
### Staff: Mentor
<MASK>
You really need to stop trying to break the equation. If you get to a point where you think you have, all it means is that you've confused yourself.
20. May 30, 2015
### Deepak K Kapur
I will not argue further on the time issue...
But,
1. if v in KE=1/2mv2 implies the motion of mass, why doesnt c in E=mc2 imply the motion of mass.
2. What does c in this equation mean?
Thanks, u hav been very helpful....
Last edited: May 30, 2015
21. May 30, 2015
### Staff: Mentor
There is no "implies" here: V is the speed of the object, C is the speed of light. Though the two equations bear some similarities, they are not the same equation.
Again, c is the speed of light. Or do you really mean what does the equation mean? The equation provides a conversion that demonstrates the equivalence of matter and energy. So, for example, if you weigh the fuel of a nuclear reactor after it is spent and find it to be less than its starting weight, this equation tells you how much energy was released by converting the lost matter into energy.
http://en.m.wikipedia.org/wiki/Mass–energy_equivalence
22. May 30, 2015
### Fredrik
Staff Emeritus
In non-relativistic classical mechanics, the total energy of a free particle of mass m moving at speed v is $\frac 1 2 mv^2$. In special relativity, it's $\frac{mc^2}{\sqrt{1-\frac{v^2}{c^2}}}$. Note that this is equal to $mc^2$ if and only if $v=0$.
23. May 30, 2015
### Deepak K Kapur
<MASK>
Matter will change into energy when...................(some relation to the speed of light)
Cant you elaborate this equation in this way?
Also plz explain the meaning of 'square' of c?
Last edited: May 30, 2015
24. May 30, 2015
### Deepak K Kapur
Also, i read on the net that one can have c=1 also. If this is the case, what meaning is left in 'conversion' then. If c=1, does it mean that energy and mass are same?
<MASK>
25. May 30, 2015
<MASK>
No, that isn't what it means. I can't really elaborate on something that isn't true.
C^2 is the conversion factor to equate matter and energy.
|
<MASK>
### Deepak K Kapur
If i put c=d/t in
E=mc2, then E=m×d2/t2
Now take m=1kg and d=1m
Does this mean that E is inversely proportional to time?
2. May 28, 2015
### Fredrik
<MASK>
3. May 28, 2015
### PeroK
<MASK>
Look up "dimensional analysis".
4. May 28, 2015
### Deepak K Kapur
But,
1. d=1m and one meter is always one meter. How does 1m depend on time?
<MASK>
Thanks.
<MASK>
### Deepak K Kapur
One more question (silly one).
Mass is something concrete, we can see it, touch it, feel it etc.
Speed is something abstract, we can't see it, touch it etc. apart from mass. It doesnt seem to have independent existence.
Then, how on earth can we multiply a concrete entity with an abstract one??
6. May 28, 2015
### Fredrik
Staff Emeritus
1 m doesn't depend on time, but this d is defined by d=ct. If it's not, then we're no longer talking about velocity c. So d "depends on" t in the sense that the number represented by the variable d can be calculated if you know the number represented by the variable t.
When d=1 m, then t is (1 m)/(299792458 m/s), so what we get is
$$E=(1~\mathrm{kg}) \frac{(1~\mathrm{m})^2}{\left(\frac{1~\mathrm{m}}{(299792458~\mathrm{m/s}}\right)^2} = (1~\mathrm{kg}) (299792458~\mathrm{m/s})^2 =(299792458)^2~\mathrm{kgm^2/s^2} =(299792458)^2~\mathrm{J}.$$
7. May 28, 2015
### Staff: Mentor
You are confusing mass with matter. You can touch and feel matter. Mass is a property that systems of matter have. It also does not have an independent existence.
Besides, there is no reason that the operation of multiplication requires "abstract" or "concrete" entities to be multiplied by themselves. The concepts of "abstract" and "concrete" don't even enter in to the operation.
8. May 28, 2015
### Deepak K Kapur
So, it means that in this equation no kind of relation between energy and time can be ever found out....is it so??
9. May 28, 2015
### Deepak K Kapur
Does it then mean that when we do mathematics ( which is something abstract), then every kind of entity, whether concrete or abstract enters the 'abstract' realm?
10. May 28, 2015
### Staff: Mentor
Mathematical operations are certainly abstract, whether the quantities represented by the math are abstract or not. However, "enters the 'abstract' realm" sounds more like something that you will find in a fantasy novel than in a scientific textbook, so I don't have a direct answer to your question.
11. May 28, 2015
### Fredrik
Staff Emeritus
Not necessarily. But it means that most candidates for such relations can be immediately discarded because they contradict experiments that don't contradict relativity.
I would say that aspects of the real world are represented by abstract mathematical things in the theory.
12. May 28, 2015
### Staff: Mentor
No we don't. The units do not disappear.
13. May 28, 2015
### Deepak K Kapur
<MASK>
14. May 28, 2015
<MASK>
Cant we say in simple language that when a mass of 1kg moves a distance of 1m then as per E=mc2, we have E=1/t2.
<MASK>
15. May 28, 2015
### ShayanJ
t can not be a variable. The point is, $c=299792458 \ \frac m s=\frac{299792458 \ m}{1 \ s}=\frac{149896229 \ m}{.5 \ s}=\frac{59958491.6 \ m}{.2 \ s}=\dots$
<MASK>
You're just substituting different representations of the same number c in the formula $E=mc^2$! Its like getting the formula $E_k=\frac 1 2 mv^2$ and writing it as $E_k=\frac{16}{32} mv^{\sqrt{4}}$!
EDIT:
More clearly, if you take d=1 m, then your t is 1/c. So the representation you're using is $E=m(\frac{1}{\frac{1}{c}})^2=m(\frac{1}{\frac{1}{299792458}})^2 \approx m(\frac{1}{3.33 \times 10^{-9}})^2$!!!
Last edited: May 28, 2015
16. May 29, 2015
### Staff: Mentor
<MASK>
Last edited: May 29, 2015
17. May 30, 2015
<MASK>
1. 'c' in this equation implies that motion of mass is involved. So, why not describe the motion of a kg of mass?
2. How is time constant here? If i put t= 2 sec, the distance gets doubled to keep the speed of light constant. So, how is time constant?
18. May 30, 2015
<MASK>
19. May 30, 2015
### Staff: Mentor
<MASK>
You really need to stop trying to break the equation. If you get to a point where you think you have, all it means is that you've confused yourself.
20. May 30, 2015
### Deepak K Kapur
I will not argue further on the time issue...
But,
1. if v in KE=1/2mv2 implies the motion of mass, why doesnt c in E=mc2 imply the motion of mass.
2. What does c in this equation mean?
Thanks, u hav been very helpful....
Last edited: May 30, 2015
21. May 30, 2015
### Staff: Mentor
There is no "implies" here: V is the speed of the object, C is the speed of light. Though the two equations bear some similarities, they are not the same equation.
Again, c is the speed of light. Or do you really mean what does the equation mean? The equation provides a conversion that demonstrates the equivalence of matter and energy. So, for example, if you weigh the fuel of a nuclear reactor after it is spent and find it to be less than its starting weight, this equation tells you how much energy was released by converting the lost matter into energy.
http://en.m.wikipedia.org/wiki/Mass–energy_equivalence
22. May 30, 2015
### Fredrik
Staff Emeritus
In non-relativistic classical mechanics, the total energy of a free particle of mass m moving at speed v is $\frac 1 2 mv^2$. In special relativity, it's $\frac{mc^2}{\sqrt{1-\frac{v^2}{c^2}}}$. Note that this is equal to $mc^2$ if and only if $v=0$.
23. May 30, 2015
### Deepak K Kapur
<MASK>
Matter will change into energy when...................(some relation to the speed of light)
Cant you elaborate this equation in this way?
Also plz explain the meaning of 'square' of c?
Last edited: May 30, 2015
24. May 30, 2015
### Deepak K Kapur
Also, i read on the net that one can have c=1 also. If this is the case, what meaning is left in 'conversion' then. If c=1, does it mean that energy and mass are same?
<MASK>
25. May 30, 2015
<MASK>
No, that isn't what it means. I can't really elaborate on something that isn't true.
C^2 is the conversion factor to equate matter and energy.
<UNMASK>
1. May 28, 2015
### Deepak K Kapur
If i put c=d/t in
E=mc2, then E=m×d2/t2
Now take m=1kg and d=1m
Does this mean that E is inversely proportional to time?
2. May 28, 2015
### Fredrik
<MASK>
3. May 28, 2015
### PeroK
<MASK>
Look up "dimensional analysis".
4. May 28, 2015
### Deepak K Kapur
But,
1. d=1m and one meter is always one meter. How does 1m depend on time?
<MASK>
Thanks.
<MASK>
### Deepak K Kapur
One more question (silly one).
Mass is something concrete, we can see it, touch it, feel it etc.
Speed is something abstract, we can't see it, touch it etc. apart from mass. It doesnt seem to have independent existence.
Then, how on earth can we multiply a concrete entity with an abstract one??
6. May 28, 2015
### Fredrik
Staff Emeritus
1 m doesn't depend on time, but this d is defined by d=ct. If it's not, then we're no longer talking about velocity c. So d "depends on" t in the sense that the number represented by the variable d can be calculated if you know the number represented by the variable t.
When d=1 m, then t is (1 m)/(299792458 m/s), so what we get is
$$E=(1~\mathrm{kg}) \frac{(1~\mathrm{m})^2}{\left(\frac{1~\mathrm{m}}{(299792458~\mathrm{m/s}}\right)^2} = (1~\mathrm{kg}) (299792458~\mathrm{m/s})^2 =(299792458)^2~\mathrm{kgm^2/s^2} =(299792458)^2~\mathrm{J}.$$
7. May 28, 2015
### Staff: Mentor
You are confusing mass with matter. You can touch and feel matter. Mass is a property that systems of matter have. It also does not have an independent existence.
Besides, there is no reason that the operation of multiplication requires "abstract" or "concrete" entities to be multiplied by themselves. The concepts of "abstract" and "concrete" don't even enter in to the operation.
8. May 28, 2015
### Deepak K Kapur
So, it means that in this equation no kind of relation between energy and time can be ever found out....is it so??
9. May 28, 2015
### Deepak K Kapur
Does it then mean that when we do mathematics ( which is something abstract), then every kind of entity, whether concrete or abstract enters the 'abstract' realm?
10. May 28, 2015
### Staff: Mentor
Mathematical operations are certainly abstract, whether the quantities represented by the math are abstract or not. However, "enters the 'abstract' realm" sounds more like something that you will find in a fantasy novel than in a scientific textbook, so I don't have a direct answer to your question.
11. May 28, 2015
### Fredrik
Staff Emeritus
Not necessarily. But it means that most candidates for such relations can be immediately discarded because they contradict experiments that don't contradict relativity.
I would say that aspects of the real world are represented by abstract mathematical things in the theory.
12. May 28, 2015
### Staff: Mentor
No we don't. The units do not disappear.
13. May 28, 2015
### Deepak K Kapur
<MASK>
14. May 28, 2015
<MASK>
Cant we say in simple language that when a mass of 1kg moves a distance of 1m then as per E=mc2, we have E=1/t2.
<MASK>
15. May 28, 2015
### ShayanJ
t can not be a variable. The point is, $c=299792458 \ \frac m s=\frac{299792458 \ m}{1 \ s}=\frac{149896229 \ m}{.5 \ s}=\frac{59958491.6 \ m}{.2 \ s}=\dots$
<MASK>
You're just substituting different representations of the same number c in the formula $E=mc^2$! Its like getting the formula $E_k=\frac 1 2 mv^2$ and writing it as $E_k=\frac{16}{32} mv^{\sqrt{4}}$!
EDIT:
More clearly, if you take d=1 m, then your t is 1/c. So the representation you're using is $E=m(\frac{1}{\frac{1}{c}})^2=m(\frac{1}{\frac{1}{299792458}})^2 \approx m(\frac{1}{3.33 \times 10^{-9}})^2$!!!
Last edited: May 28, 2015
16. May 29, 2015
### Staff: Mentor
<MASK>
Last edited: May 29, 2015
17. May 30, 2015
<MASK>
1. 'c' in this equation implies that motion of mass is involved. So, why not describe the motion of a kg of mass?
2. How is time constant here? If i put t= 2 sec, the distance gets doubled to keep the speed of light constant. So, how is time constant?
18. May 30, 2015
<MASK>
19. May 30, 2015
### Staff: Mentor
<MASK>
You really need to stop trying to break the equation. If you get to a point where you think you have, all it means is that you've confused yourself.
20. May 30, 2015
### Deepak K Kapur
I will not argue further on the time issue...
But,
1. if v in KE=1/2mv2 implies the motion of mass, why doesnt c in E=mc2 imply the motion of mass.
2. What does c in this equation mean?
Thanks, u hav been very helpful....
Last edited: May 30, 2015
21. May 30, 2015
### Staff: Mentor
There is no "implies" here: V is the speed of the object, C is the speed of light. Though the two equations bear some similarities, they are not the same equation.
Again, c is the speed of light. Or do you really mean what does the equation mean? The equation provides a conversion that demonstrates the equivalence of matter and energy. So, for example, if you weigh the fuel of a nuclear reactor after it is spent and find it to be less than its starting weight, this equation tells you how much energy was released by converting the lost matter into energy.
http://en.m.wikipedia.org/wiki/Mass–energy_equivalence
22. May 30, 2015
### Fredrik
Staff Emeritus
In non-relativistic classical mechanics, the total energy of a free particle of mass m moving at speed v is $\frac 1 2 mv^2$. In special relativity, it's $\frac{mc^2}{\sqrt{1-\frac{v^2}{c^2}}}$. Note that this is equal to $mc^2$ if and only if $v=0$.
23. May 30, 2015
### Deepak K Kapur
<MASK>
Matter will change into energy when...................(some relation to the speed of light)
Cant you elaborate this equation in this way?
Also plz explain the meaning of 'square' of c?
Last edited: May 30, 2015
24. May 30, 2015
### Deepak K Kapur
Also, i read on the net that one can have c=1 also. If this is the case, what meaning is left in 'conversion' then. If c=1, does it mean that energy and mass are same?
<MASK>
25. May 30, 2015
### Staff: Mentor
No, that isn't what it means. I can't really elaborate on something that isn't true.
C^2 is the conversion factor to equate matter and energy.
|
1. May 28, 2015
### Deepak K Kapur
If i put c=d/t in
E=mc2, then E=m×d2/t2
Now take m=1kg and d=1m
Does this mean that E is inversely proportional to time?
2. May 28, 2015
### Fredrik
<MASK>
3. May 28, 2015
### PeroK
<MASK>
Look up "dimensional analysis".
4. May 28, 2015
### Deepak K Kapur
But,
1. d=1m and one meter is always one meter. How does 1m depend on time?
<MASK>
Thanks.
<MASK>
### Deepak K Kapur
One more question (silly one).
Mass is something concrete, we can see it, touch it, feel it etc.
Speed is something abstract, we can't see it, touch it etc. apart from mass. It doesnt seem to have independent existence.
Then, how on earth can we multiply a concrete entity with an abstract one??
6. May 28, 2015
### Fredrik
Staff Emeritus
1 m doesn't depend on time, but this d is defined by d=ct. If it's not, then we're no longer talking about velocity c. So d "depends on" t in the sense that the number represented by the variable d can be calculated if you know the number represented by the variable t.
When d=1 m, then t is (1 m)/(299792458 m/s), so what we get is
$$E=(1~\mathrm{kg}) \frac{(1~\mathrm{m})^2}{\left(\frac{1~\mathrm{m}}{(299792458~\mathrm{m/s}}\right)^2} = (1~\mathrm{kg}) (299792458~\mathrm{m/s})^2 =(299792458)^2~\mathrm{kgm^2/s^2} =(299792458)^2~\mathrm{J}.$$
7. May 28, 2015
### Staff: Mentor
You are confusing mass with matter. You can touch and feel matter. Mass is a property that systems of matter have. It also does not have an independent existence.
Besides, there is no reason that the operation of multiplication requires "abstract" or "concrete" entities to be multiplied by themselves. The concepts of "abstract" and "concrete" don't even enter in to the operation.
8. May 28, 2015
### Deepak K Kapur
So, it means that in this equation no kind of relation between energy and time can be ever found out....is it so??
9. May 28, 2015
### Deepak K Kapur
Does it then mean that when we do mathematics ( which is something abstract), then every kind of entity, whether concrete or abstract enters the 'abstract' realm?
10. May 28, 2015
### Staff: Mentor
Mathematical operations are certainly abstract, whether the quantities represented by the math are abstract or not. However, "enters the 'abstract' realm" sounds more like something that you will find in a fantasy novel than in a scientific textbook, so I don't have a direct answer to your question.
11. May 28, 2015
### Fredrik
Staff Emeritus
Not necessarily. But it means that most candidates for such relations can be immediately discarded because they contradict experiments that don't contradict relativity.
I would say that aspects of the real world are represented by abstract mathematical things in the theory.
12. May 28, 2015
### Staff: Mentor
No we don't. The units do not disappear.
13. May 28, 2015
### Deepak K Kapur
<MASK>
14. May 28, 2015
<MASK>
Cant we say in simple language that when a mass of 1kg moves a distance of 1m then as per E=mc2, we have E=1/t2.
<MASK>
15. May 28, 2015
### ShayanJ
t can not be a variable. The point is, $c=299792458 \ \frac m s=\frac{299792458 \ m}{1 \ s}=\frac{149896229 \ m}{.5 \ s}=\frac{59958491.6 \ m}{.2 \ s}=\dots$
<MASK>
You're just substituting different representations of the same number c in the formula $E=mc^2$! Its like getting the formula $E_k=\frac 1 2 mv^2$ and writing it as $E_k=\frac{16}{32} mv^{\sqrt{4}}$!
EDIT:
More clearly, if you take d=1 m, then your t is 1/c. So the representation you're using is $E=m(\frac{1}{\frac{1}{c}})^2=m(\frac{1}{\frac{1}{299792458}})^2 \approx m(\frac{1}{3.33 \times 10^{-9}})^2$!!!
Last edited: May 28, 2015
16. May 29, 2015
### Staff: Mentor
<MASK>
Last edited: May 29, 2015
17. May 30, 2015
<MASK>
1. 'c' in this equation implies that motion of mass is involved. So, why not describe the motion of a kg of mass?
2. How is time constant here? If i put t= 2 sec, the distance gets doubled to keep the speed of light constant. So, how is time constant?
18. May 30, 2015
<MASK>
19. May 30, 2015
### Staff: Mentor
<MASK>
You really need to stop trying to break the equation. If you get to a point where you think you have, all it means is that you've confused yourself.
20. May 30, 2015
### Deepak K Kapur
I will not argue further on the time issue...
But,
1. if v in KE=1/2mv2 implies the motion of mass, why doesnt c in E=mc2 imply the motion of mass.
2. What does c in this equation mean?
Thanks, u hav been very helpful....
Last edited: May 30, 2015
21. May 30, 2015
### Staff: Mentor
There is no "implies" here: V is the speed of the object, C is the speed of light. Though the two equations bear some similarities, they are not the same equation.
Again, c is the speed of light. Or do you really mean what does the equation mean? The equation provides a conversion that demonstrates the equivalence of matter and energy. So, for example, if you weigh the fuel of a nuclear reactor after it is spent and find it to be less than its starting weight, this equation tells you how much energy was released by converting the lost matter into energy.
http://en.m.wikipedia.org/wiki/Mass–energy_equivalence
22. May 30, 2015
### Fredrik
Staff Emeritus
In non-relativistic classical mechanics, the total energy of a free particle of mass m moving at speed v is $\frac 1 2 mv^2$. In special relativity, it's $\frac{mc^2}{\sqrt{1-\frac{v^2}{c^2}}}$. Note that this is equal to $mc^2$ if and only if $v=0$.
23. May 30, 2015
### Deepak K Kapur
<MASK>
Matter will change into energy when...................(some relation to the speed of light)
Cant you elaborate this equation in this way?
Also plz explain the meaning of 'square' of c?
Last edited: May 30, 2015
24. May 30, 2015
### Deepak K Kapur
Also, i read on the net that one can have c=1 also. If this is the case, what meaning is left in 'conversion' then. If c=1, does it mean that energy and mass are same?
<MASK>
25. May 30, 2015
### Staff: Mentor
No, that isn't what it means. I can't really elaborate on something that isn't true.
C^2 is the conversion factor to equate matter and energy.
<UNMASK>
1. May 28, 2015
### Deepak K Kapur
If i put c=d/t in
E=mc2, then E=m×d2/t2
Now take m=1kg and d=1m
Does this mean that E is inversely proportional to time?
2. May 28, 2015
### Fredrik
Staff Emeritus
No, that would be the case if there's a constant A such that E=A/t. If there's a constant A such that E=A/t2, that we could say that E is inversely proportional to time squared. But what you found is E= md2/t2, where m is a constant and d depends on t. So md2 isn't a constant.
3. May 28, 2015
### PeroK
No, but it means that Energy has a "dimension" of $ML^2T^{-2}$.
Compare this with the classical equation $KE = \frac12 mv^2$, where kinetic energy has the same dimension as above.
Look up "dimensional analysis".
4. May 28, 2015
### Deepak K Kapur
But,
1. d=1m and one meter is always one meter. How does 1m depend on time?
2. When d=1m and m=1kg, we get E=1/t2. What is the physical significance of this reduced equation?
Thanks.
5. May 28, 2015
### Deepak K Kapur
One more question (silly one).
Mass is something concrete, we can see it, touch it, feel it etc.
Speed is something abstract, we can't see it, touch it etc. apart from mass. It doesnt seem to have independent existence.
Then, how on earth can we multiply a concrete entity with an abstract one??
6. May 28, 2015
### Fredrik
Staff Emeritus
1 m doesn't depend on time, but this d is defined by d=ct. If it's not, then we're no longer talking about velocity c. So d "depends on" t in the sense that the number represented by the variable d can be calculated if you know the number represented by the variable t.
When d=1 m, then t is (1 m)/(299792458 m/s), so what we get is
$$E=(1~\mathrm{kg}) \frac{(1~\mathrm{m})^2}{\left(\frac{1~\mathrm{m}}{(299792458~\mathrm{m/s}}\right)^2} = (1~\mathrm{kg}) (299792458~\mathrm{m/s})^2 =(299792458)^2~\mathrm{kgm^2/s^2} =(299792458)^2~\mathrm{J}.$$
7. May 28, 2015
### Staff: Mentor
You are confusing mass with matter. You can touch and feel matter. Mass is a property that systems of matter have. It also does not have an independent existence.
Besides, there is no reason that the operation of multiplication requires "abstract" or "concrete" entities to be multiplied by themselves. The concepts of "abstract" and "concrete" don't even enter in to the operation.
8. May 28, 2015
### Deepak K Kapur
So, it means that in this equation no kind of relation between energy and time can be ever found out....is it so??
9. May 28, 2015
### Deepak K Kapur
Does it then mean that when we do mathematics ( which is something abstract), then every kind of entity, whether concrete or abstract enters the 'abstract' realm?
10. May 28, 2015
### Staff: Mentor
Mathematical operations are certainly abstract, whether the quantities represented by the math are abstract or not. However, "enters the 'abstract' realm" sounds more like something that you will find in a fantasy novel than in a scientific textbook, so I don't have a direct answer to your question.
11. May 28, 2015
### Fredrik
Staff Emeritus
Not necessarily. But it means that most candidates for such relations can be immediately discarded because they contradict experiments that don't contradict relativity.
I would say that aspects of the real world are represented by abstract mathematical things in the theory.
12. May 28, 2015
### Staff: Mentor
No we don't. The units do not disappear.
13. May 28, 2015
### Deepak K Kapur
So, can it be said that when we do science or math, we are not concerned with the 'real' things but are only bothered about their properties??
14. May 28, 2015
### Deepak K Kapur
Cant we say in simple language that when a mass of 1kg moves a distance of 1m then as per E=mc2, we have E=1/t2.
Why not, this seems to be a sensible interpretation...
15. May 28, 2015
### ShayanJ
t can not be a variable. The point is, $c=299792458 \ \frac m s=\frac{299792458 \ m}{1 \ s}=\frac{149896229 \ m}{.5 \ s}=\frac{59958491.6 \ m}{.2 \ s}=\dots$
So you can only have $E=m(\frac{299792458}{1})^2=m(\frac{149896229}{.5})^2=m(\frac{59958491.6}{.2})^2=\dots$.
You're just substituting different representations of the same number c in the formula $E=mc^2$! Its like getting the formula $E_k=\frac 1 2 mv^2$ and writing it as $E_k=\frac{16}{32} mv^{\sqrt{4}}$!
EDIT:
More clearly, if you take d=1 m, then your t is 1/c. So the representation you're using is $E=m(\frac{1}{\frac{1}{c}})^2=m(\frac{1}{\frac{1}{299792458}})^2 \approx m(\frac{1}{3.33 \times 10^{-9}})^2$!!!
Last edited: May 28, 2015
16. May 29, 2015
### Staff: Mentor
Well, two reasons:
1. Because the units are still there and you are pretending they aren't. The "1" has units kg-m2. The equation should read E=1 kg-m2/t2.
2. The equation isn't describing the motion of a kilogram of mass, the piece you broke apart is describing the speed of light ("c" is the speed of light). So the 1m is the distance traveled by light in "t" time: time is a constant here, not a variable and you can't change that. So when you plug in m=1 kg, and d= 1m, t=1/300,000,000s and it all simplifies to E=9x10^16 Joules.
Science includes the implied assumption that the observed properties are real and there is nothing else to "concern with" or not "concern with". Beyond that, you're getting into philosophy, not science.
Last edited: May 29, 2015
17. May 30, 2015
### Deepak K Kapur
I dont get your point fully...
1. 'c' in this equation implies that motion of mass is involved. So, why not describe the motion of a kg of mass?
2. How is time constant here? If i put t= 2 sec, the distance gets doubled to keep the speed of light constant. So, how is time constant?
18. May 30, 2015
### ArmanCham
Time constant means this ;
c=x/t If we assume x is one then you cannot change t cause c is constant.So you are telling us E=1/t2 here you assumed x equal one.Then you assumed t =2 We cannot change time If you choose x =1.If you change time you can see easily that you are breaking fundemental physics law "Speed of light is constant"
Here the math c=x/t you said x =1 then t=1/c.Now we found t.Its a number but you are telling it can be three or four.Thats nonsense.
19. May 30, 2015
### Staff: Mentor
You are mistaken. C is just a universal constant, that happens to be the speed of light.
You stated the distance was 1m, which just makes time the new constant in the equation. No matter how you change the distance, the time must change in the opposite way in order that d/t always equals C. That's why plugging-in C=d/t is just an unnecessary complication: C is always going to be the same, so there is no point in calculating it every time you do the problem.
You really need to stop trying to break the equation. If you get to a point where you think you have, all it means is that you've confused yourself.
20. May 30, 2015
### Deepak K Kapur
I will not argue further on the time issue...
But,
1. if v in KE=1/2mv2 implies the motion of mass, why doesnt c in E=mc2 imply the motion of mass.
2. What does c in this equation mean?
Thanks, u hav been very helpful....
Last edited: May 30, 2015
21. May 30, 2015
### Staff: Mentor
There is no "implies" here: V is the speed of the object, C is the speed of light. Though the two equations bear some similarities, they are not the same equation.
Again, c is the speed of light. Or do you really mean what does the equation mean? The equation provides a conversion that demonstrates the equivalence of matter and energy. So, for example, if you weigh the fuel of a nuclear reactor after it is spent and find it to be less than its starting weight, this equation tells you how much energy was released by converting the lost matter into energy.
http://en.m.wikipedia.org/wiki/Mass–energy_equivalence
22. May 30, 2015
### Fredrik
Staff Emeritus
In non-relativistic classical mechanics, the total energy of a free particle of mass m moving at speed v is $\frac 1 2 mv^2$. In special relativity, it's $\frac{mc^2}{\sqrt{1-\frac{v^2}{c^2}}}$. Note that this is equal to $mc^2$ if and only if $v=0$.
23. May 30, 2015
### Deepak K Kapur
Actually i was expecting the meaning of this equation as follows..
Matter will change into energy when...................(some relation to the speed of light)
Cant you elaborate this equation in this way?
Also plz explain the meaning of 'square' of c?
Last edited: May 30, 2015
24. May 30, 2015
### Deepak K Kapur
Also, i read on the net that one can have c=1 also. If this is the case, what meaning is left in 'conversion' then. If c=1, does it mean that energy and mass are same?
If they are same, then do they look different to the 'observer' only or are they 'really' different?
25. May 30, 2015
### Staff: Mentor
No, that isn't what it means. I can't really elaborate on something that isn't true.
C^2 is the conversion factor to equate matter and energy.
|
<MASK>
$\theta$ $\sin(\theta)$ $0$ $0$ $\dfrac{\pi}{2}$ $1$ $\pi$ $0$ $\dfrac{3\pi}{2}$ $-1$ $2\pi$ $0$
<MASK>
A more general sine function is written as
<MASK>
$g(x) = a \sin(b x) + d$
<MASK>
<UNMASK>
<MASK>
$\theta$ $\sin(\theta)$ $0$ $0$ $\dfrac{\pi}{2}$ $1$ $\pi$ $0$ $\dfrac{3\pi}{2}$ $-1$ $2\pi$ $0$
<MASK>
A more general sine function is written as
<MASK>
$g(x) = a \sin(b x) + d$
<MASK>
$y = f(x) = a \sin(b x) + d$ , in red (with no phase shift, c = 0)
<MASK>
|
<MASK>
$\theta$ $\sin(\theta)$ $0$ $0$ $\dfrac{\pi}{2}$ $1$ $\pi$ $0$ $\dfrac{3\pi}{2}$ $-1$ $2\pi$ $0$
<MASK>
A more general sine function is written as
<MASK>
$g(x) = a \sin(b x) + d$
<MASK>
$y = f(x) = a \sin(b x) + d$ , in red (with no phase shift, c = 0)
<MASK>
<UNMASK>
# Sine Function sin x
## Definition and Graph of the Sine Function
<MASK>
The sine function is defined as
$\sin(\theta) = \dfrac{y}{r}$
where $r \$ is the distance from the origin O to any point M on the terminal side of the angle and is given by
$r = \sqrt{x^2+y^2}$
If point M on the terminal side of angle ? is such that OM = r = 1, we may use a circle with radius equal to 1 called unit circle to evaluate the sine function as follows:
$sin(\theta) = y / r = y / 1 = y$ : $\sin(\theta)$ is equal to the y coordinate of a point on the terminal side of an angle in standard position and also on a unit circle.
It is even easier and no calculator is needed to find $\sin(\theta)$ for the quadrantal angles: $0, \dfrac{\pi}{2}, \pi, ...$ as shown in the unit circle below:
The coordinates of the point corresponding to $\theta = 0$ on the unit circle are: (1,0). The y coordinate is equal to 0, hence $\sin(0) = 0$
The coordinates of the point corresponding to $\theta = \dfrac{\pi}{2}$ on the unit circle are: (0,1). The y coordinate is equal to 1, hence $\sin(\dfrac{\pi}{2}) = 1$
and so on.
Let us now put the values of the quadrantal angles angles $0, \dfrac{\pi}{2}, \pi, \dfrac{3\pi}{2} , 2\pi$ and the values of their sine on a table as shown below.
$\theta$ $\sin(\theta)$ $0$ $0$ $\dfrac{\pi}{2}$ $1$ $\pi$ $0$ $\dfrac{3\pi}{2}$ $-1$ $2\pi$ $0$
We now use a system of rectangular axes $(x,y)$ to plot the points in the above table and approximate the graph of the sine function as shown below.
NOTE however that because we are used to $x$ being the variable of a function, $x$ on the graph takes values of $\theta$ and y takes the values of $\sin(\theta)$ which is noted as $y = \sin(x)$.
After $2\pi$, and as $\theta$ increases, the values of $\sin(\theta)$ will repeat at the quadrantal angles. We say that the sine function has a period of $2\pi$ shown below in red.
## General Sine Functions
A more general sine function is written as
$f(x) = a \sin(b x + c) + d$
with its amplitude $|a|$ , period $\dfrac{2\pi}{|b|}$ and phase shift $-\dfrac{c}{b}$ are explored interactively using an html 5 applet. The investigation is carried out by changing the parameters $a, b, c$ and $d$. To deeply understand the effects of each parameter on the graph of the function, we change one parameter at the time at the start. Then later we may change more than one parameter.
Exploration and understanding of the phase shift is done by comparing the shift between the graphs of the two functions:
$f(x) = a \sin(b x + c) + d$
<MASK>
$g(x) = a \sin(b x) + d$
in red as shown in the figure below.
You may also want to consider another tutorial on the trigonometric unit circle .
## Interactive Tutorial Using Html 5 applet
$y = f(x) = a \sin(b x + c) + d$ , in blue (with phase shift)
$y = f(x) = a \sin(b x) + d$ , in red (with no phase shift, c = 0)
<MASK>
How do the 4 coefficients a, b, c and d affect the graph of f(x)?
### Amplitude
Set a = 1, b = 1, c = 0 and d = 0. Write down f(x) and take note of the amplitude, period and phase shift of f(x)? Now change a , how does it affect the graph? The amplitude is defined as $|a|$.
### Period
set a = 1, c = 0, d = 0 and change b. Find the period from the graph and compare it to $\dfrac{2\pi}{|b|}$. How does b affect the perid of the graph of f(x)?
The period is the horizontal distance (along the x-axis) between two points: one is the starting point of a cycle and the second is the end point of the same cycle; two successive maxima or minima for example.
### Phase Shift
set a = 1, b = 1, d = 0 and change parameter c starting from zero going slowly to positive larger values. Take note of the shift, is it left or right? set a = 1, b = 1, d = 0 and change parameter c starting from zero going slowly to negative smaller values. Take note of the shift, is it left or right? repeat the above for b = 2, 3 and 4, measure the shift and compare it to - c/b (the phase shift).
<MASK>
set a, b and c to non zero values and change d. What is the direction of the shift of the graph?
## More References and Links on Sine Functions
Explore interactively the Derivatives of Sine (sin x) Functions
Match Sine Functions to Graphs. Excellent activity where graphs and functions are matched.
Explore interactively the sum of a sine and a cosine functionsSum of Sine and Cosine Functions
Examples with detailed solutions and explanations on sine function problems. Tutorial on Sine Functions (1)- Problems
Tutorial on the relationship between the amplitude, the vertical shift and the maximum and minimum of the sine functionTutorial on Sine Functions (2)- Problems
Trigonometric Functions
Step by step graphing of sine functionsGraph of Sine, a*sin(bx+c), Function
Explore interactively the relationship between the graph of sine function and the coordinates of a point on the unit circle Unit Circle and Trigonometric Functions sin(x), cos(x), tan(x)
The Six Trigonometric Functions Calculator.
|
# Sine Function sin x
## Definition and Graph of the Sine Function
<MASK>
The sine function is defined as
$\sin(\theta) = \dfrac{y}{r}$
where $r \$ is the distance from the origin O to any point M on the terminal side of the angle and is given by
$r = \sqrt{x^2+y^2}$
If point M on the terminal side of angle ? is such that OM = r = 1, we may use a circle with radius equal to 1 called unit circle to evaluate the sine function as follows:
$sin(\theta) = y / r = y / 1 = y$ : $\sin(\theta)$ is equal to the y coordinate of a point on the terminal side of an angle in standard position and also on a unit circle.
It is even easier and no calculator is needed to find $\sin(\theta)$ for the quadrantal angles: $0, \dfrac{\pi}{2}, \pi, ...$ as shown in the unit circle below:
The coordinates of the point corresponding to $\theta = 0$ on the unit circle are: (1,0). The y coordinate is equal to 0, hence $\sin(0) = 0$
The coordinates of the point corresponding to $\theta = \dfrac{\pi}{2}$ on the unit circle are: (0,1). The y coordinate is equal to 1, hence $\sin(\dfrac{\pi}{2}) = 1$
and so on.
Let us now put the values of the quadrantal angles angles $0, \dfrac{\pi}{2}, \pi, \dfrac{3\pi}{2} , 2\pi$ and the values of their sine on a table as shown below.
$\theta$ $\sin(\theta)$ $0$ $0$ $\dfrac{\pi}{2}$ $1$ $\pi$ $0$ $\dfrac{3\pi}{2}$ $-1$ $2\pi$ $0$
We now use a system of rectangular axes $(x,y)$ to plot the points in the above table and approximate the graph of the sine function as shown below.
NOTE however that because we are used to $x$ being the variable of a function, $x$ on the graph takes values of $\theta$ and y takes the values of $\sin(\theta)$ which is noted as $y = \sin(x)$.
After $2\pi$, and as $\theta$ increases, the values of $\sin(\theta)$ will repeat at the quadrantal angles. We say that the sine function has a period of $2\pi$ shown below in red.
## General Sine Functions
A more general sine function is written as
$f(x) = a \sin(b x + c) + d$
with its amplitude $|a|$ , period $\dfrac{2\pi}{|b|}$ and phase shift $-\dfrac{c}{b}$ are explored interactively using an html 5 applet. The investigation is carried out by changing the parameters $a, b, c$ and $d$. To deeply understand the effects of each parameter on the graph of the function, we change one parameter at the time at the start. Then later we may change more than one parameter.
Exploration and understanding of the phase shift is done by comparing the shift between the graphs of the two functions:
$f(x) = a \sin(b x + c) + d$
<MASK>
$g(x) = a \sin(b x) + d$
in red as shown in the figure below.
You may also want to consider another tutorial on the trigonometric unit circle .
## Interactive Tutorial Using Html 5 applet
$y = f(x) = a \sin(b x + c) + d$ , in blue (with phase shift)
$y = f(x) = a \sin(b x) + d$ , in red (with no phase shift, c = 0)
<MASK>
How do the 4 coefficients a, b, c and d affect the graph of f(x)?
### Amplitude
Set a = 1, b = 1, c = 0 and d = 0. Write down f(x) and take note of the amplitude, period and phase shift of f(x)? Now change a , how does it affect the graph? The amplitude is defined as $|a|$.
### Period
set a = 1, c = 0, d = 0 and change b. Find the period from the graph and compare it to $\dfrac{2\pi}{|b|}$. How does b affect the perid of the graph of f(x)?
The period is the horizontal distance (along the x-axis) between two points: one is the starting point of a cycle and the second is the end point of the same cycle; two successive maxima or minima for example.
### Phase Shift
set a = 1, b = 1, d = 0 and change parameter c starting from zero going slowly to positive larger values. Take note of the shift, is it left or right? set a = 1, b = 1, d = 0 and change parameter c starting from zero going slowly to negative smaller values. Take note of the shift, is it left or right? repeat the above for b = 2, 3 and 4, measure the shift and compare it to - c/b (the phase shift).
<MASK>
set a, b and c to non zero values and change d. What is the direction of the shift of the graph?
## More References and Links on Sine Functions
Explore interactively the Derivatives of Sine (sin x) Functions
Match Sine Functions to Graphs. Excellent activity where graphs and functions are matched.
Explore interactively the sum of a sine and a cosine functionsSum of Sine and Cosine Functions
Examples with detailed solutions and explanations on sine function problems. Tutorial on Sine Functions (1)- Problems
Tutorial on the relationship between the amplitude, the vertical shift and the maximum and minimum of the sine functionTutorial on Sine Functions (2)- Problems
Trigonometric Functions
Step by step graphing of sine functionsGraph of Sine, a*sin(bx+c), Function
Explore interactively the relationship between the graph of sine function and the coordinates of a point on the unit circle Unit Circle and Trigonometric Functions sin(x), cos(x), tan(x)
The Six Trigonometric Functions Calculator.
<UNMASK>
# Sine Function sin x
## Definition and Graph of the Sine Function
We first consider angle ? with initial side on the positive x axis (in standard position) and terminal side OM as shown below.
The sine function is defined as
$\sin(\theta) = \dfrac{y}{r}$
where $r \$ is the distance from the origin O to any point M on the terminal side of the angle and is given by
$r = \sqrt{x^2+y^2}$
If point M on the terminal side of angle ? is such that OM = r = 1, we may use a circle with radius equal to 1 called unit circle to evaluate the sine function as follows:
$sin(\theta) = y / r = y / 1 = y$ : $\sin(\theta)$ is equal to the y coordinate of a point on the terminal side of an angle in standard position and also on a unit circle.
It is even easier and no calculator is needed to find $\sin(\theta)$ for the quadrantal angles: $0, \dfrac{\pi}{2}, \pi, ...$ as shown in the unit circle below:
The coordinates of the point corresponding to $\theta = 0$ on the unit circle are: (1,0). The y coordinate is equal to 0, hence $\sin(0) = 0$
The coordinates of the point corresponding to $\theta = \dfrac{\pi}{2}$ on the unit circle are: (0,1). The y coordinate is equal to 1, hence $\sin(\dfrac{\pi}{2}) = 1$
and so on.
Let us now put the values of the quadrantal angles angles $0, \dfrac{\pi}{2}, \pi, \dfrac{3\pi}{2} , 2\pi$ and the values of their sine on a table as shown below.
$\theta$ $\sin(\theta)$ $0$ $0$ $\dfrac{\pi}{2}$ $1$ $\pi$ $0$ $\dfrac{3\pi}{2}$ $-1$ $2\pi$ $0$
We now use a system of rectangular axes $(x,y)$ to plot the points in the above table and approximate the graph of the sine function as shown below.
NOTE however that because we are used to $x$ being the variable of a function, $x$ on the graph takes values of $\theta$ and y takes the values of $\sin(\theta)$ which is noted as $y = \sin(x)$.
After $2\pi$, and as $\theta$ increases, the values of $\sin(\theta)$ will repeat at the quadrantal angles. We say that the sine function has a period of $2\pi$ shown below in red.
## General Sine Functions
A more general sine function is written as
$f(x) = a \sin(b x + c) + d$
with its amplitude $|a|$ , period $\dfrac{2\pi}{|b|}$ and phase shift $-\dfrac{c}{b}$ are explored interactively using an html 5 applet. The investigation is carried out by changing the parameters $a, b, c$ and $d$. To deeply understand the effects of each parameter on the graph of the function, we change one parameter at the time at the start. Then later we may change more than one parameter.
Exploration and understanding of the phase shift is done by comparing the shift between the graphs of the two functions:
$f(x) = a \sin(b x + c) + d$
<MASK>
$g(x) = a \sin(b x) + d$
in red as shown in the figure below.
You may also want to consider another tutorial on the trigonometric unit circle .
## Interactive Tutorial Using Html 5 applet
$y = f(x) = a \sin(b x + c) + d$ , in blue (with phase shift)
$y = f(x) = a \sin(b x) + d$ , in red (with no phase shift, c = 0)
<MASK>
How do the 4 coefficients a, b, c and d affect the graph of f(x)?
### Amplitude
Set a = 1, b = 1, c = 0 and d = 0. Write down f(x) and take note of the amplitude, period and phase shift of f(x)? Now change a , how does it affect the graph? The amplitude is defined as $|a|$.
### Period
set a = 1, c = 0, d = 0 and change b. Find the period from the graph and compare it to $\dfrac{2\pi}{|b|}$. How does b affect the perid of the graph of f(x)?
The period is the horizontal distance (along the x-axis) between two points: one is the starting point of a cycle and the second is the end point of the same cycle; two successive maxima or minima for example.
### Phase Shift
set a = 1, b = 1, d = 0 and change parameter c starting from zero going slowly to positive larger values. Take note of the shift, is it left or right? set a = 1, b = 1, d = 0 and change parameter c starting from zero going slowly to negative smaller values. Take note of the shift, is it left or right? repeat the above for b = 2, 3 and 4, measure the shift and compare it to - c/b (the phase shift).
### Vertical Shift
set a, b and c to non zero values and change d. What is the direction of the shift of the graph?
## More References and Links on Sine Functions
Explore interactively the Derivatives of Sine (sin x) Functions
Match Sine Functions to Graphs. Excellent activity where graphs and functions are matched.
Explore interactively the sum of a sine and a cosine functionsSum of Sine and Cosine Functions
Examples with detailed solutions and explanations on sine function problems. Tutorial on Sine Functions (1)- Problems
Tutorial on the relationship between the amplitude, the vertical shift and the maximum and minimum of the sine functionTutorial on Sine Functions (2)- Problems
Trigonometric Functions
Step by step graphing of sine functionsGraph of Sine, a*sin(bx+c), Function
Explore interactively the relationship between the graph of sine function and the coordinates of a point on the unit circle Unit Circle and Trigonometric Functions sin(x), cos(x), tan(x)
The Six Trigonometric Functions Calculator.
|
# Sine Function sin x
## Definition and Graph of the Sine Function
We first consider angle ? with initial side on the positive x axis (in standard position) and terminal side OM as shown below.
The sine function is defined as
$\sin(\theta) = \dfrac{y}{r}$
where $r \$ is the distance from the origin O to any point M on the terminal side of the angle and is given by
$r = \sqrt{x^2+y^2}$
If point M on the terminal side of angle ? is such that OM = r = 1, we may use a circle with radius equal to 1 called unit circle to evaluate the sine function as follows:
$sin(\theta) = y / r = y / 1 = y$ : $\sin(\theta)$ is equal to the y coordinate of a point on the terminal side of an angle in standard position and also on a unit circle.
It is even easier and no calculator is needed to find $\sin(\theta)$ for the quadrantal angles: $0, \dfrac{\pi}{2}, \pi, ...$ as shown in the unit circle below:
The coordinates of the point corresponding to $\theta = 0$ on the unit circle are: (1,0). The y coordinate is equal to 0, hence $\sin(0) = 0$
The coordinates of the point corresponding to $\theta = \dfrac{\pi}{2}$ on the unit circle are: (0,1). The y coordinate is equal to 1, hence $\sin(\dfrac{\pi}{2}) = 1$
and so on.
Let us now put the values of the quadrantal angles angles $0, \dfrac{\pi}{2}, \pi, \dfrac{3\pi}{2} , 2\pi$ and the values of their sine on a table as shown below.
$\theta$ $\sin(\theta)$ $0$ $0$ $\dfrac{\pi}{2}$ $1$ $\pi$ $0$ $\dfrac{3\pi}{2}$ $-1$ $2\pi$ $0$
We now use a system of rectangular axes $(x,y)$ to plot the points in the above table and approximate the graph of the sine function as shown below.
NOTE however that because we are used to $x$ being the variable of a function, $x$ on the graph takes values of $\theta$ and y takes the values of $\sin(\theta)$ which is noted as $y = \sin(x)$.
After $2\pi$, and as $\theta$ increases, the values of $\sin(\theta)$ will repeat at the quadrantal angles. We say that the sine function has a period of $2\pi$ shown below in red.
## General Sine Functions
A more general sine function is written as
$f(x) = a \sin(b x + c) + d$
with its amplitude $|a|$ , period $\dfrac{2\pi}{|b|}$ and phase shift $-\dfrac{c}{b}$ are explored interactively using an html 5 applet. The investigation is carried out by changing the parameters $a, b, c$ and $d$. To deeply understand the effects of each parameter on the graph of the function, we change one parameter at the time at the start. Then later we may change more than one parameter.
Exploration and understanding of the phase shift is done by comparing the shift between the graphs of the two functions:
$f(x) = a \sin(b x + c) + d$
<MASK>
$g(x) = a \sin(b x) + d$
in red as shown in the figure below.
You may also want to consider another tutorial on the trigonometric unit circle .
## Interactive Tutorial Using Html 5 applet
$y = f(x) = a \sin(b x + c) + d$ , in blue (with phase shift)
$y = f(x) = a \sin(b x) + d$ , in red (with no phase shift, c = 0)
<MASK>
How do the 4 coefficients a, b, c and d affect the graph of f(x)?
### Amplitude
Set a = 1, b = 1, c = 0 and d = 0. Write down f(x) and take note of the amplitude, period and phase shift of f(x)? Now change a , how does it affect the graph? The amplitude is defined as $|a|$.
### Period
set a = 1, c = 0, d = 0 and change b. Find the period from the graph and compare it to $\dfrac{2\pi}{|b|}$. How does b affect the perid of the graph of f(x)?
The period is the horizontal distance (along the x-axis) between two points: one is the starting point of a cycle and the second is the end point of the same cycle; two successive maxima or minima for example.
### Phase Shift
set a = 1, b = 1, d = 0 and change parameter c starting from zero going slowly to positive larger values. Take note of the shift, is it left or right? set a = 1, b = 1, d = 0 and change parameter c starting from zero going slowly to negative smaller values. Take note of the shift, is it left or right? repeat the above for b = 2, 3 and 4, measure the shift and compare it to - c/b (the phase shift).
### Vertical Shift
set a, b and c to non zero values and change d. What is the direction of the shift of the graph?
## More References and Links on Sine Functions
Explore interactively the Derivatives of Sine (sin x) Functions
Match Sine Functions to Graphs. Excellent activity where graphs and functions are matched.
Explore interactively the sum of a sine and a cosine functionsSum of Sine and Cosine Functions
Examples with detailed solutions and explanations on sine function problems. Tutorial on Sine Functions (1)- Problems
Tutorial on the relationship between the amplitude, the vertical shift and the maximum and minimum of the sine functionTutorial on Sine Functions (2)- Problems
Trigonometric Functions
Step by step graphing of sine functionsGraph of Sine, a*sin(bx+c), Function
Explore interactively the relationship between the graph of sine function and the coordinates of a point on the unit circle Unit Circle and Trigonometric Functions sin(x), cos(x), tan(x)
The Six Trigonometric Functions Calculator.
<UNMASK>
# Sine Function sin x
## Definition and Graph of the Sine Function
We first consider angle ? with initial side on the positive x axis (in standard position) and terminal side OM as shown below.
The sine function is defined as
$\sin(\theta) = \dfrac{y}{r}$
where $r \$ is the distance from the origin O to any point M on the terminal side of the angle and is given by
$r = \sqrt{x^2+y^2}$
If point M on the terminal side of angle ? is such that OM = r = 1, we may use a circle with radius equal to 1 called unit circle to evaluate the sine function as follows:
$sin(\theta) = y / r = y / 1 = y$ : $\sin(\theta)$ is equal to the y coordinate of a point on the terminal side of an angle in standard position and also on a unit circle.
It is even easier and no calculator is needed to find $\sin(\theta)$ for the quadrantal angles: $0, \dfrac{\pi}{2}, \pi, ...$ as shown in the unit circle below:
The coordinates of the point corresponding to $\theta = 0$ on the unit circle are: (1,0). The y coordinate is equal to 0, hence $\sin(0) = 0$
The coordinates of the point corresponding to $\theta = \dfrac{\pi}{2}$ on the unit circle are: (0,1). The y coordinate is equal to 1, hence $\sin(\dfrac{\pi}{2}) = 1$
and so on.
Let us now put the values of the quadrantal angles angles $0, \dfrac{\pi}{2}, \pi, \dfrac{3\pi}{2} , 2\pi$ and the values of their sine on a table as shown below.
$\theta$ $\sin(\theta)$ $0$ $0$ $\dfrac{\pi}{2}$ $1$ $\pi$ $0$ $\dfrac{3\pi}{2}$ $-1$ $2\pi$ $0$
We now use a system of rectangular axes $(x,y)$ to plot the points in the above table and approximate the graph of the sine function as shown below.
NOTE however that because we are used to $x$ being the variable of a function, $x$ on the graph takes values of $\theta$ and y takes the values of $\sin(\theta)$ which is noted as $y = \sin(x)$.
After $2\pi$, and as $\theta$ increases, the values of $\sin(\theta)$ will repeat at the quadrantal angles. We say that the sine function has a period of $2\pi$ shown below in red.
## General Sine Functions
A more general sine function is written as
$f(x) = a \sin(b x + c) + d$
with its amplitude $|a|$ , period $\dfrac{2\pi}{|b|}$ and phase shift $-\dfrac{c}{b}$ are explored interactively using an html 5 applet. The investigation is carried out by changing the parameters $a, b, c$ and $d$. To deeply understand the effects of each parameter on the graph of the function, we change one parameter at the time at the start. Then later we may change more than one parameter.
Exploration and understanding of the phase shift is done by comparing the shift between the graphs of the two functions:
$f(x) = a \sin(b x + c) + d$
in blue and
$g(x) = a \sin(b x) + d$
in red as shown in the figure below.
You may also want to consider another tutorial on the trigonometric unit circle .
## Interactive Tutorial Using Html 5 applet
$y = f(x) = a \sin(b x + c) + d$ , in blue (with phase shift)
$y = f(x) = a \sin(b x) + d$ , in red (with no phase shift, c = 0)
a = 1 b = 1 c = 0.5 d = 0
>
How do the 4 coefficients a, b, c and d affect the graph of f(x)?
### Amplitude
Set a = 1, b = 1, c = 0 and d = 0. Write down f(x) and take note of the amplitude, period and phase shift of f(x)? Now change a , how does it affect the graph? The amplitude is defined as $|a|$.
### Period
set a = 1, c = 0, d = 0 and change b. Find the period from the graph and compare it to $\dfrac{2\pi}{|b|}$. How does b affect the perid of the graph of f(x)?
The period is the horizontal distance (along the x-axis) between two points: one is the starting point of a cycle and the second is the end point of the same cycle; two successive maxima or minima for example.
### Phase Shift
set a = 1, b = 1, d = 0 and change parameter c starting from zero going slowly to positive larger values. Take note of the shift, is it left or right? set a = 1, b = 1, d = 0 and change parameter c starting from zero going slowly to negative smaller values. Take note of the shift, is it left or right? repeat the above for b = 2, 3 and 4, measure the shift and compare it to - c/b (the phase shift).
### Vertical Shift
set a, b and c to non zero values and change d. What is the direction of the shift of the graph?
## More References and Links on Sine Functions
Explore interactively the Derivatives of Sine (sin x) Functions
Match Sine Functions to Graphs. Excellent activity where graphs and functions are matched.
Explore interactively the sum of a sine and a cosine functionsSum of Sine and Cosine Functions
Examples with detailed solutions and explanations on sine function problems. Tutorial on Sine Functions (1)- Problems
Tutorial on the relationship between the amplitude, the vertical shift and the maximum and minimum of the sine functionTutorial on Sine Functions (2)- Problems
Trigonometric Functions
Step by step graphing of sine functionsGraph of Sine, a*sin(bx+c), Function
Explore interactively the relationship between the graph of sine function and the coordinates of a point on the unit circle Unit Circle and Trigonometric Functions sin(x), cos(x), tan(x)
The Six Trigonometric Functions Calculator.
|
<MASK>
• Animator.
• Mathematics teacher.
• Fashion designer.
• Plumber.
• Game developer.
• Interior designer.
• Surveyor.
<MASK>
Geometry is a kind of mathematics that deals with shapes and figures. Geometry explains how to build or draw shapes, measure them, and compare them. People use geometry in many kinds of work, from building houses and bridges to planning space travel.
<MASK>
## How far apart should volleyball posts be?
<MASK>
<UNMASK>
<MASK>
A plane figure is two-dimensional, and a solid figure is three-dimensional. The difference between plane and solid figures is in their dimensions. The same shape takes on extra dimension by adding additional points and lines to give the shape height, width and depth.
<MASK>
## What is a solid figure?
<MASK>
• Animator.
• Mathematics teacher.
• Fashion designer.
• Plumber.
• Game developer.
• Interior designer.
• Surveyor.
## Can we live without geometry?
<MASK>
Geometry is a kind of mathematics that deals with shapes and figures. Geometry explains how to build or draw shapes, measure them, and compare them. People use geometry in many kinds of work, from building houses and bridges to planning space travel.
<MASK>
## How far apart should volleyball posts be?
<MASK>
|
<MASK>
A plane figure is two-dimensional, and a solid figure is three-dimensional. The difference between plane and solid figures is in their dimensions. The same shape takes on extra dimension by adding additional points and lines to give the shape height, width and depth.
<MASK>
## What is a solid figure?
<MASK>
• Animator.
• Mathematics teacher.
• Fashion designer.
• Plumber.
• Game developer.
• Interior designer.
• Surveyor.
## Can we live without geometry?
<MASK>
Geometry is a kind of mathematics that deals with shapes and figures. Geometry explains how to build or draw shapes, measure them, and compare them. People use geometry in many kinds of work, from building houses and bridges to planning space travel.
<MASK>
## How far apart should volleyball posts be?
<MASK>
<UNMASK>
<MASK>
## What is the difference between a space figure and a solid figure?
A plane figure is two-dimensional, and a solid figure is three-dimensional. The difference between plane and solid figures is in their dimensions. The same shape takes on extra dimension by adding additional points and lines to give the shape height, width and depth.
<MASK>
## Is a figure in space or space figure?
<MASK>
A triangular prism has a triangle as its base, a rectangular prism has a rectangle as its base, and a cube is a rectangular prism with all its sides of equal length. A cylinder is another type of right prism which has a circle as its base.
## What is a solid figure?
<MASK>
## What jobs do you need geometry for?
<MASK>
• Animator.
• Mathematics teacher.
• Fashion designer.
• Plumber.
• Game developer.
• Interior designer.
• Surveyor.
## Can we live without geometry?
<MASK>
Geometry is a kind of mathematics that deals with shapes and figures. Geometry explains how to build or draw shapes, measure them, and compare them. People use geometry in many kinds of work, from building houses and bridges to planning space travel.
<MASK>
## How far apart should volleyball posts be?
<MASK>
The volleyball court is surrounded by a free zone. The free zone is the area outside the court that players may enter to make a play on the ball. The free zone should be at least 3 meters wide from the court. If any part of the ball crosses the net directly above or outside the antenna, the ball is out of play.
|
<MASK>
## What is the difference between a space figure and a solid figure?
A plane figure is two-dimensional, and a solid figure is three-dimensional. The difference between plane and solid figures is in their dimensions. The same shape takes on extra dimension by adding additional points and lines to give the shape height, width and depth.
<MASK>
## Is a figure in space or space figure?
<MASK>
A triangular prism has a triangle as its base, a rectangular prism has a rectangle as its base, and a cube is a rectangular prism with all its sides of equal length. A cylinder is another type of right prism which has a circle as its base.
## What is a solid figure?
<MASK>
## What jobs do you need geometry for?
<MASK>
• Animator.
• Mathematics teacher.
• Fashion designer.
• Plumber.
• Game developer.
• Interior designer.
• Surveyor.
## Can we live without geometry?
<MASK>
Geometry is a kind of mathematics that deals with shapes and figures. Geometry explains how to build or draw shapes, measure them, and compare them. People use geometry in many kinds of work, from building houses and bridges to planning space travel.
<MASK>
## How far apart should volleyball posts be?
<MASK>
The volleyball court is surrounded by a free zone. The free zone is the area outside the court that players may enter to make a play on the ball. The free zone should be at least 3 meters wide from the court. If any part of the ball crosses the net directly above or outside the antenna, the ball is out of play.
<UNMASK>
<MASK>
## What is the difference between a space figure and a solid figure?
A plane figure is two-dimensional, and a solid figure is three-dimensional. The difference between plane and solid figures is in their dimensions. The same shape takes on extra dimension by adding additional points and lines to give the shape height, width and depth.
## What are the example of spatial figures?
<MASK>
## What is the space figure of volleyball?
<MASK>
## Is a figure in space or space figure?
<MASK>
You might be interested: What large body of water borders mexico on the east
<MASK>
A triangular prism has a triangle as its base, a rectangular prism has a rectangle as its base, and a cube is a rectangular prism with all its sides of equal length. A cylinder is another type of right prism which has a circle as its base.
## What is a solid figure?
<MASK>
## What do you call a space figure that has two circular bases and a curved surface?
<MASK>
## What jobs do you need geometry for?
<MASK>
• Animator.
• Mathematics teacher.
• Fashion designer.
• Plumber.
• Game developer.
• Interior designer.
• Surveyor.
## Can we live without geometry?
Without Geometry things would have been very challenging in Day to day life as well in various technological fields. Lines, Angles, Shapes, 2d & 3d designs plays a vital role in designing of home and commercial infra, mechanical and engineering design. This is feasible only because of Geometry.
## How do you explain geometry to a child?
Geometry is a kind of mathematics that deals with shapes and figures. Geometry explains how to build or draw shapes, measure them, and compare them. People use geometry in many kinds of work, from building houses and bridges to planning space travel.
<MASK>
## How far apart should volleyball posts be?
<MASK>
Volleyball has come a long way from the dusty-old YMCA gymnasium of Holyoke, Massachusetts, USA, where the visionary William G. Morgan invented the sport back in 1895.
<MASK>
The volleyball court is surrounded by a free zone. The free zone is the area outside the court that players may enter to make a play on the ball. The free zone should be at least 3 meters wide from the court. If any part of the ball crosses the net directly above or outside the antenna, the ball is out of play.
|
<MASK>
## What is the difference between a space figure and a solid figure?
A plane figure is two-dimensional, and a solid figure is three-dimensional. The difference between plane and solid figures is in their dimensions. The same shape takes on extra dimension by adding additional points and lines to give the shape height, width and depth.
## What are the example of spatial figures?
<MASK>
## What is the space figure of volleyball?
<MASK>
## Is a figure in space or space figure?
<MASK>
You might be interested: What large body of water borders mexico on the east
<MASK>
A triangular prism has a triangle as its base, a rectangular prism has a rectangle as its base, and a cube is a rectangular prism with all its sides of equal length. A cylinder is another type of right prism which has a circle as its base.
## What is a solid figure?
<MASK>
## What do you call a space figure that has two circular bases and a curved surface?
<MASK>
## What jobs do you need geometry for?
<MASK>
• Animator.
• Mathematics teacher.
• Fashion designer.
• Plumber.
• Game developer.
• Interior designer.
• Surveyor.
## Can we live without geometry?
Without Geometry things would have been very challenging in Day to day life as well in various technological fields. Lines, Angles, Shapes, 2d & 3d designs plays a vital role in designing of home and commercial infra, mechanical and engineering design. This is feasible only because of Geometry.
## How do you explain geometry to a child?
Geometry is a kind of mathematics that deals with shapes and figures. Geometry explains how to build or draw shapes, measure them, and compare them. People use geometry in many kinds of work, from building houses and bridges to planning space travel.
<MASK>
## How far apart should volleyball posts be?
<MASK>
Volleyball has come a long way from the dusty-old YMCA gymnasium of Holyoke, Massachusetts, USA, where the visionary William G. Morgan invented the sport back in 1895.
<MASK>
The volleyball court is surrounded by a free zone. The free zone is the area outside the court that players may enter to make a play on the ball. The free zone should be at least 3 meters wide from the court. If any part of the ball crosses the net directly above or outside the antenna, the ball is out of play.
<UNMASK>
# Often asked: What Is A Space Figure?
A space figure or three-dimensional figure is a figure that has depth in addition to width and height. Some common simple space figures include cubes, spheres, cylinders, prisms, cones, and pyramids. A space figure having all flat faces is called a polyhedron.
## What is the difference between a space figure and a solid figure?
A plane figure is two-dimensional, and a solid figure is three-dimensional. The difference between plane and solid figures is in their dimensions. The same shape takes on extra dimension by adding additional points and lines to give the shape height, width and depth.
## What are the example of spatial figures?
In mathematics, spatial figures are defined simply as three-dimensional objects. For example, a basketball or a cardboard box are spatial figures that we have likely encountered in our lives.
## How are space figures useful?
Three-dimensional geometry, or space geometry, is used to describe the buildings we live and work in, the tools we work with, and the objects we create. They made important discoveries and consequently they got to name the objects they discovered. That’s why geometric figures usually have Greek names!
## What is the space figure of volleyball?
Dimensions. The playing court is 18m long and 9m wide and is surrounded by a free zone 3m wide on all sides. The space above the playing area is known as the free playing space and is a minimum of 7m high from the playing surface.
## Is a figure in space or space figure?
Space figures are figures whose points do not all lie in the same plane. In this unit, we’ll study the polyhedron, the cylinder, the cone, and the sphere. Polyhedrons are space figures with flat surfaces, called faces, which are made of polygons. Prisms and pyramids are examples of polyhedrons.
You might be interested: What large body of water borders mexico on the east
## Can prisms have circular bases?
A triangular prism has a triangle as its base, a rectangular prism has a rectangle as its base, and a cube is a rectangular prism with all its sides of equal length. A cylinder is another type of right prism which has a circle as its base.
## What is a solid figure?
Solid figures are basically three-dimensional objects, which means that they have length, height and width. Because the solid figures have three dimensions, they have depth and take up space in our universe. Solid figures are identified according to the features that are unique to each type of solid.
## What do you call a space figure that has two circular bases and a curved surface?
A cylinder is similar to a prism, but its two bases are circles, not polygons. Also, the sides of a cylinder are curved, not flat.
## What jobs do you need geometry for?
Jobs that use geometry
• Animator.
• Mathematics teacher.
• Fashion designer.
• Plumber.
• Game developer.
• Interior designer.
• Surveyor.
## Can we live without geometry?
Without Geometry things would have been very challenging in Day to day life as well in various technological fields. Lines, Angles, Shapes, 2d & 3d designs plays a vital role in designing of home and commercial infra, mechanical and engineering design. This is feasible only because of Geometry.
## How do you explain geometry to a child?
Geometry is a kind of mathematics that deals with shapes and figures. Geometry explains how to build or draw shapes, measure them, and compare them. People use geometry in many kinds of work, from building houses and bridges to planning space travel.
You might be interested: Who is a famous person from mexico
## How far apart should volleyball posts be?
Posts should be placed 1m (3′-4”) from each side line, 36′-8” from each other. A recommended free or clearance zone of at least 10 ft is recommended. For the most versatile facility, it is recommended to install poles 36′-8” from each other to allow for both competition and recreational play.
## Who created volleyball?
Volleyball has come a long way from the dusty-old YMCA gymnasium of Holyoke, Massachusetts, USA, where the visionary William G. Morgan invented the sport back in 1895.
## Why does a volleyball court need a free zone?
The volleyball court is surrounded by a free zone. The free zone is the area outside the court that players may enter to make a play on the ball. The free zone should be at least 3 meters wide from the court. If any part of the ball crosses the net directly above or outside the antenna, the ball is out of play.
|
<MASK>
<UNMASK>
<MASK>
Fig. 1 Single-layer 2-pole 6-slot integer-slot winding with a winding factor of 1.0.
<MASK>
|
<MASK>
Fig. 1 Single-layer 2-pole 6-slot integer-slot winding with a winding factor of 1.0.
<MASK>
<UNMASK>
<MASK>
Fig. 1 Single-layer 2-pole 6-slot integer-slot winding with a winding factor of 1.0.
<MASK>
Read about another glossary term
|
<MASK>
Fig. 1 Single-layer 2-pole 6-slot integer-slot winding with a winding factor of 1.0.
<MASK>
Read about another glossary term
<UNMASK>
# Reduction of coil span
This value expresses the reduction of the coil span compared to full pitch. The higher the reduction of coil span, the shorter the end-turns of your winding.
In the Emetor winding calculator, the reduction of coil span compared to full pitch is expressed in number of slot pitches.
<MASK>
Single-layer integer-slot windings have always a coil span that is equal to full pitch, ie. the reduction of coil span gets zero. For example, the 2-pole 6-slot winding in Fig. 1 has a coil span of 3, a full pitch of $\frac{Q_s}{p} = \frac{6}{2} = 3$, and consequently a reduction of coil span of 0:
A c B a C b
Fig. 1 Single-layer 2-pole 6-slot integer-slot winding with a winding factor of 1.0.
In order to reduce the coil span, which has the advantage of reducing the length of the end-turns in your winding, a two-layer winding becomes necessary, see Fig. 2. In this case, the 2-pole 6-slot winding has a coil span of only 2, an unchanged full pitch of $\frac{Q_s}{p} = \frac{6}{2} = 3$, and consequently a reduction of coil span of 1 slot pitch:
A c B a C b b A c B a C
<MASK>
Fractional-slot windings and concentrated windings have full pitches that are not integers, as opposed to integer-slot windings. In addition, concentrated windings have always a coil span of one, due to their nature of being wound around a single tooth.
Examples:
Full pitch of 2-pole 9-slot fractional-slot winding: $\frac{Q_s}{p} = \frac{9}{2} = 4.5$
Full pitch of 4-pole 9-slot concentrated winding: $\frac{Q_s}{p} = \frac{9}{4} = 2.25$
Read about another glossary term
|
# Reduction of coil span
This value expresses the reduction of the coil span compared to full pitch. The higher the reduction of coil span, the shorter the end-turns of your winding.
In the Emetor winding calculator, the reduction of coil span compared to full pitch is expressed in number of slot pitches.
<MASK>
Single-layer integer-slot windings have always a coil span that is equal to full pitch, ie. the reduction of coil span gets zero. For example, the 2-pole 6-slot winding in Fig. 1 has a coil span of 3, a full pitch of $\frac{Q_s}{p} = \frac{6}{2} = 3$, and consequently a reduction of coil span of 0:
A c B a C b
Fig. 1 Single-layer 2-pole 6-slot integer-slot winding with a winding factor of 1.0.
In order to reduce the coil span, which has the advantage of reducing the length of the end-turns in your winding, a two-layer winding becomes necessary, see Fig. 2. In this case, the 2-pole 6-slot winding has a coil span of only 2, an unchanged full pitch of $\frac{Q_s}{p} = \frac{6}{2} = 3$, and consequently a reduction of coil span of 1 slot pitch:
A c B a C b b A c B a C
<MASK>
Fractional-slot windings and concentrated windings have full pitches that are not integers, as opposed to integer-slot windings. In addition, concentrated windings have always a coil span of one, due to their nature of being wound around a single tooth.
Examples:
Full pitch of 2-pole 9-slot fractional-slot winding: $\frac{Q_s}{p} = \frac{9}{2} = 4.5$
Full pitch of 4-pole 9-slot concentrated winding: $\frac{Q_s}{p} = \frac{9}{4} = 2.25$
Read about another glossary term
<UNMASK>
# Reduction of coil span
This value expresses the reduction of the coil span compared to full pitch. The higher the reduction of coil span, the shorter the end-turns of your winding.
In the Emetor winding calculator, the reduction of coil span compared to full pitch is expressed in number of slot pitches.
The reduction of coil span can be expressed as: $$\frac{Q_s}{p} - \mbox{coil span,}$$ where $\frac{Q_s}{p}$ is the coil span for full pitch.
Single-layer integer-slot windings have always a coil span that is equal to full pitch, ie. the reduction of coil span gets zero. For example, the 2-pole 6-slot winding in Fig. 1 has a coil span of 3, a full pitch of $\frac{Q_s}{p} = \frac{6}{2} = 3$, and consequently a reduction of coil span of 0:
A c B a C b
Fig. 1 Single-layer 2-pole 6-slot integer-slot winding with a winding factor of 1.0.
In order to reduce the coil span, which has the advantage of reducing the length of the end-turns in your winding, a two-layer winding becomes necessary, see Fig. 2. In this case, the 2-pole 6-slot winding has a coil span of only 2, an unchanged full pitch of $\frac{Q_s}{p} = \frac{6}{2} = 3$, and consequently a reduction of coil span of 1 slot pitch:
A c B a C b b A c B a C
Fig. 2 Double-layer 2-pole 6-slot integer-slot winding with reduced coil span and a winding factor of 0.866.
Fractional-slot windings and concentrated windings have full pitches that are not integers, as opposed to integer-slot windings. In addition, concentrated windings have always a coil span of one, due to their nature of being wound around a single tooth.
Examples:
Full pitch of 2-pole 9-slot fractional-slot winding: $\frac{Q_s}{p} = \frac{9}{2} = 4.5$
Full pitch of 4-pole 9-slot concentrated winding: $\frac{Q_s}{p} = \frac{9}{4} = 2.25$
Read about another glossary term
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Subject: Java Programming - Quadratic Equations Category: Computers > Programming Asked by: java_design-ga List Price: \$15.00 Posted: 29 Nov 2005 05:34 PST Expires: 30 Nov 2005 09:47 PST Question ID: 598931
```Design and develop a Java program that continuously computes and displays value(s) for x, given quadratic equations (i.e. a second-order polynomials) of the form: ax2 + bx + c = 0 where the values for the coefficients a, b and c are supplied by the user, and are assumed to be integers within the range of -100 to 100. To control the loop use a menu interface. The menu should include two options: "Calculate quadratic" and "End". Note that to solve a quadratic equation we must calculate the roots. This can be done using the quadratic formula: root 1 = (-b + sqrt(b2-4ac)) / 2a root2 = (-b - sqrt(b2-4ac)) / 2a Example: x2 + 2x - 8 = 0 a= 1, b = 2, c = -8 roots = (-2 +or- sqrt(22-4x1x-8)) / 2x1 = (-2 +or- sqrt(4+32)) / 2 root1 = (-2 + 6)/2 = 4/2 = 2.0 root2 = (-2 - 6)/2 = -8/2 = -4.0 x = 2.0 or -4.0 However, there are certain special consideration to be taken into account: If a and b are both zero there is no solution (this is referred to as the degenerate case): -8 = 0? a= 0, b = 0, c = -8 (degenerate case) If a is zero and b is non zero the equation becomes a linear equation. 2x - 8 = 0 a= 0, b = 2, c = -8 (Linear equation) root = -c/b = 8/2 = 4.0 x = 4.0 If the value for the term b2 - 4ac (the discriminant) is negative there is no solution (conventionally we cannot find the square root of a negative number!): x2 + 2x + 8 = 0 a= 1, b = 2, c = 8 roots = (-2 +or- sqrt(22-4x1x8)) / 2x1 = (-2 +or- sqrt(4-32)) / 2 = (-2 +or- sqrt(-28)) Negative discriminant therefore no solution. If the discriminant is 0 then there are two identical solutions, i.e. only one solution (root) need be calculated: x2 + 4x + 4 = 0 a= 1, b = 4, c = 4 roots = (-4 +or- sqrt(42-4x1x4)) / 2x1 = (-4 +or- sqrt(16-16)) / 2 (Discriminant = 0, there fore only one solution) root = -4/2 = -2 x = -2.0 Output, where appropriate, should be accurate to at least several decimal places. Please try to include explanations where appropriate.``` Clarification of Question by java_design-ga on 29 Nov 2005 05:39 PST `The program must be written in Java 1.5 !!!`
`I'd do it for \$200. I wouldn't do it for \$15.`
```import java.io.*; public class quadratic { public int a = 0,b = 0,c = 0; public int flag=0; public double r1=0,r2=0; public quadratic() { do { System.out.println("\n\n\n\n\nType 1 to Calculate quadratic equation"); System.out.println("Type 3 to END"); int choice = getChoice(); switch(choice) { case 1: a = inputABC("a"); b = inputABC("b"); c = inputABC("c"); if((Math.pow(b,2)-4*a*c)<0) {System.out.println("\nNegative discriminant therefore no solution!");} else if((Math.pow(b,2)-4*a*c)==0) { r1 = getRoot1(a,b,c); System.out.print("x= "+r1); } else if(a==0&&b==0) {System.out.println("\nDegenerate Case");} else if(a==0) {double u = -c/b; System.out.println("The root is "+u);} else { r1 = getRoot1(a,b,c); r2 = getRoot2(a,b,c); System.out.print("x= "+r1+" or "+r2); } break; case 3: flag = 1; break; default: System.out.println("\nThat is not an option! Try Again."); break; } }while(flag==0); } public int inputABC(String s2) { InputStreamReader stdin = new InputStreamReader(System.in); BufferedReader console = new BufferedReader(stdin); int i1=0; String s1; try { System.out.print("Enter interger for "+s2+": "); s1 = console.readLine(); i1 = Integer.parseInt(s1); } catch(IOException ioex) { System.out.println("\nInput error"); System.exit(1); } catch(NumberFormatException nfex) { System.out.println("\"" + nfex.getMessage() + "\" is not numeric"); System.exit(1); } return(i1); } public int getChoice() { InputStreamReader stdin = new InputStreamReader(System.in); BufferedReader console = new BufferedReader(stdin); int i2=0; String s2; try { System.out.print("User's Choice: "); s2 = console.readLine(); i2 = Integer.parseInt(s2); } catch(IOException ioex) { System.out.println("Input error"); System.exit(1); } catch(NumberFormatException nfex) { System.out.println("\"" + nfex.getMessage() + "\" is not numeric"); System.exit(1); } return(i2); } public double getRoot1(int x, int y, int z) { double root1 = (-y + Math.sqrt((Math.pow(b,2))-(4*x*z)))/(2*x); return(root1); } public double getRoot2(int x, int y, int z) { double root2 = (-y - Math.sqrt((Math.pow(b,2))-(4*x*z)))/(2*x); return(root2); } public static void main(String[] args) { quadratic qd = new quadratic(); } }```
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A121505 Hit triangle for unit circle area (Pi) approximation problem described in A121500. 1
1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0 (list; table; graph; refs; listen; history; text; internal format)
OFFSET 3,1 COMMENTS Record for n=3,4,... only those (n, A121500(n)) pairs which have relative error E(n, A121500(n)) smaller than all errors with smaller n. This produces the table a(n,m). The unit circle area is approximated by the arithmetic mean of the areas of an inscribed regular n-gon and a circumscribed regular m-gon. For each row n>=3 the minimal relative error E(n,m):= ((Fin(n) + Fout(m))/2-Pi)/ Pi) appears for m= A121500(n). The same hit triangle is obtained when one considers the minimal relative errors for the columns m>=3 and collects the sequence with decreasing errors, starting with m=3. LINKS W. Lang: First rows. FORMULA a(n,m) = 1 if m = A121500(n) and E(n,m) < min(E(k,A121500(k)), k=3..n-1), n>=4. a(3,3) = 1, else a(n,m) = 0. EXAMPLE [1], [0,0], [0,1,0], [0, 0, 1, 0], [0, 0, 0, 0, 0], [0, 0, 0, 1, 0, 0],... CROSSREFS Sequence in context: A120524 A014177 A014129 * A014289 A015297 A015073 Adjacent sequences: A121502 A121503 A121504 * A121506 A121507 A121508 KEYWORD nonn,tabl,easy AUTHOR Wolfdieter Lang, Aug 16 2006 STATUS approved
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A121505 Hit triangle for unit circle area (Pi) approximation problem described in A121500. 1
1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0 (list; table; graph; refs; listen; history; text; internal format)
OFFSET 3,1 COMMENTS Record for n=3,4,... only those (n, A121500(n)) pairs which have relative error E(n, A121500(n)) smaller than all errors with smaller n. This produces the table a(n,m). The unit circle area is approximated by the arithmetic mean of the areas of an inscribed regular n-gon and a circumscribed regular m-gon. For each row n>=3 the minimal relative error E(n,m):= ((Fin(n) + Fout(m))/2-Pi)/ Pi) appears for m= A121500(n). The same hit triangle is obtained when one considers the minimal relative errors for the columns m>=3 and collects the sequence with decreasing errors, starting with m=3. LINKS W. Lang: First rows. FORMULA a(n,m) = 1 if m = A121500(n) and E(n,m) < min(E(k,A121500(k)), k=3..n-1), n>=4. a(3,3) = 1, else a(n,m) = 0. EXAMPLE [1], [0,0], [0,1,0], [0, 0, 1, 0], [0, 0, 0, 0, 0], [0, 0, 0, 1, 0, 0],... CROSSREFS Sequence in context: A120524 A014177 A014129 * A014289 A015297 A015073 Adjacent sequences: A121502 A121503 A121504 * A121506 A121507 A121508 KEYWORD nonn,tabl,easy AUTHOR Wolfdieter Lang, Aug 16 2006 STATUS approved
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A121505 Hit triangle for unit circle area (Pi) approximation problem described in A121500. 1
1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0 (list; table; graph; refs; listen; history; text; internal format)
OFFSET 3,1 COMMENTS Record for n=3,4,... only those (n, A121500(n)) pairs which have relative error E(n, A121500(n)) smaller than all errors with smaller n. This produces the table a(n,m). The unit circle area is approximated by the arithmetic mean of the areas of an inscribed regular n-gon and a circumscribed regular m-gon. For each row n>=3 the minimal relative error E(n,m):= ((Fin(n) + Fout(m))/2-Pi)/ Pi) appears for m= A121500(n). The same hit triangle is obtained when one considers the minimal relative errors for the columns m>=3 and collects the sequence with decreasing errors, starting with m=3. LINKS W. Lang: First rows. FORMULA a(n,m) = 1 if m = A121500(n) and E(n,m) < min(E(k,A121500(k)), k=3..n-1), n>=4. a(3,3) = 1, else a(n,m) = 0. EXAMPLE [1], [0,0], [0,1,0], [0, 0, 1, 0], [0, 0, 0, 0, 0], [0, 0, 0, 1, 0, 0],... CROSSREFS Sequence in context: A120524 A014177 A014129 * A014289 A015297 A015073 Adjacent sequences: A121502 A121503 A121504 * A121506 A121507 A121508 KEYWORD nonn,tabl,easy AUTHOR Wolfdieter Lang, Aug 16 2006 STATUS approved
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Last modified March 26 00:37 EDT 2019. Contains 321479 sequences. (Running on oeis4.)
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Using MINITAB perform the regression and correlation analysis for the data on CREDIT BALANCE (Y) and SIZE (X) by answering the following.
1. Generate a scatterplot for CREDIT BALANCE vs. SIZE, including the graph of the "best fit" line. Interpret.
2. Determine the equation of the "best fit" line, which describes the relationship between CREDIT BALANCE and SIZE.
3. Determine the coefficient of correlation. Interpret.
4. Determine the coefficient of determination. Interpret.
5. Test the utility of this regression model (use a two tail test with ? =.05). Interpret your results, including the p-value.
6. Based on your findings in 1-5, what is your opinion about using SIZE to predict CREDIT BALANCE? Explain.
7. Compute the 95% confidence interval for . Interpret this interval.
8. Using an interval, estimate the average credit balance for customers that have household size of 5. Interpret this interval.
9. Using an interval, predict the credit balance for a customer that has a household size of 5. Interpret this interval.
10. What can we say about the credit balance for a customer that has a household size of 10? Explain your answer.
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2. Determine the equation of the best fit line, which describes the relationship between CREDIT BALANCE and SIZE.
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10. What can we say about the credit balance for a customer that has a household size of 10? Explain your answer.
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<MASK>
Using MINITAB perform the regression and correlation analysis for the data on CREDIT BALANCE (Y) and SIZE (X) by answering the following.
1. Generate a scatterplot for CREDIT BALANCE vs. SIZE, including the graph of the "best fit" line. Interpret.
2. Determine the equation of the "best fit" line, which describes the relationship between CREDIT BALANCE and SIZE.
3. Determine the coefficient of correlation. Interpret.
4. Determine the coefficient of determination. Interpret.
5. Test the utility of this regression model (use a two tail test with ? =.05). Interpret your results, including the p-value.
6. Based on your findings in 1-5, what is your opinion about using SIZE to predict CREDIT BALANCE? Explain.
7. Compute the 95% confidence interval for . Interpret this interval.
8. Using an interval, estimate the average credit balance for customers that have household size of 5. Interpret this interval.
9. Using an interval, predict the credit balance for a customer that has a household size of 5. Interpret this interval.
10. What can we say about the credit balance for a customer that has a household size of 10? Explain your answer.
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2. Determine the equation of the best fit line, which describes the relationship between CREDIT BALANCE and SIZE.
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10. What can we say about the credit balance for a customer that has a household size of 10? Explain your answer.
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# Using MINITAB, Perform the Regression and Correlation Analysis
<MASK>
Using MINITAB perform the regression and correlation analysis for the data on CREDIT BALANCE (Y) and SIZE (X) by answering the following.
1. Generate a scatterplot for CREDIT BALANCE vs. SIZE, including the graph of the "best fit" line. Interpret.
2. Determine the equation of the "best fit" line, which describes the relationship between CREDIT BALANCE and SIZE.
3. Determine the coefficient of correlation. Interpret.
4. Determine the coefficient of determination. Interpret.
5. Test the utility of this regression model (use a two tail test with ? =.05). Interpret your results, including the p-value.
6. Based on your findings in 1-5, what is your opinion about using SIZE to predict CREDIT BALANCE? Explain.
7. Compute the 95% confidence interval for . Interpret this interval.
8. Using an interval, estimate the average credit balance for customers that have household size of 5. Interpret this interval.
9. Using an interval, predict the credit balance for a customer that has a household size of 5. Interpret this interval.
10. What can we say about the credit balance for a customer that has a household size of 10? Explain your answer.
<MASK>
Using MINITAB perform the regression and correlation analysis for the data on CREDIT BALANCE (Y) and SIZE (X) by answering the following.
<MASK>
2. Determine the equation of the best fit line, which describes the relationship between CREDIT BALANCE and SIZE.
<MASK>
10. What can we say about the credit balance for a customer that has a household size of 10? Explain your answer.
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12. Perform the Global Test for Utility (F-Test). Explain your conclusion.
<MASK>
Summarize your results from 1-14 in a report that is three pages or less in length and explains and interprets the results in ways that are understandable to someone who does not know statistics.
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# Using MINITAB, Perform the Regression and Correlation Analysis
<MASK>
Using MINITAB perform the regression and correlation analysis for the data on CREDIT BALANCE (Y) and SIZE (X) by answering the following.
1. Generate a scatterplot for CREDIT BALANCE vs. SIZE, including the graph of the "best fit" line. Interpret.
2. Determine the equation of the "best fit" line, which describes the relationship between CREDIT BALANCE and SIZE.
3. Determine the coefficient of correlation. Interpret.
4. Determine the coefficient of determination. Interpret.
5. Test the utility of this regression model (use a two tail test with ? =.05). Interpret your results, including the p-value.
6. Based on your findings in 1-5, what is your opinion about using SIZE to predict CREDIT BALANCE? Explain.
7. Compute the 95% confidence interval for . Interpret this interval.
8. Using an interval, estimate the average credit balance for customers that have household size of 5. Interpret this interval.
9. Using an interval, predict the credit balance for a customer that has a household size of 5. Interpret this interval.
10. What can we say about the credit balance for a customer that has a household size of 10? Explain your answer.
<MASK>
Using MINITAB perform the regression and correlation analysis for the data on CREDIT BALANCE (Y) and SIZE (X) by answering the following.
<MASK>
2. Determine the equation of the best fit line, which describes the relationship between CREDIT BALANCE and SIZE.
<MASK>
10. What can we say about the credit balance for a customer that has a household size of 10? Explain your answer.
<MASK>
12. Perform the Global Test for Utility (F-Test). Explain your conclusion.
<MASK>
Summarize your results from 1-14 in a report that is three pages or less in length and explains and interprets the results in ways that are understandable to someone who does not know statistics.
<MASK>
View Full Posting Details
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# Using MINITAB, Perform the Regression and Correlation Analysis
<MASK>
Using MINITAB perform the regression and correlation analysis for the data on CREDIT BALANCE (Y) and SIZE (X) by answering the following.
1. Generate a scatterplot for CREDIT BALANCE vs. SIZE, including the graph of the "best fit" line. Interpret.
2. Determine the equation of the "best fit" line, which describes the relationship between CREDIT BALANCE and SIZE.
3. Determine the coefficient of correlation. Interpret.
4. Determine the coefficient of determination. Interpret.
5. Test the utility of this regression model (use a two tail test with ? =.05). Interpret your results, including the p-value.
6. Based on your findings in 1-5, what is your opinion about using SIZE to predict CREDIT BALANCE? Explain.
7. Compute the 95% confidence interval for . Interpret this interval.
8. Using an interval, estimate the average credit balance for customers that have household size of 5. Interpret this interval.
9. Using an interval, predict the credit balance for a customer that has a household size of 5. Interpret this interval.
10. What can we say about the credit balance for a customer that has a household size of 10? Explain your answer.
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<MASK>
Using MINITAB perform the regression and correlation analysis for the data on CREDIT BALANCE (Y) and SIZE (X) by answering the following.
<MASK>
2. Determine the equation of the best fit line, which describes the relationship between CREDIT BALANCE and SIZE.
<MASK>
8. Using an interval, estimate the average credit balance for customers that have household size of 5. Interpret this interval.
<MASK>
10. What can we say about the credit balance for a customer that has a household size of 10? Explain your answer.
<MASK>
12. Perform the Global Test for Utility (F-Test). Explain your conclusion.
<MASK>
Summarize your results from 1-14 in a report that is three pages or less in length and explains and interprets the results in ways that are understandable to someone who does not know statistics.
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View Full Posting Details
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# Using MINITAB, Perform the Regression and Correlation Analysis
<MASK>
Using MINITAB perform the regression and correlation analysis for the data on CREDIT BALANCE (Y) and SIZE (X) by answering the following.
1. Generate a scatterplot for CREDIT BALANCE vs. SIZE, including the graph of the "best fit" line. Interpret.
2. Determine the equation of the "best fit" line, which describes the relationship between CREDIT BALANCE and SIZE.
3. Determine the coefficient of correlation. Interpret.
4. Determine the coefficient of determination. Interpret.
5. Test the utility of this regression model (use a two tail test with ? =.05). Interpret your results, including the p-value.
6. Based on your findings in 1-5, what is your opinion about using SIZE to predict CREDIT BALANCE? Explain.
7. Compute the 95% confidence interval for . Interpret this interval.
8. Using an interval, estimate the average credit balance for customers that have household size of 5. Interpret this interval.
9. Using an interval, predict the credit balance for a customer that has a household size of 5. Interpret this interval.
10. What can we say about the credit balance for a customer that has a household size of 10? Explain your answer.
<MASK>
\$2.19
<MASK>
Using MINITAB perform the regression and correlation analysis for the data on CREDIT BALANCE (Y) and SIZE (X) by answering the following.
<MASK>
2. Determine the equation of the best fit line, which describes the relationship between CREDIT BALANCE and SIZE.
<MASK>
8. Using an interval, estimate the average credit balance for customers that have household size of 5. Interpret this interval.
<MASK>
10. What can we say about the credit balance for a customer that has a household size of 10? Explain your answer.
<MASK>
12. Perform the Global Test for Utility (F-Test). Explain your conclusion.
<MASK>
Summarize your results from 1-14 in a report that is three pages or less in length and explains and interprets the results in ways that are understandable to someone who does not know statistics.
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# Using MINITAB, Perform the Regression and Correlation Analysis
This content was STOLEN from BrainMass.com - View the original, and get the already-completed solution here!
Using MINITAB perform the regression and correlation analysis for the data on CREDIT BALANCE (Y) and SIZE (X) by answering the following.
1. Generate a scatterplot for CREDIT BALANCE vs. SIZE, including the graph of the "best fit" line. Interpret.
2. Determine the equation of the "best fit" line, which describes the relationship between CREDIT BALANCE and SIZE.
3. Determine the coefficient of correlation. Interpret.
4. Determine the coefficient of determination. Interpret.
5. Test the utility of this regression model (use a two tail test with ? =.05). Interpret your results, including the p-value.
6. Based on your findings in 1-5, what is your opinion about using SIZE to predict CREDIT BALANCE? Explain.
7. Compute the 95% confidence interval for . Interpret this interval.
8. Using an interval, estimate the average credit balance for customers that have household size of 5. Interpret this interval.
9. Using an interval, predict the credit balance for a customer that has a household size of 5. Interpret this interval.
10. What can we say about the credit balance for a customer that has a household size of 10? Explain your answer.
In an attempt to improve the model, we attempt to do a multiple regression model predicting CREDIT BALANCE based on INCOME, SIZE and YEARS.
11. Using MINITAB run the multiple regression analysis using the variables INCOME, SIZE and YEARS to predict CREDIT BALANCE. State the equation for this multiple regression model.
12. Perform the Global Test for Utility (F-Test). Explain your conclusion.
13. Perform the t-test on each independent variable. Explain your conclusions and clearly state how you should proceed. In particular, which independent variables should we keep and which should be discarded.
14. Is this multiple regression model better than the linear model that we generated in parts 1-10? Explain.
Summarize your results report that is three pages or less in length and explains and interprets the results in ways that are understandable to someone who does not know statistics. Submission: The summary report + (Minitab Output + interpretations) as an appendix.
Suggested format:
A. Summary Report
B. Addressed with appropriate output, graphs and interpretations.
https://brainmass.com/statistics/regression-analysis/using-minitab-perform-the-regression-and-correlation-analysis-508281
#### Solution Summary
The answers (except Q7 which is missing information) are attached in a Word, Excel and .MPJ format including the necessary graphs and charts.
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## Regression and Correlation Analysis.
Using MINITAB perform the regression and correlation analysis for the data on CREDIT BALANCE (Y) and SIZE (X) by answering the following.
1. Generate a scatterplot for CREDIT BALANCE vs. SIZE, including the graph of the best fit line. Interpret.
2. Determine the equation of the best fit line, which describes the relationship between CREDIT BALANCE and SIZE.
3. Determine the coefficient of correlation. Interpret.
4. Determine the index of determination. Interpret.
5. Test the utility of this regression model (use a two tailed test with alpha =.05). Interpret your results, including the p-value.
6. Based on your findings in 1-5, what is your opinion about using SIZE to predict CREDIT BALANCE? Explain.
7. Compute the 95% confidence interval for mean. Interpret this interval.
8. Using an interval, estimate the average credit balance for customers that have household size of 5. Interpret this interval.
9. Using an interval, predict the credit balance for a customer that has a household size of 5. Interpret this interval.
10. What can we say about the credit balance for a customer that has a household size of 10? Explain your answer.
In an attempt to improve the model, we attempt to do a multiple regression model predicting CREDIT BALANCE based on INCOME, SIZE and YEARS.
11. Using MINITAB run the multiple regression analysis using the variables INCOME, SIZE and YEARS to predict CREDIT BALANCE. State the equation for this multiple regression model.
12. Perform the Global Test for Utility (F-Test). Explain your conclusion.
13. Perform the t-test on each independent variable. Explain your conclusions and clearly state how you should proceed. In particular, which independent variables should we keep and which should be discarded.
14. Is this multiple regression model better than the linear model that we generated in parts 1-10? Explain.
Summarize your results from 1-14 in a report that is three pages or less in length and explains and interprets the results in ways that are understandable to someone who does not know statistics.
Please explain in Microsoft Word format. No MINITAB.
View Full Posting Details
|
Convert Yard/second to Picometer/second | pm/s to yd/s
# document.write(document.title);
## yd/s to pm/s Converter
<MASK>
From pm/s to yd/s: 1 pm/s = 1.0936132983333E-12 yd/s;
## How to Convert Yard/second to Picometer/second?
<MASK>
25 yd/s = 25 × 914400000003.66 pm/s = pm/s
To convert Picometer/second to Yard/second, divide your pm/s figure by 914400000003.66.
Example : convert 25 pm/s to yd/s:
25 pm/s = 25 ÷ 914400000003.66 yd/s = yd/s
<MASK>
45 pm/s = 45 × 1.0936132983333E-12 yd/s = yd/s
To convert Yard/second to Picometer/second, divide your yd/s figure by 1.0936132983333E-12.
<MASK>
## Convert Yard/second or Picometer/second to Other Speed and Velocity Units
Yard/second Conversion Table
yd/s to m/s 1 yd/s = 0.9144 m/s yd/s to km/h 1 yd/s = 3.29184 km/h yd/s to mi/h 1 yd/s = 2.0454545455 mi/h yd/s to m/h 1 yd/s = 3291.84 m/h yd/s to m/min 1 yd/s = 54.864 m/min yd/s to km/min 1 yd/s = 0.054864 km/min yd/s to km/s 1 yd/s = 0.0009144 km/s yd/s to cm/h 1 yd/s = 329184 cm/h yd/s to cm/min 1 yd/s = 5486.4 cm/min yd/s to cm/s 1 yd/s = 91.44 cm/s yd/s to mm/h 1 yd/s = 3291840 mm/h yd/s to mm/min 1 yd/s = 54864 mm/min yd/s to mm/s 1 yd/s = 914.4 mm/s yd/s to ft/h 1 yd/s = 10800 ft/h yd/s to ft/min 1 yd/s = 180 ft/min yd/s to ft/s 1 yd/s = 3 ft/s yd/s to yd/h 1 yd/s = 3600 yd/h yd/s to yd/min 1 yd/s = 60 yd/min yd/s to mi/min 1 yd/s = 0.0340909091 mi/min yd/s to mi/s 1 yd/s = 0.0005681818 mi/s yd/s to kt 1 yd/s = 1.7774514039 kt yd/s to kt (UK) 1 yd/s = 1.7763157895 kt (UK) yd/s to c 1 yd/s = 3.050110086E-9 c yd/s to Cosmic velocity - first 1 yd/s = 0.0001157468 Cosmic velocity - first yd/s to Cosmic velocity - second 1 yd/s = 0.0000816429 Cosmic velocity - second yd/s to Cosmic velocity - third 1 yd/s = 0.000054853 Cosmic velocity - third yd/s to Earth's velocity 1 yd/s = 0.0000307206 Earth's velocity yd/s to Velocity of sound in pure water 1 yd/s = 0.0006167128 Velocity of sound in pure water yd/s to Velocity of sound in sea water (20°C, 10 meter deep) 1 yd/s = 0.0006009464 Velocity of sound in sea water (20°C, 10 meter deep) yd/s to Mach (20°C, 1 atm) 1 yd/s = 0.002661234 Mach (20°C, 1 atm) yd/s to Mach (SI standard) 1 yd/s = 0.0030991736 Mach (SI standard) Created @ o.vg Free Unit Converters
Yard/second Conversion Table
yd/s to m/ms 1 yd/s = 0.00091440000000366 m/ms yd/s to m/us 1 yd/s = 9.1440000000366E-7 m/us yd/s to m/ns 1 yd/s = 9.1440000000366E-10 m/ns yd/s to km/ms 1 yd/s = 9.1440000000366E-7 km/ms yd/s to km/us 1 yd/s = 9.1440000000366E-10 km/us yd/s to km/ns 1 yd/s = 9.1440000000366E-13 km/ns yd/s to km/d 1 yd/s = 79.004160000316 km/d yd/s to hm/s 1 yd/s = 0.0091440000000366 hm/s yd/s to dm/s 1 yd/s = 9.1440000000366 dm/s yd/s to um/s 1 yd/s = 914400.00000366 um/s yd/s to nm/s 1 yd/s = 914400000.00366 nm/s yd/s to pm/s 1 yd/s = 914400000003.66 pm/s yd/s to mi/d 1 yd/s = 49.091208091396 mi/d yd/s to nmi/h 1 yd/s = 1.7774514016631 nmi/h yd/s to nautical league (int.)/h 1 yd/s = 0.59248380055437 nautical league (int.)/h yd/s to kyd/s 1 yd/s = 0.01 kyd/s yd/s to in/s 1 yd/s = 36 in/s Created @ o.vg Free Unit Converters
## FAQ
### What is 9 Yard/second in Picometer/second?
<MASK>
### yd/s to pm/s converter in batch
<MASK>
"" at https://www.o.vg/unit/speed/yard-second-to-picometer-second.php from www.o.vg Inc,06/14/2024. https://www.o.vg - Instant, Quick, Free Online Unit Converters
<UNMASK>
Convert Yard/second to Picometer/second | pm/s to yd/s
# document.write(document.title);
## yd/s to pm/s Converter
<MASK>
From pm/s to yd/s: 1 pm/s = 1.0936132983333E-12 yd/s;
## How to Convert Yard/second to Picometer/second?
As we know One yd/s is equal to 914400000003.66 pm/s (1 yd/s = 914400000003.66 pm/s).
To convert Yard/second to Picometer/second, multiply your yd/s figure by 914400000003.66.
<MASK>
25 yd/s = 25 × 914400000003.66 pm/s = pm/s
To convert Picometer/second to Yard/second, divide your pm/s figure by 914400000003.66.
Example : convert 25 pm/s to yd/s:
25 pm/s = 25 ÷ 914400000003.66 yd/s = yd/s
<MASK>
As we know One pm/s is equal to 1.0936132983333E-12 yd/s (1 pm/s = 1.0936132983333E-12 yd/s).
To convert Picometer/second to Yard/second, multiply your pm/s figure by 1.0936132983333E-12.
<MASK>
45 pm/s = 45 × 1.0936132983333E-12 yd/s = yd/s
To convert Yard/second to Picometer/second, divide your yd/s figure by 1.0936132983333E-12.
Example : convert 45 yd/s to pm/s:
<MASK>
## Convert Yard/second or Picometer/second to Other Speed and Velocity Units
Yard/second Conversion Table
yd/s to m/s 1 yd/s = 0.9144 m/s yd/s to km/h 1 yd/s = 3.29184 km/h yd/s to mi/h 1 yd/s = 2.0454545455 mi/h yd/s to m/h 1 yd/s = 3291.84 m/h yd/s to m/min 1 yd/s = 54.864 m/min yd/s to km/min 1 yd/s = 0.054864 km/min yd/s to km/s 1 yd/s = 0.0009144 km/s yd/s to cm/h 1 yd/s = 329184 cm/h yd/s to cm/min 1 yd/s = 5486.4 cm/min yd/s to cm/s 1 yd/s = 91.44 cm/s yd/s to mm/h 1 yd/s = 3291840 mm/h yd/s to mm/min 1 yd/s = 54864 mm/min yd/s to mm/s 1 yd/s = 914.4 mm/s yd/s to ft/h 1 yd/s = 10800 ft/h yd/s to ft/min 1 yd/s = 180 ft/min yd/s to ft/s 1 yd/s = 3 ft/s yd/s to yd/h 1 yd/s = 3600 yd/h yd/s to yd/min 1 yd/s = 60 yd/min yd/s to mi/min 1 yd/s = 0.0340909091 mi/min yd/s to mi/s 1 yd/s = 0.0005681818 mi/s yd/s to kt 1 yd/s = 1.7774514039 kt yd/s to kt (UK) 1 yd/s = 1.7763157895 kt (UK) yd/s to c 1 yd/s = 3.050110086E-9 c yd/s to Cosmic velocity - first 1 yd/s = 0.0001157468 Cosmic velocity - first yd/s to Cosmic velocity - second 1 yd/s = 0.0000816429 Cosmic velocity - second yd/s to Cosmic velocity - third 1 yd/s = 0.000054853 Cosmic velocity - third yd/s to Earth's velocity 1 yd/s = 0.0000307206 Earth's velocity yd/s to Velocity of sound in pure water 1 yd/s = 0.0006167128 Velocity of sound in pure water yd/s to Velocity of sound in sea water (20°C, 10 meter deep) 1 yd/s = 0.0006009464 Velocity of sound in sea water (20°C, 10 meter deep) yd/s to Mach (20°C, 1 atm) 1 yd/s = 0.002661234 Mach (20°C, 1 atm) yd/s to Mach (SI standard) 1 yd/s = 0.0030991736 Mach (SI standard) Created @ o.vg Free Unit Converters
Yard/second Conversion Table
yd/s to m/ms 1 yd/s = 0.00091440000000366 m/ms yd/s to m/us 1 yd/s = 9.1440000000366E-7 m/us yd/s to m/ns 1 yd/s = 9.1440000000366E-10 m/ns yd/s to km/ms 1 yd/s = 9.1440000000366E-7 km/ms yd/s to km/us 1 yd/s = 9.1440000000366E-10 km/us yd/s to km/ns 1 yd/s = 9.1440000000366E-13 km/ns yd/s to km/d 1 yd/s = 79.004160000316 km/d yd/s to hm/s 1 yd/s = 0.0091440000000366 hm/s yd/s to dm/s 1 yd/s = 9.1440000000366 dm/s yd/s to um/s 1 yd/s = 914400.00000366 um/s yd/s to nm/s 1 yd/s = 914400000.00366 nm/s yd/s to pm/s 1 yd/s = 914400000003.66 pm/s yd/s to mi/d 1 yd/s = 49.091208091396 mi/d yd/s to nmi/h 1 yd/s = 1.7774514016631 nmi/h yd/s to nautical league (int.)/h 1 yd/s = 0.59248380055437 nautical league (int.)/h yd/s to kyd/s 1 yd/s = 0.01 kyd/s yd/s to in/s 1 yd/s = 36 in/s Created @ o.vg Free Unit Converters
## FAQ
### What is 9 Yard/second in Picometer/second?
pm/s. Since one yd/s equals 914400000003.66 pm/s, 9 yd/s in pm/s will be pm/s.
<MASK>
1 pm/s is approximately equal to 1.0936132983333E-12 yd/s.
<MASK>
### yd/s to pm/s converter in batch
<MASK>
"" at https://www.o.vg/unit/speed/yard-second-to-picometer-second.php from www.o.vg Inc,06/14/2024. https://www.o.vg - Instant, Quick, Free Online Unit Converters
|
Convert Yard/second to Picometer/second | pm/s to yd/s
# document.write(document.title);
## yd/s to pm/s Converter
<MASK>
From pm/s to yd/s: 1 pm/s = 1.0936132983333E-12 yd/s;
## How to Convert Yard/second to Picometer/second?
As we know One yd/s is equal to 914400000003.66 pm/s (1 yd/s = 914400000003.66 pm/s).
To convert Yard/second to Picometer/second, multiply your yd/s figure by 914400000003.66.
<MASK>
25 yd/s = 25 × 914400000003.66 pm/s = pm/s
To convert Picometer/second to Yard/second, divide your pm/s figure by 914400000003.66.
Example : convert 25 pm/s to yd/s:
25 pm/s = 25 ÷ 914400000003.66 yd/s = yd/s
<MASK>
As we know One pm/s is equal to 1.0936132983333E-12 yd/s (1 pm/s = 1.0936132983333E-12 yd/s).
To convert Picometer/second to Yard/second, multiply your pm/s figure by 1.0936132983333E-12.
<MASK>
45 pm/s = 45 × 1.0936132983333E-12 yd/s = yd/s
To convert Yard/second to Picometer/second, divide your yd/s figure by 1.0936132983333E-12.
Example : convert 45 yd/s to pm/s:
<MASK>
## Convert Yard/second or Picometer/second to Other Speed and Velocity Units
Yard/second Conversion Table
yd/s to m/s 1 yd/s = 0.9144 m/s yd/s to km/h 1 yd/s = 3.29184 km/h yd/s to mi/h 1 yd/s = 2.0454545455 mi/h yd/s to m/h 1 yd/s = 3291.84 m/h yd/s to m/min 1 yd/s = 54.864 m/min yd/s to km/min 1 yd/s = 0.054864 km/min yd/s to km/s 1 yd/s = 0.0009144 km/s yd/s to cm/h 1 yd/s = 329184 cm/h yd/s to cm/min 1 yd/s = 5486.4 cm/min yd/s to cm/s 1 yd/s = 91.44 cm/s yd/s to mm/h 1 yd/s = 3291840 mm/h yd/s to mm/min 1 yd/s = 54864 mm/min yd/s to mm/s 1 yd/s = 914.4 mm/s yd/s to ft/h 1 yd/s = 10800 ft/h yd/s to ft/min 1 yd/s = 180 ft/min yd/s to ft/s 1 yd/s = 3 ft/s yd/s to yd/h 1 yd/s = 3600 yd/h yd/s to yd/min 1 yd/s = 60 yd/min yd/s to mi/min 1 yd/s = 0.0340909091 mi/min yd/s to mi/s 1 yd/s = 0.0005681818 mi/s yd/s to kt 1 yd/s = 1.7774514039 kt yd/s to kt (UK) 1 yd/s = 1.7763157895 kt (UK) yd/s to c 1 yd/s = 3.050110086E-9 c yd/s to Cosmic velocity - first 1 yd/s = 0.0001157468 Cosmic velocity - first yd/s to Cosmic velocity - second 1 yd/s = 0.0000816429 Cosmic velocity - second yd/s to Cosmic velocity - third 1 yd/s = 0.000054853 Cosmic velocity - third yd/s to Earth's velocity 1 yd/s = 0.0000307206 Earth's velocity yd/s to Velocity of sound in pure water 1 yd/s = 0.0006167128 Velocity of sound in pure water yd/s to Velocity of sound in sea water (20°C, 10 meter deep) 1 yd/s = 0.0006009464 Velocity of sound in sea water (20°C, 10 meter deep) yd/s to Mach (20°C, 1 atm) 1 yd/s = 0.002661234 Mach (20°C, 1 atm) yd/s to Mach (SI standard) 1 yd/s = 0.0030991736 Mach (SI standard) Created @ o.vg Free Unit Converters
Yard/second Conversion Table
yd/s to m/ms 1 yd/s = 0.00091440000000366 m/ms yd/s to m/us 1 yd/s = 9.1440000000366E-7 m/us yd/s to m/ns 1 yd/s = 9.1440000000366E-10 m/ns yd/s to km/ms 1 yd/s = 9.1440000000366E-7 km/ms yd/s to km/us 1 yd/s = 9.1440000000366E-10 km/us yd/s to km/ns 1 yd/s = 9.1440000000366E-13 km/ns yd/s to km/d 1 yd/s = 79.004160000316 km/d yd/s to hm/s 1 yd/s = 0.0091440000000366 hm/s yd/s to dm/s 1 yd/s = 9.1440000000366 dm/s yd/s to um/s 1 yd/s = 914400.00000366 um/s yd/s to nm/s 1 yd/s = 914400000.00366 nm/s yd/s to pm/s 1 yd/s = 914400000003.66 pm/s yd/s to mi/d 1 yd/s = 49.091208091396 mi/d yd/s to nmi/h 1 yd/s = 1.7774514016631 nmi/h yd/s to nautical league (int.)/h 1 yd/s = 0.59248380055437 nautical league (int.)/h yd/s to kyd/s 1 yd/s = 0.01 kyd/s yd/s to in/s 1 yd/s = 36 in/s Created @ o.vg Free Unit Converters
## FAQ
### What is 9 Yard/second in Picometer/second?
pm/s. Since one yd/s equals 914400000003.66 pm/s, 9 yd/s in pm/s will be pm/s.
<MASK>
1 pm/s is approximately equal to 1.0936132983333E-12 yd/s.
<MASK>
### yd/s to pm/s converter in batch
<MASK>
"" at https://www.o.vg/unit/speed/yard-second-to-picometer-second.php from www.o.vg Inc,06/14/2024. https://www.o.vg - Instant, Quick, Free Online Unit Converters
<UNMASK>
Convert Yard/second to Picometer/second | pm/s to yd/s
# document.write(document.title);
## yd/s to pm/s Converter
<MASK>
From pm/s to yd/s: 1 pm/s = 1.0936132983333E-12 yd/s;
## How to Convert Yard/second to Picometer/second?
As we know One yd/s is equal to 914400000003.66 pm/s (1 yd/s = 914400000003.66 pm/s).
To convert Yard/second to Picometer/second, multiply your yd/s figure by 914400000003.66.
<MASK>
25 yd/s = 25 × 914400000003.66 pm/s = pm/s
To convert Picometer/second to Yard/second, divide your pm/s figure by 914400000003.66.
Example : convert 25 pm/s to yd/s:
25 pm/s = 25 ÷ 914400000003.66 yd/s = yd/s
## How to Convert Picometer/second to Yard/second?
As we know One pm/s is equal to 1.0936132983333E-12 yd/s (1 pm/s = 1.0936132983333E-12 yd/s).
To convert Picometer/second to Yard/second, multiply your pm/s figure by 1.0936132983333E-12.
Example : convert 45 pm/s to yd/s:
45 pm/s = 45 × 1.0936132983333E-12 yd/s = yd/s
To convert Yard/second to Picometer/second, divide your yd/s figure by 1.0936132983333E-12.
Example : convert 45 yd/s to pm/s:
<MASK>
## Convert Yard/second or Picometer/second to Other Speed and Velocity Units
Yard/second Conversion Table
yd/s to m/s 1 yd/s = 0.9144 m/s yd/s to km/h 1 yd/s = 3.29184 km/h yd/s to mi/h 1 yd/s = 2.0454545455 mi/h yd/s to m/h 1 yd/s = 3291.84 m/h yd/s to m/min 1 yd/s = 54.864 m/min yd/s to km/min 1 yd/s = 0.054864 km/min yd/s to km/s 1 yd/s = 0.0009144 km/s yd/s to cm/h 1 yd/s = 329184 cm/h yd/s to cm/min 1 yd/s = 5486.4 cm/min yd/s to cm/s 1 yd/s = 91.44 cm/s yd/s to mm/h 1 yd/s = 3291840 mm/h yd/s to mm/min 1 yd/s = 54864 mm/min yd/s to mm/s 1 yd/s = 914.4 mm/s yd/s to ft/h 1 yd/s = 10800 ft/h yd/s to ft/min 1 yd/s = 180 ft/min yd/s to ft/s 1 yd/s = 3 ft/s yd/s to yd/h 1 yd/s = 3600 yd/h yd/s to yd/min 1 yd/s = 60 yd/min yd/s to mi/min 1 yd/s = 0.0340909091 mi/min yd/s to mi/s 1 yd/s = 0.0005681818 mi/s yd/s to kt 1 yd/s = 1.7774514039 kt yd/s to kt (UK) 1 yd/s = 1.7763157895 kt (UK) yd/s to c 1 yd/s = 3.050110086E-9 c yd/s to Cosmic velocity - first 1 yd/s = 0.0001157468 Cosmic velocity - first yd/s to Cosmic velocity - second 1 yd/s = 0.0000816429 Cosmic velocity - second yd/s to Cosmic velocity - third 1 yd/s = 0.000054853 Cosmic velocity - third yd/s to Earth's velocity 1 yd/s = 0.0000307206 Earth's velocity yd/s to Velocity of sound in pure water 1 yd/s = 0.0006167128 Velocity of sound in pure water yd/s to Velocity of sound in sea water (20°C, 10 meter deep) 1 yd/s = 0.0006009464 Velocity of sound in sea water (20°C, 10 meter deep) yd/s to Mach (20°C, 1 atm) 1 yd/s = 0.002661234 Mach (20°C, 1 atm) yd/s to Mach (SI standard) 1 yd/s = 0.0030991736 Mach (SI standard) Created @ o.vg Free Unit Converters
Yard/second Conversion Table
yd/s to m/ms 1 yd/s = 0.00091440000000366 m/ms yd/s to m/us 1 yd/s = 9.1440000000366E-7 m/us yd/s to m/ns 1 yd/s = 9.1440000000366E-10 m/ns yd/s to km/ms 1 yd/s = 9.1440000000366E-7 km/ms yd/s to km/us 1 yd/s = 9.1440000000366E-10 km/us yd/s to km/ns 1 yd/s = 9.1440000000366E-13 km/ns yd/s to km/d 1 yd/s = 79.004160000316 km/d yd/s to hm/s 1 yd/s = 0.0091440000000366 hm/s yd/s to dm/s 1 yd/s = 9.1440000000366 dm/s yd/s to um/s 1 yd/s = 914400.00000366 um/s yd/s to nm/s 1 yd/s = 914400000.00366 nm/s yd/s to pm/s 1 yd/s = 914400000003.66 pm/s yd/s to mi/d 1 yd/s = 49.091208091396 mi/d yd/s to nmi/h 1 yd/s = 1.7774514016631 nmi/h yd/s to nautical league (int.)/h 1 yd/s = 0.59248380055437 nautical league (int.)/h yd/s to kyd/s 1 yd/s = 0.01 kyd/s yd/s to in/s 1 yd/s = 36 in/s Created @ o.vg Free Unit Converters
## FAQ
### What is 9 Yard/second in Picometer/second?
pm/s. Since one yd/s equals 914400000003.66 pm/s, 9 yd/s in pm/s will be pm/s.
### How many Picometer/second are in a Yard/second?
<MASK>
### How many yd/s is equal to 1 pm/s?
1 pm/s is approximately equal to 1.0936132983333E-12 yd/s.
### What is the yd/s value of 8 pm/s?
<MASK>
### yd/s to pm/s converter in batch
<MASK>
"" at https://www.o.vg/unit/speed/yard-second-to-picometer-second.php from www.o.vg Inc,06/14/2024. https://www.o.vg - Instant, Quick, Free Online Unit Converters
|
Convert Yard/second to Picometer/second | pm/s to yd/s
# document.write(document.title);
## yd/s to pm/s Converter
<MASK>
From pm/s to yd/s: 1 pm/s = 1.0936132983333E-12 yd/s;
## How to Convert Yard/second to Picometer/second?
As we know One yd/s is equal to 914400000003.66 pm/s (1 yd/s = 914400000003.66 pm/s).
To convert Yard/second to Picometer/second, multiply your yd/s figure by 914400000003.66.
<MASK>
25 yd/s = 25 × 914400000003.66 pm/s = pm/s
To convert Picometer/second to Yard/second, divide your pm/s figure by 914400000003.66.
Example : convert 25 pm/s to yd/s:
25 pm/s = 25 ÷ 914400000003.66 yd/s = yd/s
## How to Convert Picometer/second to Yard/second?
As we know One pm/s is equal to 1.0936132983333E-12 yd/s (1 pm/s = 1.0936132983333E-12 yd/s).
To convert Picometer/second to Yard/second, multiply your pm/s figure by 1.0936132983333E-12.
Example : convert 45 pm/s to yd/s:
45 pm/s = 45 × 1.0936132983333E-12 yd/s = yd/s
To convert Yard/second to Picometer/second, divide your yd/s figure by 1.0936132983333E-12.
Example : convert 45 yd/s to pm/s:
<MASK>
## Convert Yard/second or Picometer/second to Other Speed and Velocity Units
Yard/second Conversion Table
yd/s to m/s 1 yd/s = 0.9144 m/s yd/s to km/h 1 yd/s = 3.29184 km/h yd/s to mi/h 1 yd/s = 2.0454545455 mi/h yd/s to m/h 1 yd/s = 3291.84 m/h yd/s to m/min 1 yd/s = 54.864 m/min yd/s to km/min 1 yd/s = 0.054864 km/min yd/s to km/s 1 yd/s = 0.0009144 km/s yd/s to cm/h 1 yd/s = 329184 cm/h yd/s to cm/min 1 yd/s = 5486.4 cm/min yd/s to cm/s 1 yd/s = 91.44 cm/s yd/s to mm/h 1 yd/s = 3291840 mm/h yd/s to mm/min 1 yd/s = 54864 mm/min yd/s to mm/s 1 yd/s = 914.4 mm/s yd/s to ft/h 1 yd/s = 10800 ft/h yd/s to ft/min 1 yd/s = 180 ft/min yd/s to ft/s 1 yd/s = 3 ft/s yd/s to yd/h 1 yd/s = 3600 yd/h yd/s to yd/min 1 yd/s = 60 yd/min yd/s to mi/min 1 yd/s = 0.0340909091 mi/min yd/s to mi/s 1 yd/s = 0.0005681818 mi/s yd/s to kt 1 yd/s = 1.7774514039 kt yd/s to kt (UK) 1 yd/s = 1.7763157895 kt (UK) yd/s to c 1 yd/s = 3.050110086E-9 c yd/s to Cosmic velocity - first 1 yd/s = 0.0001157468 Cosmic velocity - first yd/s to Cosmic velocity - second 1 yd/s = 0.0000816429 Cosmic velocity - second yd/s to Cosmic velocity - third 1 yd/s = 0.000054853 Cosmic velocity - third yd/s to Earth's velocity 1 yd/s = 0.0000307206 Earth's velocity yd/s to Velocity of sound in pure water 1 yd/s = 0.0006167128 Velocity of sound in pure water yd/s to Velocity of sound in sea water (20°C, 10 meter deep) 1 yd/s = 0.0006009464 Velocity of sound in sea water (20°C, 10 meter deep) yd/s to Mach (20°C, 1 atm) 1 yd/s = 0.002661234 Mach (20°C, 1 atm) yd/s to Mach (SI standard) 1 yd/s = 0.0030991736 Mach (SI standard) Created @ o.vg Free Unit Converters
Yard/second Conversion Table
yd/s to m/ms 1 yd/s = 0.00091440000000366 m/ms yd/s to m/us 1 yd/s = 9.1440000000366E-7 m/us yd/s to m/ns 1 yd/s = 9.1440000000366E-10 m/ns yd/s to km/ms 1 yd/s = 9.1440000000366E-7 km/ms yd/s to km/us 1 yd/s = 9.1440000000366E-10 km/us yd/s to km/ns 1 yd/s = 9.1440000000366E-13 km/ns yd/s to km/d 1 yd/s = 79.004160000316 km/d yd/s to hm/s 1 yd/s = 0.0091440000000366 hm/s yd/s to dm/s 1 yd/s = 9.1440000000366 dm/s yd/s to um/s 1 yd/s = 914400.00000366 um/s yd/s to nm/s 1 yd/s = 914400000.00366 nm/s yd/s to pm/s 1 yd/s = 914400000003.66 pm/s yd/s to mi/d 1 yd/s = 49.091208091396 mi/d yd/s to nmi/h 1 yd/s = 1.7774514016631 nmi/h yd/s to nautical league (int.)/h 1 yd/s = 0.59248380055437 nautical league (int.)/h yd/s to kyd/s 1 yd/s = 0.01 kyd/s yd/s to in/s 1 yd/s = 36 in/s Created @ o.vg Free Unit Converters
## FAQ
### What is 9 Yard/second in Picometer/second?
pm/s. Since one yd/s equals 914400000003.66 pm/s, 9 yd/s in pm/s will be pm/s.
### How many Picometer/second are in a Yard/second?
<MASK>
### How many yd/s is equal to 1 pm/s?
1 pm/s is approximately equal to 1.0936132983333E-12 yd/s.
### What is the yd/s value of 8 pm/s?
<MASK>
### yd/s to pm/s converter in batch
<MASK>
"" at https://www.o.vg/unit/speed/yard-second-to-picometer-second.php from www.o.vg Inc,06/14/2024. https://www.o.vg - Instant, Quick, Free Online Unit Converters
<UNMASK>
Convert Yard/second to Picometer/second | pm/s to yd/s
# document.write(document.title);
## yd/s to pm/s Converter
<MASK>
From pm/s to yd/s: 1 pm/s = 1.0936132983333E-12 yd/s;
## How to Convert Yard/second to Picometer/second?
As we know One yd/s is equal to 914400000003.66 pm/s (1 yd/s = 914400000003.66 pm/s).
To convert Yard/second to Picometer/second, multiply your yd/s figure by 914400000003.66.
Example : convert 25 yd/s to pm/s:
25 yd/s = 25 × 914400000003.66 pm/s = pm/s
To convert Picometer/second to Yard/second, divide your pm/s figure by 914400000003.66.
Example : convert 25 pm/s to yd/s:
25 pm/s = 25 ÷ 914400000003.66 yd/s = yd/s
## How to Convert Picometer/second to Yard/second?
As we know One pm/s is equal to 1.0936132983333E-12 yd/s (1 pm/s = 1.0936132983333E-12 yd/s).
To convert Picometer/second to Yard/second, multiply your pm/s figure by 1.0936132983333E-12.
Example : convert 45 pm/s to yd/s:
45 pm/s = 45 × 1.0936132983333E-12 yd/s = yd/s
To convert Yard/second to Picometer/second, divide your yd/s figure by 1.0936132983333E-12.
Example : convert 45 yd/s to pm/s:
45 yd/s = 45 ÷ 1.0936132983333E-12 pm/s = pm/s
## Convert Yard/second or Picometer/second to Other Speed and Velocity Units
Yard/second Conversion Table
yd/s to m/s 1 yd/s = 0.9144 m/s yd/s to km/h 1 yd/s = 3.29184 km/h yd/s to mi/h 1 yd/s = 2.0454545455 mi/h yd/s to m/h 1 yd/s = 3291.84 m/h yd/s to m/min 1 yd/s = 54.864 m/min yd/s to km/min 1 yd/s = 0.054864 km/min yd/s to km/s 1 yd/s = 0.0009144 km/s yd/s to cm/h 1 yd/s = 329184 cm/h yd/s to cm/min 1 yd/s = 5486.4 cm/min yd/s to cm/s 1 yd/s = 91.44 cm/s yd/s to mm/h 1 yd/s = 3291840 mm/h yd/s to mm/min 1 yd/s = 54864 mm/min yd/s to mm/s 1 yd/s = 914.4 mm/s yd/s to ft/h 1 yd/s = 10800 ft/h yd/s to ft/min 1 yd/s = 180 ft/min yd/s to ft/s 1 yd/s = 3 ft/s yd/s to yd/h 1 yd/s = 3600 yd/h yd/s to yd/min 1 yd/s = 60 yd/min yd/s to mi/min 1 yd/s = 0.0340909091 mi/min yd/s to mi/s 1 yd/s = 0.0005681818 mi/s yd/s to kt 1 yd/s = 1.7774514039 kt yd/s to kt (UK) 1 yd/s = 1.7763157895 kt (UK) yd/s to c 1 yd/s = 3.050110086E-9 c yd/s to Cosmic velocity - first 1 yd/s = 0.0001157468 Cosmic velocity - first yd/s to Cosmic velocity - second 1 yd/s = 0.0000816429 Cosmic velocity - second yd/s to Cosmic velocity - third 1 yd/s = 0.000054853 Cosmic velocity - third yd/s to Earth's velocity 1 yd/s = 0.0000307206 Earth's velocity yd/s to Velocity of sound in pure water 1 yd/s = 0.0006167128 Velocity of sound in pure water yd/s to Velocity of sound in sea water (20°C, 10 meter deep) 1 yd/s = 0.0006009464 Velocity of sound in sea water (20°C, 10 meter deep) yd/s to Mach (20°C, 1 atm) 1 yd/s = 0.002661234 Mach (20°C, 1 atm) yd/s to Mach (SI standard) 1 yd/s = 0.0030991736 Mach (SI standard) Created @ o.vg Free Unit Converters
Yard/second Conversion Table
yd/s to m/ms 1 yd/s = 0.00091440000000366 m/ms yd/s to m/us 1 yd/s = 9.1440000000366E-7 m/us yd/s to m/ns 1 yd/s = 9.1440000000366E-10 m/ns yd/s to km/ms 1 yd/s = 9.1440000000366E-7 km/ms yd/s to km/us 1 yd/s = 9.1440000000366E-10 km/us yd/s to km/ns 1 yd/s = 9.1440000000366E-13 km/ns yd/s to km/d 1 yd/s = 79.004160000316 km/d yd/s to hm/s 1 yd/s = 0.0091440000000366 hm/s yd/s to dm/s 1 yd/s = 9.1440000000366 dm/s yd/s to um/s 1 yd/s = 914400.00000366 um/s yd/s to nm/s 1 yd/s = 914400000.00366 nm/s yd/s to pm/s 1 yd/s = 914400000003.66 pm/s yd/s to mi/d 1 yd/s = 49.091208091396 mi/d yd/s to nmi/h 1 yd/s = 1.7774514016631 nmi/h yd/s to nautical league (int.)/h 1 yd/s = 0.59248380055437 nautical league (int.)/h yd/s to kyd/s 1 yd/s = 0.01 kyd/s yd/s to in/s 1 yd/s = 36 in/s Created @ o.vg Free Unit Converters
## FAQ
### What is 9 Yard/second in Picometer/second?
pm/s. Since one yd/s equals 914400000003.66 pm/s, 9 yd/s in pm/s will be pm/s.
### How many Picometer/second are in a Yard/second?
<MASK>
### How many yd/s is equal to 1 pm/s?
1 pm/s is approximately equal to 1.0936132983333E-12 yd/s.
### What is the yd/s value of 8 pm/s?
<MASK>
### yd/s to pm/s converter in batch
<MASK>
"" at https://www.o.vg/unit/speed/yard-second-to-picometer-second.php from www.o.vg Inc,06/14/2024. https://www.o.vg - Instant, Quick, Free Online Unit Converters
|
Convert Yard/second to Picometer/second | pm/s to yd/s
# document.write(document.title);
## yd/s to pm/s Converter
<MASK>
From pm/s to yd/s: 1 pm/s = 1.0936132983333E-12 yd/s;
## How to Convert Yard/second to Picometer/second?
As we know One yd/s is equal to 914400000003.66 pm/s (1 yd/s = 914400000003.66 pm/s).
To convert Yard/second to Picometer/second, multiply your yd/s figure by 914400000003.66.
Example : convert 25 yd/s to pm/s:
25 yd/s = 25 × 914400000003.66 pm/s = pm/s
To convert Picometer/second to Yard/second, divide your pm/s figure by 914400000003.66.
Example : convert 25 pm/s to yd/s:
25 pm/s = 25 ÷ 914400000003.66 yd/s = yd/s
## How to Convert Picometer/second to Yard/second?
As we know One pm/s is equal to 1.0936132983333E-12 yd/s (1 pm/s = 1.0936132983333E-12 yd/s).
To convert Picometer/second to Yard/second, multiply your pm/s figure by 1.0936132983333E-12.
Example : convert 45 pm/s to yd/s:
45 pm/s = 45 × 1.0936132983333E-12 yd/s = yd/s
To convert Yard/second to Picometer/second, divide your yd/s figure by 1.0936132983333E-12.
Example : convert 45 yd/s to pm/s:
45 yd/s = 45 ÷ 1.0936132983333E-12 pm/s = pm/s
## Convert Yard/second or Picometer/second to Other Speed and Velocity Units
Yard/second Conversion Table
yd/s to m/s 1 yd/s = 0.9144 m/s yd/s to km/h 1 yd/s = 3.29184 km/h yd/s to mi/h 1 yd/s = 2.0454545455 mi/h yd/s to m/h 1 yd/s = 3291.84 m/h yd/s to m/min 1 yd/s = 54.864 m/min yd/s to km/min 1 yd/s = 0.054864 km/min yd/s to km/s 1 yd/s = 0.0009144 km/s yd/s to cm/h 1 yd/s = 329184 cm/h yd/s to cm/min 1 yd/s = 5486.4 cm/min yd/s to cm/s 1 yd/s = 91.44 cm/s yd/s to mm/h 1 yd/s = 3291840 mm/h yd/s to mm/min 1 yd/s = 54864 mm/min yd/s to mm/s 1 yd/s = 914.4 mm/s yd/s to ft/h 1 yd/s = 10800 ft/h yd/s to ft/min 1 yd/s = 180 ft/min yd/s to ft/s 1 yd/s = 3 ft/s yd/s to yd/h 1 yd/s = 3600 yd/h yd/s to yd/min 1 yd/s = 60 yd/min yd/s to mi/min 1 yd/s = 0.0340909091 mi/min yd/s to mi/s 1 yd/s = 0.0005681818 mi/s yd/s to kt 1 yd/s = 1.7774514039 kt yd/s to kt (UK) 1 yd/s = 1.7763157895 kt (UK) yd/s to c 1 yd/s = 3.050110086E-9 c yd/s to Cosmic velocity - first 1 yd/s = 0.0001157468 Cosmic velocity - first yd/s to Cosmic velocity - second 1 yd/s = 0.0000816429 Cosmic velocity - second yd/s to Cosmic velocity - third 1 yd/s = 0.000054853 Cosmic velocity - third yd/s to Earth's velocity 1 yd/s = 0.0000307206 Earth's velocity yd/s to Velocity of sound in pure water 1 yd/s = 0.0006167128 Velocity of sound in pure water yd/s to Velocity of sound in sea water (20°C, 10 meter deep) 1 yd/s = 0.0006009464 Velocity of sound in sea water (20°C, 10 meter deep) yd/s to Mach (20°C, 1 atm) 1 yd/s = 0.002661234 Mach (20°C, 1 atm) yd/s to Mach (SI standard) 1 yd/s = 0.0030991736 Mach (SI standard) Created @ o.vg Free Unit Converters
Yard/second Conversion Table
yd/s to m/ms 1 yd/s = 0.00091440000000366 m/ms yd/s to m/us 1 yd/s = 9.1440000000366E-7 m/us yd/s to m/ns 1 yd/s = 9.1440000000366E-10 m/ns yd/s to km/ms 1 yd/s = 9.1440000000366E-7 km/ms yd/s to km/us 1 yd/s = 9.1440000000366E-10 km/us yd/s to km/ns 1 yd/s = 9.1440000000366E-13 km/ns yd/s to km/d 1 yd/s = 79.004160000316 km/d yd/s to hm/s 1 yd/s = 0.0091440000000366 hm/s yd/s to dm/s 1 yd/s = 9.1440000000366 dm/s yd/s to um/s 1 yd/s = 914400.00000366 um/s yd/s to nm/s 1 yd/s = 914400000.00366 nm/s yd/s to pm/s 1 yd/s = 914400000003.66 pm/s yd/s to mi/d 1 yd/s = 49.091208091396 mi/d yd/s to nmi/h 1 yd/s = 1.7774514016631 nmi/h yd/s to nautical league (int.)/h 1 yd/s = 0.59248380055437 nautical league (int.)/h yd/s to kyd/s 1 yd/s = 0.01 kyd/s yd/s to in/s 1 yd/s = 36 in/s Created @ o.vg Free Unit Converters
## FAQ
### What is 9 Yard/second in Picometer/second?
pm/s. Since one yd/s equals 914400000003.66 pm/s, 9 yd/s in pm/s will be pm/s.
### How many Picometer/second are in a Yard/second?
<MASK>
### How many yd/s is equal to 1 pm/s?
1 pm/s is approximately equal to 1.0936132983333E-12 yd/s.
### What is the yd/s value of 8 pm/s?
<MASK>
### yd/s to pm/s converter in batch
<MASK>
"" at https://www.o.vg/unit/speed/yard-second-to-picometer-second.php from www.o.vg Inc,06/14/2024. https://www.o.vg - Instant, Quick, Free Online Unit Converters
<UNMASK>
Convert Yard/second to Picometer/second | pm/s to yd/s
# document.write(document.title);
## yd/s to pm/s Converter
From yd/s to pm/s: 1 yd/s = 914400000003.66 pm/s;
From pm/s to yd/s: 1 pm/s = 1.0936132983333E-12 yd/s;
## How to Convert Yard/second to Picometer/second?
As we know One yd/s is equal to 914400000003.66 pm/s (1 yd/s = 914400000003.66 pm/s).
To convert Yard/second to Picometer/second, multiply your yd/s figure by 914400000003.66.
Example : convert 25 yd/s to pm/s:
25 yd/s = 25 × 914400000003.66 pm/s = pm/s
To convert Picometer/second to Yard/second, divide your pm/s figure by 914400000003.66.
Example : convert 25 pm/s to yd/s:
25 pm/s = 25 ÷ 914400000003.66 yd/s = yd/s
## How to Convert Picometer/second to Yard/second?
As we know One pm/s is equal to 1.0936132983333E-12 yd/s (1 pm/s = 1.0936132983333E-12 yd/s).
To convert Picometer/second to Yard/second, multiply your pm/s figure by 1.0936132983333E-12.
Example : convert 45 pm/s to yd/s:
45 pm/s = 45 × 1.0936132983333E-12 yd/s = yd/s
To convert Yard/second to Picometer/second, divide your yd/s figure by 1.0936132983333E-12.
Example : convert 45 yd/s to pm/s:
45 yd/s = 45 ÷ 1.0936132983333E-12 pm/s = pm/s
## Convert Yard/second or Picometer/second to Other Speed and Velocity Units
Yard/second Conversion Table
yd/s to m/s 1 yd/s = 0.9144 m/s yd/s to km/h 1 yd/s = 3.29184 km/h yd/s to mi/h 1 yd/s = 2.0454545455 mi/h yd/s to m/h 1 yd/s = 3291.84 m/h yd/s to m/min 1 yd/s = 54.864 m/min yd/s to km/min 1 yd/s = 0.054864 km/min yd/s to km/s 1 yd/s = 0.0009144 km/s yd/s to cm/h 1 yd/s = 329184 cm/h yd/s to cm/min 1 yd/s = 5486.4 cm/min yd/s to cm/s 1 yd/s = 91.44 cm/s yd/s to mm/h 1 yd/s = 3291840 mm/h yd/s to mm/min 1 yd/s = 54864 mm/min yd/s to mm/s 1 yd/s = 914.4 mm/s yd/s to ft/h 1 yd/s = 10800 ft/h yd/s to ft/min 1 yd/s = 180 ft/min yd/s to ft/s 1 yd/s = 3 ft/s yd/s to yd/h 1 yd/s = 3600 yd/h yd/s to yd/min 1 yd/s = 60 yd/min yd/s to mi/min 1 yd/s = 0.0340909091 mi/min yd/s to mi/s 1 yd/s = 0.0005681818 mi/s yd/s to kt 1 yd/s = 1.7774514039 kt yd/s to kt (UK) 1 yd/s = 1.7763157895 kt (UK) yd/s to c 1 yd/s = 3.050110086E-9 c yd/s to Cosmic velocity - first 1 yd/s = 0.0001157468 Cosmic velocity - first yd/s to Cosmic velocity - second 1 yd/s = 0.0000816429 Cosmic velocity - second yd/s to Cosmic velocity - third 1 yd/s = 0.000054853 Cosmic velocity - third yd/s to Earth's velocity 1 yd/s = 0.0000307206 Earth's velocity yd/s to Velocity of sound in pure water 1 yd/s = 0.0006167128 Velocity of sound in pure water yd/s to Velocity of sound in sea water (20°C, 10 meter deep) 1 yd/s = 0.0006009464 Velocity of sound in sea water (20°C, 10 meter deep) yd/s to Mach (20°C, 1 atm) 1 yd/s = 0.002661234 Mach (20°C, 1 atm) yd/s to Mach (SI standard) 1 yd/s = 0.0030991736 Mach (SI standard) Created @ o.vg Free Unit Converters
Yard/second Conversion Table
yd/s to m/ms 1 yd/s = 0.00091440000000366 m/ms yd/s to m/us 1 yd/s = 9.1440000000366E-7 m/us yd/s to m/ns 1 yd/s = 9.1440000000366E-10 m/ns yd/s to km/ms 1 yd/s = 9.1440000000366E-7 km/ms yd/s to km/us 1 yd/s = 9.1440000000366E-10 km/us yd/s to km/ns 1 yd/s = 9.1440000000366E-13 km/ns yd/s to km/d 1 yd/s = 79.004160000316 km/d yd/s to hm/s 1 yd/s = 0.0091440000000366 hm/s yd/s to dm/s 1 yd/s = 9.1440000000366 dm/s yd/s to um/s 1 yd/s = 914400.00000366 um/s yd/s to nm/s 1 yd/s = 914400000.00366 nm/s yd/s to pm/s 1 yd/s = 914400000003.66 pm/s yd/s to mi/d 1 yd/s = 49.091208091396 mi/d yd/s to nmi/h 1 yd/s = 1.7774514016631 nmi/h yd/s to nautical league (int.)/h 1 yd/s = 0.59248380055437 nautical league (int.)/h yd/s to kyd/s 1 yd/s = 0.01 kyd/s yd/s to in/s 1 yd/s = 36 in/s Created @ o.vg Free Unit Converters
## FAQ
### What is 9 Yard/second in Picometer/second?
pm/s. Since one yd/s equals 914400000003.66 pm/s, 9 yd/s in pm/s will be pm/s.
### How many Picometer/second are in a Yard/second?
There are 914400000003.66 pm/s in one yd/s. In turn, one pm/s is equal to 1.0936132983333E-12 yd/s.
### How many yd/s is equal to 1 pm/s?
1 pm/s is approximately equal to 1.0936132983333E-12 yd/s.
### What is the yd/s value of 8 pm/s?
The Yard/second value of 8 pm/s is yd/s. (i.e.,) 8 x 1.0936132983333E-12 = yd/s.
### yd/s to pm/s converter in batch
Cite this Converter, Content or Page as:
"" at https://www.o.vg/unit/speed/yard-second-to-picometer-second.php from www.o.vg Inc,06/14/2024. https://www.o.vg - Instant, Quick, Free Online Unit Converters
|
<MASK>
blue on Mar 21, 2010
0
<MASK>
<UNMASK>
<MASK>
So we have:
40 * (40 + 1) / 2 = 820
blue on Mar 21, 2010
0
<MASK>
Evandro on Dec 30, 2010
3
<MASK>
an87 on Jun 13, 2011
|
<MASK>
So we have:
40 * (40 + 1) / 2 = 820
blue on Mar 21, 2010
0
<MASK>
Evandro on Dec 30, 2010
3
<MASK>
an87 on Jun 13, 2011
<UNMASK>
<MASK>
So we have:
40 * (40 + 1) / 2 = 820
blue on Mar 21, 2010
0
<MASK>
sum of 1 and 40 = 41
OR
sum of 1 + -40 = -39
Evandro on Dec 30, 2010
3
<MASK>
an87 on Jun 13, 2011
|
<MASK>
So we have:
40 * (40 + 1) / 2 = 820
blue on Mar 21, 2010
0
<MASK>
sum of 1 and 40 = 41
OR
sum of 1 + -40 = -39
Evandro on Dec 30, 2010
3
<MASK>
an87 on Jun 13, 2011
<UNMASK>
<MASK>
# What is the sum of 1-40?
<MASK>
Interview Candidate on Feb 11, 2010
2
<MASK>
So we have:
40 * (40 + 1) / 2 = 820
blue on Mar 21, 2010
0
<MASK>
sum of 1 and 40 = 41
OR
sum of 1 + -40 = -39
Evandro on Dec 30, 2010
3
Any sum like this is easy. take the lowest and highest... 1 +40 = 41, then the next highest and the next lowest. 2 + 39 = 41. Realize that there are 20 such pairs. 20 * 41 = 820.
For the numbers 1 - 100, 1 + 100 = 101, 2 + 99 = 101; there are 50 such pairs, so 50 * 101 = 5050
scienceguy on Jan 11, 2011
1
http://brainteaserbible.com/interview-brainteaser-sum-of-the-numbers-from-1-to-50
an87 on Jun 13, 2011
|
<MASK>
# What is the sum of 1-40?
<MASK>
Interview Candidate on Feb 11, 2010
2
<MASK>
So we have:
40 * (40 + 1) / 2 = 820
blue on Mar 21, 2010
0
<MASK>
sum of 1 and 40 = 41
OR
sum of 1 + -40 = -39
Evandro on Dec 30, 2010
3
Any sum like this is easy. take the lowest and highest... 1 +40 = 41, then the next highest and the next lowest. 2 + 39 = 41. Realize that there are 20 such pairs. 20 * 41 = 820.
For the numbers 1 - 100, 1 + 100 = 101, 2 + 99 = 101; there are 50 such pairs, so 50 * 101 = 5050
scienceguy on Jan 11, 2011
1
http://brainteaserbible.com/interview-brainteaser-sum-of-the-numbers-from-1-to-50
an87 on Jun 13, 2011
<UNMASK>
UBS Interview Question: What is the sum of 1-40?... | Glassdoor
# What is the sum of 1-40?
0
820
Interview Candidate on Feb 11, 2010
2
We can have the sum by using the following formula
N * (N + 1) / 2
So we have:
40 * (40 + 1) / 2 = 820
blue on Mar 21, 2010
0
Since he isn't asking to sum all the numbers from 1 to 40:
sum of 1 and 40 = 41
OR
sum of 1 + -40 = -39
Evandro on Dec 30, 2010
3
Any sum like this is easy. take the lowest and highest... 1 +40 = 41, then the next highest and the next lowest. 2 + 39 = 41. Realize that there are 20 such pairs. 20 * 41 = 820.
For the numbers 1 - 100, 1 + 100 = 101, 2 + 99 = 101; there are 50 such pairs, so 50 * 101 = 5050
scienceguy on Jan 11, 2011
1
http://brainteaserbible.com/interview-brainteaser-sum-of-the-numbers-from-1-to-50
an87 on Jun 13, 2011
|
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who was the murderer ?
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* They have only one torch and the river is too risky to cross without the torch.
* If all people cross simultaneously then torch light wont be sufficient.
* Speed of each person of crossing the river is different.cross time for each person is 1 min, 2 minutes, 7 minutes and 10 minutes.
What is the shortest time needed for all four of them to cross the river ?
<MASK>
# Mind Teasers : Logic Question In Maths
Difficulty Popularity
Two trains under a controlled experiment begin at a speed of 100 mph in the opposite direction in a tunnel. A supersonic bee is left in the tunnel which can fly at a speed of 1000 mph. The tunnel is 200 miles long. When the trains start running on a constant speed of 100 mph, the supersonic bee starts flying from one train towards the other. As soon as the bee reaches the second train, it starts flying back towards the first train.
If the bee keeps flying to and fro in the tunnel till the trains collide, how much distance will it have covered in total?
• Views : 80k+
• Sol Viewed : 20k+
# Mind Teasers : Popular Logic Brain Teaser
Difficulty Popularity
A girl liked to collect money in a piggy bank. She bought pink colored piggy bank when she was 10 years old. She put \$250 in the box on each of her birthday. Her younger sister took \$50 out of her piggy bank on her birthday. The girl died when she was 50 years old due to an incurable disease.
<MASK>
• Views : 80k+
• Sol Viewed : 20k+
# Mind Teasers : Sherlock Holmes Murderer Mystery Puzzle
Difficulty Popularity
There are five people. One of them shot and killed one of the other five.
1. Dan ran in the NY City marathon yesterday with one of the innocent men.
2. Mike considered being a farmer before he moved to the city.
3. Jeff is a topnotch computer consultant and wants to install Ben's new computer next week.
4. The murderer had his leg amputated last month.
5. Ben met Jack for the first time six months ago.
6. Jack has been in seclusion since the crime.
7. Dan used to drink heavily.
8. Ben and Jeff built their last computers together.
9. The murderer is Jack's brother. They grew up together in Seattle.
Sherlock holmes arrived at the scene and immediately find the murderer.
who was the murderer ?
• Views : 70k+
• Sol Viewed : 20k+
# Mind Teasers : Math Trick Riddle
Difficulty Popularity
Removing one from eleven makes it ten and removing one from nine makes it ten.
Is it even possible?
• Views : 80k+
• Sol Viewed : 20k+
# Mind Teasers : Best Maths Brain Teaser
Difficulty Popularity
A competitive exam was held in which five students took part namely Billy, Gerry, Clark, Peeta and Jonathan. In this exam, they had to answer five questions each out of which, three had multiple choices as a, b or c and two were simple true and false questions. The five of them gave different answers to the questions and the details are as given below.
Name 1 2 3 4 5
Gerry c b True True False
Billy c c True True True
Clark a c False True True
<MASK>
Jonathan a b True False True
<MASK>
• Views : 60k+
• Sol Viewed : 20k+
# Mind Teasers : Crack the code Riddle
Difficulty Popularity
<MASK>
• Views : 70k+
• Sol Viewed : 20k+
# Mind Teasers : Hard Logic Puzzle
Difficulty Popularity
A great meeting is held by a great logician where all the other logicians are called upon. The master logician takes them in a room and makes them sit in circle. A hat is placed on each of their heads. Now all of them can see the color of hats others are wearing but can’t see his own. They are told that there different colors of hats.
<MASK>
### Latest Puzzles
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<MASK>
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<MASK>
19 February
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<MASK>
18 February
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<MASK>
<UNMASK>
• Views : 40k+
• Sol Viewed : 10k+
# Mind Teasers : Funny New Year Riddle
Difficulty Popularity
What are these ?
January first 2010 => 800 X 600
January first 2011 => 1280 X 1024
January first 2012 => 600 X 1200
Discussion
Suggestions
• Views : 70k+
• Sol Viewed : 20k+
# Mind Teasers : Tricky Math Race Riddle
Difficulty Popularity
Rooney, Hernandez, and Robin race each other in a 100 meters race. All of them run at a constant speed throughout the race.
Rooney beats Hernandez by 20 meters.
Hernandez beats Robin by 20 meters.
How many meters does Rooney beat Robin by ?
• Views : 50k+
• Sol Viewed : 20k+
# Mind Teasers : Probability To Choose Own Toy Brain Teaser
Difficulty Popularity
I bought three toys for my triplet boys (one for each). I placed the toys in the dark store. One by one each boy went to the store and pick the toy. What is the probability that no boy will choose his own toy?
• Views : 70k+
• Sol Viewed : 20k+
# Mind Teasers : Crossing The River In Minimum Time Puzzle
Difficulty Popularity
Four people need to cross a dark river at night.
* They have only one torch and the river is too risky to cross without the torch.
* If all people cross simultaneously then torch light wont be sufficient.
* Speed of each person of crossing the river is different.cross time for each person is 1 min, 2 minutes, 7 minutes and 10 minutes.
What is the shortest time needed for all four of them to cross the river ?
• Views : 40k+
• Sol Viewed : 10k+
# Mind Teasers : Logic Question In Maths
Difficulty Popularity
Two trains under a controlled experiment begin at a speed of 100 mph in the opposite direction in a tunnel. A supersonic bee is left in the tunnel which can fly at a speed of 1000 mph. The tunnel is 200 miles long. When the trains start running on a constant speed of 100 mph, the supersonic bee starts flying from one train towards the other. As soon as the bee reaches the second train, it starts flying back towards the first train.
If the bee keeps flying to and fro in the tunnel till the trains collide, how much distance will it have covered in total?
• Views : 80k+
• Sol Viewed : 20k+
# Mind Teasers : Popular Logic Brain Teaser
Difficulty Popularity
A girl liked to collect money in a piggy bank. She bought pink colored piggy bank when she was 10 years old. She put \$250 in the box on each of her birthday. Her younger sister took \$50 out of her piggy bank on her birthday. The girl died when she was 50 years old due to an incurable disease.
When the piggy bank was opened, it had just \$500. How can that be possible?
• Views : 80k+
• Sol Viewed : 20k+
# Mind Teasers : Sherlock Holmes Murderer Mystery Puzzle
Difficulty Popularity
There are five people. One of them shot and killed one of the other five.
1. Dan ran in the NY City marathon yesterday with one of the innocent men.
2. Mike considered being a farmer before he moved to the city.
3. Jeff is a topnotch computer consultant and wants to install Ben's new computer next week.
4. The murderer had his leg amputated last month.
5. Ben met Jack for the first time six months ago.
6. Jack has been in seclusion since the crime.
7. Dan used to drink heavily.
8. Ben and Jeff built their last computers together.
9. The murderer is Jack's brother. They grew up together in Seattle.
Sherlock holmes arrived at the scene and immediately find the murderer.
who was the murderer ?
• Views : 70k+
• Sol Viewed : 20k+
# Mind Teasers : Math Trick Riddle
Difficulty Popularity
Removing one from eleven makes it ten and removing one from nine makes it ten.
Is it even possible?
• Views : 80k+
• Sol Viewed : 20k+
# Mind Teasers : Best Maths Brain Teaser
Difficulty Popularity
A competitive exam was held in which five students took part namely Billy, Gerry, Clark, Peeta and Jonathan. In this exam, they had to answer five questions each out of which, three had multiple choices as a, b or c and two were simple true and false questions. The five of them gave different answers to the questions and the details are as given below.
Name 1 2 3 4 5
Gerry c b True True False
Billy c c True True True
Clark a c False True True
Peeta b a True True False
Jonathan a b True False True
None of the two students gave the same number of correct answers. Can you find out the correct answers to the question? Also, calculate the individual score of all the five guys.
• Views : 60k+
• Sol Viewed : 20k+
# Mind Teasers : Crack the code Riddle
Difficulty Popularity
A man desired to get into his work building, however he had forgotten his code.
However, he did recollect five pieces of information
-> Sum of 5th number and 3rd number is 14.
-> Difference of 4th and 2nd number is 1.
-> The 1st number is one less than twice the 2nd number.
->The 2nd number and the 3rd number equals 10.
->The sum of all digits is 30.
Crack the code ?
• Views : 70k+
• Sol Viewed : 20k+
# Mind Teasers : Hard Logic Puzzle
Difficulty Popularity
A great meeting is held by a great logician where all the other logicians are called upon. The master logician takes them in a room and makes them sit in circle. A hat is placed on each of their heads. Now all of them can see the color of hats others are wearing but can’t see his own. They are told that there different colors of hats.
The master logician explains that a bell will be rung at regular intervals and the moment when a logician knows the color of his hat, he will leave on the next bell. If anyone leaves at the wrong bell, he will be disqualified and sent home.
All of them are assured of one thing that the puzzle will not be impossible for anyone of them. How will they manage the situation?
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James celebrated his 23rd Birthday on 20...
|
<MASK>
<UNMASK>
<MASK>
$$\begin{equation*} \bigl(\frac{a}{b}\bigr)^4 + \bigl(\frac{b}{c}\bigr)^4 + \bigl(\frac{c}{d}\bigr)^4 + \bigl(\frac{d}{e}\bigr)^4 + \bigl(\frac{e}{a}\bigr)^4 \ge \frac{b}{a} + \frac{c}{b} + \frac{d}{c} + \frac{e}{d} + \frac{a}{e} \end{equation*}$$
There's not really a clear-cut way to use AM-GM on this problem. I've been thinking of maybe using the Power Mean Inequality, but I don't exactly see a way to do that. Maybe we could use harmonic mean for the RHS?
<MASK>
|
<MASK>
$$\begin{equation*} \bigl(\frac{a}{b}\bigr)^4 + \bigl(\frac{b}{c}\bigr)^4 + \bigl(\frac{c}{d}\bigr)^4 + \bigl(\frac{d}{e}\bigr)^4 + \bigl(\frac{e}{a}\bigr)^4 \ge \frac{b}{a} + \frac{c}{b} + \frac{d}{c} + \frac{e}{d} + \frac{a}{e} \end{equation*}$$
There's not really a clear-cut way to use AM-GM on this problem. I've been thinking of maybe using the Power Mean Inequality, but I don't exactly see a way to do that. Maybe we could use harmonic mean for the RHS?
<MASK>
<UNMASK>
<MASK>
$$\begin{equation*} \bigl(\frac{a}{b}\bigr)^4 + \bigl(\frac{b}{c}\bigr)^4 + \bigl(\frac{c}{d}\bigr)^4 + \bigl(\frac{d}{e}\bigr)^4 + \bigl(\frac{e}{a}\bigr)^4 \ge \frac{b}{a} + \frac{c}{b} + \frac{d}{c} + \frac{e}{d} + \frac{a}{e} \end{equation*}$$
There's not really a clear-cut way to use AM-GM on this problem. I've been thinking of maybe using the Power Mean Inequality, but I don't exactly see a way to do that. Maybe we could use harmonic mean for the RHS?
<MASK>
What is the minumum of the function $$f(x_1,x_2,x_3,x_4,x_5)=\sum_{i=1}^{5}(x_i^4-x_i^{-1})$$ with domain $$\Bbb{R}^{5+}$$, subject to the constraint equation $$x_1x_2x_3x_4x_5=1$$?
<MASK>
|
<MASK>
$$\begin{equation*} \bigl(\frac{a}{b}\bigr)^4 + \bigl(\frac{b}{c}\bigr)^4 + \bigl(\frac{c}{d}\bigr)^4 + \bigl(\frac{d}{e}\bigr)^4 + \bigl(\frac{e}{a}\bigr)^4 \ge \frac{b}{a} + \frac{c}{b} + \frac{d}{c} + \frac{e}{d} + \frac{a}{e} \end{equation*}$$
There's not really a clear-cut way to use AM-GM on this problem. I've been thinking of maybe using the Power Mean Inequality, but I don't exactly see a way to do that. Maybe we could use harmonic mean for the RHS?
<MASK>
What is the minumum of the function $$f(x_1,x_2,x_3,x_4,x_5)=\sum_{i=1}^{5}(x_i^4-x_i^{-1})$$ with domain $$\Bbb{R}^{5+}$$, subject to the constraint equation $$x_1x_2x_3x_4x_5=1$$?
<MASK>
<UNMASK>
<MASK>
$$\begin{equation*} \bigl(\frac{a}{b}\bigr)^4 + \bigl(\frac{b}{c}\bigr)^4 + \bigl(\frac{c}{d}\bigr)^4 + \bigl(\frac{d}{e}\bigr)^4 + \bigl(\frac{e}{a}\bigr)^4 \ge \frac{b}{a} + \frac{c}{b} + \frac{d}{c} + \frac{e}{d} + \frac{a}{e} \end{equation*}$$
There's not really a clear-cut way to use AM-GM on this problem. I've been thinking of maybe using the Power Mean Inequality, but I don't exactly see a way to do that. Maybe we could use harmonic mean for the RHS?
<MASK>
• I think in the end you should have $5LHS-LHS\geq 4RHS$, since you repeat the procedure 5 times, not 4. Then everything works. :) Commented Dec 5, 2022 at 9:33
• @Freshman'sDream You're right, I just corrected this typo. Thanks!
– NN2
Commented Dec 5, 2022 at 9:34
• ohhhhh ok thanks! Commented Dec 9, 2022 at 23:08
<MASK>
What is the minumum of the function $$f(x_1,x_2,x_3,x_4,x_5)=\sum_{i=1}^{5}(x_i^4-x_i^{-1})$$ with domain $$\Bbb{R}^{5+}$$, subject to the constraint equation $$x_1x_2x_3x_4x_5=1$$?
<MASK>
|
<MASK>
$$\begin{equation*} \bigl(\frac{a}{b}\bigr)^4 + \bigl(\frac{b}{c}\bigr)^4 + \bigl(\frac{c}{d}\bigr)^4 + \bigl(\frac{d}{e}\bigr)^4 + \bigl(\frac{e}{a}\bigr)^4 \ge \frac{b}{a} + \frac{c}{b} + \frac{d}{c} + \frac{e}{d} + \frac{a}{e} \end{equation*}$$
There's not really a clear-cut way to use AM-GM on this problem. I've been thinking of maybe using the Power Mean Inequality, but I don't exactly see a way to do that. Maybe we could use harmonic mean for the RHS?
<MASK>
• I think in the end you should have $5LHS-LHS\geq 4RHS$, since you repeat the procedure 5 times, not 4. Then everything works. :) Commented Dec 5, 2022 at 9:33
• @Freshman'sDream You're right, I just corrected this typo. Thanks!
– NN2
Commented Dec 5, 2022 at 9:34
• ohhhhh ok thanks! Commented Dec 9, 2022 at 23:08
<MASK>
What is the minumum of the function $$f(x_1,x_2,x_3,x_4,x_5)=\sum_{i=1}^{5}(x_i^4-x_i^{-1})$$ with domain $$\Bbb{R}^{5+}$$, subject to the constraint equation $$x_1x_2x_3x_4x_5=1$$?
<MASK>
<UNMASK>
# $(\frac{a}{b})^4+(\frac{b}{c})^4+(\frac{c}{d})^4+(\frac{d}{e})^4+(\frac{e}{a})^4\ge\frac{b}{a}+\frac{c}{b}+\frac{d}{c}+\frac{e}{d}+\frac{a}{e}$
How exactly do I solve this problem? (Source: 1984 British Math Olympiad #3 part II)
$$\begin{equation*} \bigl(\frac{a}{b}\bigr)^4 + \bigl(\frac{b}{c}\bigr)^4 + \bigl(\frac{c}{d}\bigr)^4 + \bigl(\frac{d}{e}\bigr)^4 + \bigl(\frac{e}{a}\bigr)^4 \ge \frac{b}{a} + \frac{c}{b} + \frac{d}{c} + \frac{e}{d} + \frac{a}{e} \end{equation*}$$
There's not really a clear-cut way to use AM-GM on this problem. I've been thinking of maybe using the Power Mean Inequality, but I don't exactly see a way to do that. Maybe we could use harmonic mean for the RHS?
• someone please explain why this is closed. I think I have adequately explained some strategies that I've tried. I believe I've provided enough context. Commented Dec 10, 2022 at 19:54
• I'm kinda new around here, but I was also surprised to see it closed. Also I found the accepted solution to be very nice. Commented Dec 10, 2022 at 21:48
Applying the AM-GM $$LHS - \bigl(\frac{e}{a}\bigr)^4 = \bigl(\frac{a}{b}\bigr)^4 + \bigl(\frac{b}{c}\bigr)^4 + \bigl(\frac{c}{d}\bigr)^4 + \bigl(\frac{d}{e}\bigr)^4 \ge4 \cdot \frac{a}{b} \cdot \frac{b}{c} \cdot \frac{c}{d} \cdot\frac{d}{e} = 4\cdot\frac{a}{e}$$ Do the same thing for these 4 others terms, and make the sum $$5 LHS - LHS \ge 4 RHS$$ $$\Longleftrightarrow LHS \ge RHS$$ The equality occurs when $$a=b=c=d=e$$
• I think in the end you should have $5LHS-LHS\geq 4RHS$, since you repeat the procedure 5 times, not 4. Then everything works. :) Commented Dec 5, 2022 at 9:33
• @Freshman'sDream You're right, I just corrected this typo. Thanks!
– NN2
Commented Dec 5, 2022 at 9:34
• ohhhhh ok thanks! Commented Dec 9, 2022 at 23:08
NN2 gave a simple and very elegamt proof. I tried another way.
What is the minumum of the function $$f(x_1,x_2,x_3,x_4,x_5)=\sum_{i=1}^{5}(x_i^4-x_i^{-1})$$ with domain $$\Bbb{R}^{5+}$$, subject to the constraint equation $$x_1x_2x_3x_4x_5=1$$?
The system of a Lagrange multplier $$\lambda$$ gives the equations $$4x_i^3+x_i^{-2}=\lambda x_i^{-1}$$ for all $$i=1,2,3,4,5$$. From these equations we have $$4x_ix_j(x_i^4-x_j^4)=x_i-x_j$$ for all $$i,j.$$ I am stuck. Any ideas?
|
<MASK>
<UNMASK>
## Oct 13, 2015
<MASK>
Having been in Java world for around 5 years, I believe Bubble Sort is the probably simplest and widely used sorting algorithm that works by repeatedly swapping the adjacent elements if they are not in the right order. However, with just a little tweaking, it’s possible to make it more efficient. So, we will first see what Bubble Sort algorithm is and then move to its optimization part.
To sort the entire array, the array is traversed n-1 time (array having n elements). These are called passes. In the first pass the largest element moves to the last position (sorting in ascending order). So if the original (unsorted) array is:
55 33 99 77 44
During the first iteration, adjacent elements will be compared, and swapped if larger element is on left side), as shown below:
After first iteration, the largest element is at the last position.
33 55 77 44 99
Now, in the 2nd pass, we will consider the first (n-1) elements only (because last position already has largest element). After 2nd pass the array will be
33 55 44 77 99
i.e., 77 will be moved to the (n-1)th position.
After (n-1) iterations, (n-1) elements will be moved to their proper positions, the last element has to be the smallest. So the array will be sorted after n-1 passes.
Algorithm:
<MASK>
Time Complexity: O(n*n)
Auxiliary Space: O(1)
Boundary Cases: Bubble sort takes minimum time (Order of n) when elements are already sorted.
Sorting In Place: Yes
Stable: Yes
<MASK>
|
## Oct 13, 2015
<MASK>
Having been in Java world for around 5 years, I believe Bubble Sort is the probably simplest and widely used sorting algorithm that works by repeatedly swapping the adjacent elements if they are not in the right order. However, with just a little tweaking, it’s possible to make it more efficient. So, we will first see what Bubble Sort algorithm is and then move to its optimization part.
To sort the entire array, the array is traversed n-1 time (array having n elements). These are called passes. In the first pass the largest element moves to the last position (sorting in ascending order). So if the original (unsorted) array is:
55 33 99 77 44
During the first iteration, adjacent elements will be compared, and swapped if larger element is on left side), as shown below:
After first iteration, the largest element is at the last position.
33 55 77 44 99
Now, in the 2nd pass, we will consider the first (n-1) elements only (because last position already has largest element). After 2nd pass the array will be
33 55 44 77 99
i.e., 77 will be moved to the (n-1)th position.
After (n-1) iterations, (n-1) elements will be moved to their proper positions, the last element has to be the smallest. So the array will be sorted after n-1 passes.
Algorithm:
<MASK>
Time Complexity: O(n*n)
Auxiliary Space: O(1)
Boundary Cases: Bubble sort takes minimum time (Order of n) when elements are already sorted.
Sorting In Place: Yes
Stable: Yes
<MASK>
<UNMASK>
## Oct 13, 2015
<MASK>
Having been in Java world for around 5 years, I believe Bubble Sort is the probably simplest and widely used sorting algorithm that works by repeatedly swapping the adjacent elements if they are not in the right order. However, with just a little tweaking, it’s possible to make it more efficient. So, we will first see what Bubble Sort algorithm is and then move to its optimization part.
To sort the entire array, the array is traversed n-1 time (array having n elements). These are called passes. In the first pass the largest element moves to the last position (sorting in ascending order). So if the original (unsorted) array is:
55 33 99 77 44
During the first iteration, adjacent elements will be compared, and swapped if larger element is on left side), as shown below:
After first iteration, the largest element is at the last position.
33 55 77 44 99
Now, in the 2nd pass, we will consider the first (n-1) elements only (because last position already has largest element). After 2nd pass the array will be
33 55 44 77 99
i.e., 77 will be moved to the (n-1)th position.
After (n-1) iterations, (n-1) elements will be moved to their proper positions, the last element has to be the smallest. So the array will be sorted after n-1 passes.
Algorithm:
void bubbleSort(int array[], int n) {
for (int i = 0; i < n; i++) {
for (int j = 0; j < n - i - 1; j++) {
if (array[j] > array[j + 1]) {
int temp = array[j + 1];
array[j + 1] = array[j];
array[j] = temp;
}
}
}
}
Time Complexity: O(n*n)
Auxiliary Space: O(1)
Boundary Cases: Bubble sort takes minimum time (Order of n) when elements are already sorted.
Sorting In Place: Yes
Stable: Yes
Improvement (Optimization):
In the above example, the array got sorted after 2nd pass, but we will still continue with the 3rd, 4th pass. Suppose if the array is already sorted, then there will be no swapping (because adjacent elements are always in order), but still we will continue with the passes and there will still be (n-1) passes.
If we can identify, that the array is sorted, then we should stop execution of further passes. This is the optimization over the original bubble sort algorithm.
If there is no swapping in an iteration pass, it means the array has become sorted, so algorithm should be smart enough to skip further iterations.
For this we can have a flag variable which is set to true before each pass and is made false when a swapping is performed.
<MASK>
|
## Oct 13, 2015
<MASK>
Having been in Java world for around 5 years, I believe Bubble Sort is the probably simplest and widely used sorting algorithm that works by repeatedly swapping the adjacent elements if they are not in the right order. However, with just a little tweaking, it’s possible to make it more efficient. So, we will first see what Bubble Sort algorithm is and then move to its optimization part.
To sort the entire array, the array is traversed n-1 time (array having n elements). These are called passes. In the first pass the largest element moves to the last position (sorting in ascending order). So if the original (unsorted) array is:
55 33 99 77 44
During the first iteration, adjacent elements will be compared, and swapped if larger element is on left side), as shown below:
After first iteration, the largest element is at the last position.
33 55 77 44 99
Now, in the 2nd pass, we will consider the first (n-1) elements only (because last position already has largest element). After 2nd pass the array will be
33 55 44 77 99
i.e., 77 will be moved to the (n-1)th position.
After (n-1) iterations, (n-1) elements will be moved to their proper positions, the last element has to be the smallest. So the array will be sorted after n-1 passes.
Algorithm:
void bubbleSort(int array[], int n) {
for (int i = 0; i < n; i++) {
for (int j = 0; j < n - i - 1; j++) {
if (array[j] > array[j + 1]) {
int temp = array[j + 1];
array[j + 1] = array[j];
array[j] = temp;
}
}
}
}
Time Complexity: O(n*n)
Auxiliary Space: O(1)
Boundary Cases: Bubble sort takes minimum time (Order of n) when elements are already sorted.
Sorting In Place: Yes
Stable: Yes
Improvement (Optimization):
In the above example, the array got sorted after 2nd pass, but we will still continue with the 3rd, 4th pass. Suppose if the array is already sorted, then there will be no swapping (because adjacent elements are always in order), but still we will continue with the passes and there will still be (n-1) passes.
If we can identify, that the array is sorted, then we should stop execution of further passes. This is the optimization over the original bubble sort algorithm.
If there is no swapping in an iteration pass, it means the array has become sorted, so algorithm should be smart enough to skip further iterations.
For this we can have a flag variable which is set to true before each pass and is made false when a swapping is performed.
<MASK>
<UNMASK>
## Oct 13, 2015
### Optimizing Bubble Sort Algorithm
Having been in Java world for around 5 years, I believe Bubble Sort is the probably simplest and widely used sorting algorithm that works by repeatedly swapping the adjacent elements if they are not in the right order. However, with just a little tweaking, it’s possible to make it more efficient. So, we will first see what Bubble Sort algorithm is and then move to its optimization part.
To sort the entire array, the array is traversed n-1 time (array having n elements). These are called passes. In the first pass the largest element moves to the last position (sorting in ascending order). So if the original (unsorted) array is:
55 33 99 77 44
During the first iteration, adjacent elements will be compared, and swapped if larger element is on left side), as shown below:
After first iteration, the largest element is at the last position.
33 55 77 44 99
Now, in the 2nd pass, we will consider the first (n-1) elements only (because last position already has largest element). After 2nd pass the array will be
33 55 44 77 99
i.e., 77 will be moved to the (n-1)th position.
After (n-1) iterations, (n-1) elements will be moved to their proper positions, the last element has to be the smallest. So the array will be sorted after n-1 passes.
Algorithm:
void bubbleSort(int array[], int n) {
for (int i = 0; i < n; i++) {
for (int j = 0; j < n - i - 1; j++) {
if (array[j] > array[j + 1]) {
int temp = array[j + 1];
array[j + 1] = array[j];
array[j] = temp;
}
}
}
}
Time Complexity: O(n*n)
Auxiliary Space: O(1)
Boundary Cases: Bubble sort takes minimum time (Order of n) when elements are already sorted.
Sorting In Place: Yes
Stable: Yes
Improvement (Optimization):
In the above example, the array got sorted after 2nd pass, but we will still continue with the 3rd, 4th pass. Suppose if the array is already sorted, then there will be no swapping (because adjacent elements are always in order), but still we will continue with the passes and there will still be (n-1) passes.
If we can identify, that the array is sorted, then we should stop execution of further passes. This is the optimization over the original bubble sort algorithm.
If there is no swapping in an iteration pass, it means the array has become sorted, so algorithm should be smart enough to skip further iterations.
For this we can have a flag variable which is set to true before each pass and is made false when a swapping is performed.
void optimizedBubbleSort(int array[], int n) {
for (int i = 0; i < n; i++) {
boolean flag = true;
for (int j = 0; j < n - i - 1; j++) {
if (array[j] > array[j + 1]) {
flag = false;
int temp = array[j + 1];
array[j + 1] = array[j];
array[j] = temp;
}
}
<MASK>
|
## Oct 13, 2015
### Optimizing Bubble Sort Algorithm
Having been in Java world for around 5 years, I believe Bubble Sort is the probably simplest and widely used sorting algorithm that works by repeatedly swapping the adjacent elements if they are not in the right order. However, with just a little tweaking, it’s possible to make it more efficient. So, we will first see what Bubble Sort algorithm is and then move to its optimization part.
To sort the entire array, the array is traversed n-1 time (array having n elements). These are called passes. In the first pass the largest element moves to the last position (sorting in ascending order). So if the original (unsorted) array is:
55 33 99 77 44
During the first iteration, adjacent elements will be compared, and swapped if larger element is on left side), as shown below:
After first iteration, the largest element is at the last position.
33 55 77 44 99
Now, in the 2nd pass, we will consider the first (n-1) elements only (because last position already has largest element). After 2nd pass the array will be
33 55 44 77 99
i.e., 77 will be moved to the (n-1)th position.
After (n-1) iterations, (n-1) elements will be moved to their proper positions, the last element has to be the smallest. So the array will be sorted after n-1 passes.
Algorithm:
void bubbleSort(int array[], int n) {
for (int i = 0; i < n; i++) {
for (int j = 0; j < n - i - 1; j++) {
if (array[j] > array[j + 1]) {
int temp = array[j + 1];
array[j + 1] = array[j];
array[j] = temp;
}
}
}
}
Time Complexity: O(n*n)
Auxiliary Space: O(1)
Boundary Cases: Bubble sort takes minimum time (Order of n) when elements are already sorted.
Sorting In Place: Yes
Stable: Yes
Improvement (Optimization):
In the above example, the array got sorted after 2nd pass, but we will still continue with the 3rd, 4th pass. Suppose if the array is already sorted, then there will be no swapping (because adjacent elements are always in order), but still we will continue with the passes and there will still be (n-1) passes.
If we can identify, that the array is sorted, then we should stop execution of further passes. This is the optimization over the original bubble sort algorithm.
If there is no swapping in an iteration pass, it means the array has become sorted, so algorithm should be smart enough to skip further iterations.
For this we can have a flag variable which is set to true before each pass and is made false when a swapping is performed.
void optimizedBubbleSort(int array[], int n) {
for (int i = 0; i < n; i++) {
boolean flag = true;
for (int j = 0; j < n - i - 1; j++) {
if (array[j] > array[j + 1]) {
flag = false;
int temp = array[j + 1];
array[j + 1] = array[j];
array[j] = temp;
}
}
<MASK>
<UNMASK>
## Oct 13, 2015
### Optimizing Bubble Sort Algorithm
Having been in Java world for around 5 years, I believe Bubble Sort is the probably simplest and widely used sorting algorithm that works by repeatedly swapping the adjacent elements if they are not in the right order. However, with just a little tweaking, it’s possible to make it more efficient. So, we will first see what Bubble Sort algorithm is and then move to its optimization part.
To sort the entire array, the array is traversed n-1 time (array having n elements). These are called passes. In the first pass the largest element moves to the last position (sorting in ascending order). So if the original (unsorted) array is:
55 33 99 77 44
During the first iteration, adjacent elements will be compared, and swapped if larger element is on left side), as shown below:
After first iteration, the largest element is at the last position.
33 55 77 44 99
Now, in the 2nd pass, we will consider the first (n-1) elements only (because last position already has largest element). After 2nd pass the array will be
33 55 44 77 99
i.e., 77 will be moved to the (n-1)th position.
After (n-1) iterations, (n-1) elements will be moved to their proper positions, the last element has to be the smallest. So the array will be sorted after n-1 passes.
Algorithm:
void bubbleSort(int array[], int n) {
for (int i = 0; i < n; i++) {
for (int j = 0; j < n - i - 1; j++) {
if (array[j] > array[j + 1]) {
int temp = array[j + 1];
array[j + 1] = array[j];
array[j] = temp;
}
}
}
}
Time Complexity: O(n*n)
Auxiliary Space: O(1)
Boundary Cases: Bubble sort takes minimum time (Order of n) when elements are already sorted.
Sorting In Place: Yes
Stable: Yes
Improvement (Optimization):
In the above example, the array got sorted after 2nd pass, but we will still continue with the 3rd, 4th pass. Suppose if the array is already sorted, then there will be no swapping (because adjacent elements are always in order), but still we will continue with the passes and there will still be (n-1) passes.
If we can identify, that the array is sorted, then we should stop execution of further passes. This is the optimization over the original bubble sort algorithm.
If there is no swapping in an iteration pass, it means the array has become sorted, so algorithm should be smart enough to skip further iterations.
For this we can have a flag variable which is set to true before each pass and is made false when a swapping is performed.
void optimizedBubbleSort(int array[], int n) {
for (int i = 0; i < n; i++) {
boolean flag = true;
for (int j = 0; j < n - i - 1; j++) {
if (array[j] > array[j + 1]) {
flag = false;
int temp = array[j + 1];
array[j + 1] = array[j];
array[j] = temp;
}
}
// No Swapping happened, array is sorted
if (flag) {
return;
}
}
}
For the best case (array already sorted) the optimized algorithm will be O(n).
For average case also, the performance will see an improvement. Whereas the original algorithm was O(n2) for all the cases.
|
<MASK>
The quadratic formula calculator is an online calculator to solve a standard quadratic equation using the Quadratic Formula, which is -
Roots(x1, x2) = −b ± √ b2 − 4ac 2a
<MASK>
If b2 – 4ac = 0, then √b2 − 4ac is zero; in this case, Quadratic formula calculator gives us real and equal roots.
<MASK>
If b2 – 4ac is a perfect square, then √b2 − 4ac is a rational number; in this case, Quadratic formula calculator gives us rational roots, else Quadratic formula calculator gives us irrational roots
<MASK>
For example, let' suppose, we have to solve a quadratic equation which is in the form, x2 - 9x = -20. So in this case, we will convert the quadratic equation into Standard Form of Quadratic Equation, ax2 + bx +c = 0
<MASK>
Roots(x1, x2) = −b ± √ b2 − 4ac 2a
<MASK>
= −(-9) + √ (-9)2 − 4(1)(20) = 5 2(1)
x2 = −b - √ b2 − 4ac 2a
<MASK>
<UNMASK>
<MASK>
The quadratic formula calculator is an online calculator to solve a standard quadratic equation using the Quadratic Formula, which is -
Roots(x1, x2) = −b ± √ b2 − 4ac 2a
<MASK>
If b2 – 4ac = 0, then √b2 − 4ac is zero; in this case, Quadratic formula calculator gives us real and equal roots.
<MASK>
If b2 – 4ac is a perfect square, then √b2 − 4ac is a rational number; in this case, Quadratic formula calculator gives us rational roots, else Quadratic formula calculator gives us irrational roots
<MASK>
For example, let' suppose, we have to solve a quadratic equation which is in the form, x2 - 9x = -20. So in this case, we will convert the quadratic equation into Standard Form of Quadratic Equation, ax2 + bx +c = 0
<MASK>
Step 2 - Now, we have to compare the derived quadratic equation in standard form which is x2 - 9x + 20 = 0 with Standard Form of Quadratic Equation, ax2 + bx +c = 0 and find the value of a, b and c.
<MASK>
Roots(x1, x2) = −b ± √ b2 − 4ac 2a
<MASK>
= −(-9) + √ (-9)2 − 4(1)(20) = 5 2(1)
x2 = −b - √ b2 − 4ac 2a
= −(-9) - √ (-9)2 − 4(1)(20) = 4 2(1)
|
<MASK>
The quadratic formula calculator is an online calculator to solve a standard quadratic equation using the Quadratic Formula, which is -
Roots(x1, x2) = −b ± √ b2 − 4ac 2a
<MASK>
If b2 – 4ac = 0, then √b2 − 4ac is zero; in this case, Quadratic formula calculator gives us real and equal roots.
<MASK>
If b2 – 4ac is a perfect square, then √b2 − 4ac is a rational number; in this case, Quadratic formula calculator gives us rational roots, else Quadratic formula calculator gives us irrational roots
<MASK>
For example, let' suppose, we have to solve a quadratic equation which is in the form, x2 - 9x = -20. So in this case, we will convert the quadratic equation into Standard Form of Quadratic Equation, ax2 + bx +c = 0
<MASK>
Step 2 - Now, we have to compare the derived quadratic equation in standard form which is x2 - 9x + 20 = 0 with Standard Form of Quadratic Equation, ax2 + bx +c = 0 and find the value of a, b and c.
<MASK>
Roots(x1, x2) = −b ± √ b2 − 4ac 2a
<MASK>
= −(-9) + √ (-9)2 − 4(1)(20) = 5 2(1)
x2 = −b - √ b2 − 4ac 2a
= −(-9) - √ (-9)2 − 4(1)(20) = 4 2(1)
<UNMASK>
## ax2 + bx + c = 0
The quadratic formula calculator is an online calculator to solve a standard quadratic equation using the Quadratic Formula, which is -
Roots(x1, x2) = −b ± √ b2 − 4ac 2a
After solving the quadratic equation using a quadratic formula calculator, we get two values:
x1 = −b + √ b2 − 4ac 2a
<MASK>
These two values of x, x1 and x2, which are calculated from Quadratic Formula Calculator, are also known as roots of quadratic equation.
Nature of roots of Quadratic Equation
If b2 – 4ac > 0, then √b2 − 4ac is real; in this case, Quadratic formula calculator gives us two real and distinct roots.
If b2 – 4ac = 0, then √b2 − 4ac is zero; in this case, Quadratic formula calculator gives us real and equal roots.
<MASK>
If b2 – 4ac is a perfect square, then √b2 − 4ac is a rational number; in this case, Quadratic formula calculator gives us rational roots, else Quadratic formula calculator gives us irrational roots
How to use Quadratic formula calculator
Step 1 - To use the Quadratic formula calculator, we have to convert the Quadratic Equation to be calculated into Standard Form of Quadratic Equation, ax2 + bx + c = 0
For example, let' suppose, we have to solve a quadratic equation which is in the form, x2 - 9x = -20. So in this case, we will convert the quadratic equation into Standard Form of Quadratic Equation, ax2 + bx +c = 0
<MASK>
Step 2 - Now, we have to compare the derived quadratic equation in standard form which is x2 - 9x + 20 = 0 with Standard Form of Quadratic Equation, ax2 + bx +c = 0 and find the value of a, b and c.
For example, comparing x2 - 9x + 20 = 0 with ax2 + bx + c = 0, we get
<MASK>
Roots(x1, x2) = −b ± √ b2 − 4ac 2a
x1 = −b + √ b2 − 4ac 2a
= −(-9) + √ (-9)2 − 4(1)(20) = 5 2(1)
x2 = −b - √ b2 − 4ac 2a
= −(-9) - √ (-9)2 − 4(1)(20) = 4 2(1)
|
## ax2 + bx + c = 0
The quadratic formula calculator is an online calculator to solve a standard quadratic equation using the Quadratic Formula, which is -
Roots(x1, x2) = −b ± √ b2 − 4ac 2a
After solving the quadratic equation using a quadratic formula calculator, we get two values:
x1 = −b + √ b2 − 4ac 2a
<MASK>
These two values of x, x1 and x2, which are calculated from Quadratic Formula Calculator, are also known as roots of quadratic equation.
Nature of roots of Quadratic Equation
If b2 – 4ac > 0, then √b2 − 4ac is real; in this case, Quadratic formula calculator gives us two real and distinct roots.
If b2 – 4ac = 0, then √b2 − 4ac is zero; in this case, Quadratic formula calculator gives us real and equal roots.
<MASK>
If b2 – 4ac is a perfect square, then √b2 − 4ac is a rational number; in this case, Quadratic formula calculator gives us rational roots, else Quadratic formula calculator gives us irrational roots
How to use Quadratic formula calculator
Step 1 - To use the Quadratic formula calculator, we have to convert the Quadratic Equation to be calculated into Standard Form of Quadratic Equation, ax2 + bx + c = 0
For example, let' suppose, we have to solve a quadratic equation which is in the form, x2 - 9x = -20. So in this case, we will convert the quadratic equation into Standard Form of Quadratic Equation, ax2 + bx +c = 0
<MASK>
Step 2 - Now, we have to compare the derived quadratic equation in standard form which is x2 - 9x + 20 = 0 with Standard Form of Quadratic Equation, ax2 + bx +c = 0 and find the value of a, b and c.
For example, comparing x2 - 9x + 20 = 0 with ax2 + bx + c = 0, we get
<MASK>
Roots(x1, x2) = −b ± √ b2 − 4ac 2a
x1 = −b + √ b2 − 4ac 2a
= −(-9) + √ (-9)2 − 4(1)(20) = 5 2(1)
x2 = −b - √ b2 − 4ac 2a
= −(-9) - √ (-9)2 − 4(1)(20) = 4 2(1)
<UNMASK>
## ax2 + bx + c = 0
The quadratic formula calculator is an online calculator to solve a standard quadratic equation using the Quadratic Formula, which is -
Roots(x1, x2) = −b ± √ b2 − 4ac 2a
After solving the quadratic equation using a quadratic formula calculator, we get two values:
x1 = −b + √ b2 − 4ac 2a
<MASK>
These two values of x, x1 and x2, which are calculated from Quadratic Formula Calculator, are also known as roots of quadratic equation.
Nature of roots of Quadratic Equation
If b2 – 4ac > 0, then √b2 − 4ac is real; in this case, Quadratic formula calculator gives us two real and distinct roots.
If b2 – 4ac = 0, then √b2 − 4ac is zero; in this case, Quadratic formula calculator gives us real and equal roots.
<MASK>
If b2 – 4ac is a perfect square, then √b2 − 4ac is a rational number; in this case, Quadratic formula calculator gives us rational roots, else Quadratic formula calculator gives us irrational roots
How to use Quadratic formula calculator
Step 1 - To use the Quadratic formula calculator, we have to convert the Quadratic Equation to be calculated into Standard Form of Quadratic Equation, ax2 + bx + c = 0
For example, let' suppose, we have to solve a quadratic equation which is in the form, x2 - 9x = -20. So in this case, we will convert the quadratic equation into Standard Form of Quadratic Equation, ax2 + bx +c = 0
Hence, x2 - 9x = -20 in standard form would be x2 - 9x + 20 = 0
Step 2 - Now, we have to compare the derived quadratic equation in standard form which is x2 - 9x + 20 = 0 with Standard Form of Quadratic Equation, ax2 + bx +c = 0 and find the value of a, b and c.
For example, comparing x2 - 9x + 20 = 0 with ax2 + bx + c = 0, we get
a = 1,
b = -9,
c = 20
Step 3 - Once we have derived the value of coefficients a, b and c, then all we need is to insert these values in the quadratic formula, which is
Roots(x1, x2) = −b ± √ b2 − 4ac 2a
x1 = −b + √ b2 − 4ac 2a
= −(-9) + √ (-9)2 − 4(1)(20) = 5 2(1)
x2 = −b - √ b2 − 4ac 2a
= −(-9) - √ (-9)2 − 4(1)(20) = 4 2(1)
|
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