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Applied mathematics is one which is used in day-to-day life, in solving tensions (problems) or in business purposes. Let me write an example:
George had some money. He gave 14 Dollars to Matthew. Now he has 27 dollars. How much money had he?
If you are familiar with day-to-day calculations –you must say that George had 41 dollars, and since he had 41, gave 14 to Matthew saving 27 dollars. That’s right? Off course! This is a general(layman) approach. ‘How will we achieve it mathematically?’ –we shall restate the above problem as another statement (meaning the same):
George had some money $x$ dollars. He gave 14 dollars to Matthew. Now he has 27 dollars. How much money he had? Find the value of $x$ .$This is equivalent to the problem asked above. I have just replaced ‘some money’ by ‘x dollars’. As ‘some’ senses as unknown quantity—$ x$does the same. Now all we need to get the value of x. When solving for$ x$, we should have a plan like this: George had$ x$dollars. He gave to Matthew 14 dollars Now he must have$ x-14$dollars But problem says that he has 27 dollars left. This implies that$ x-14$dollars are equal to 27 dollars. i.e.,$ x-14=27 x-14=27$contains an alphabet x which we assumed to be unknown–can have any certain value. Statements (like$ x-14=27$) containing unknown quantities and an equality are called Equations. The unknown quantities used in equations are called variables, usually represented by bottom letters in English alphabet (e.g.,$ x,y,z$). Top letters of alphabet ($ a,b,c,d$..) are usually used to represent constants (one whose value is known, but not shown). Now let we concentrate on the problem again. We have the equation x-14=27. Now adding 14 to both sides of the equal sign:$ x-14 +14 =27 +14$or,$ x-0 = 41 $(-14+14=0) or,$ x= 41$. So,$ x$is 41. This means George had 41 dollars. And this answer is equal to the answer we found practically. Solving problems practically are not always possible, specially when complicated problems encountered —we use theory of equations. To solve equations, you need to know only four basic operations viz., Addition, Subtraction, Multiplication and Division; and also about the properties of equality sign. We could also deal above problem as this way:$ x-14= 27$or,$ x= 27+14 =41$-14 transfers to another side, which makes the change in sign of the value, i.e., +14. When we transport a number from left side to right of the equal sign, the sign of the number changes and vice-versa. As here -14 converts into +14; +18 converts into -18 in example below:$ x+18 =32$or,$ x=32 -18 =14$. Please note, any number not having a sign before its value is deemed to be positive—e.g., 179 and +179 are the same, in theory of equations. Before we proceed, why not take another example? Marry had seven sheep. Marry’s uncle gifted her some more sheep. She has eighteen sheep now. How many sheep did her uncle gift? First of all, how would you state it as an equation?$ 7 + x = 18$or,$ +7 +x =18$(just to illustrate that 7=+7) or,$ x= 18-7 =9$. So, Marry’s uncle gifted her 9 sheep. /// Now tackle this problem, Monty had some cricket balls. Graham had double number of balls as compared to Monty. Adam had also 6 cricket balls. They all collected their balls and found that total number of cricket balls was 27. How many balls had Monty and Graham? As usual our first step to solve this problem must be to restate it as an equation. We do it like this: Monty had (let) x balls. Then Graham must had$ x \times 2=2x$balls. Adam had 6 balls. The total sum=$ x+2x+6=3x+6$But that is 27 according to our question. Hence,$ 3x+6=27$or,$ 3x=27-6 =21$or,$ x=21 /3 =7$. Here multiplication sign converts into division sign, when transferred. Since$ x=7$, we can say that Monty had 7 balls (instead of x balls) and Graham had 14 (instead of$ 2x$). /// # Types of Equations They are many types of algebraic equations (we suffix ‘Algebraic’ because it includes variables which are part of algebra) depending on their properties. In common we classify them into two main parts: 1. Equations with one variable (univariable algebraic equations, or just Univariables) 2. Equations with more than one variables (multivariable algebraic equations, or just Multivariables) Univariable Equations Equations consisting of only one variable are called univariable equations. All of the equations we solved above are univariables since they contain only one variable (x). Other examples are:$ 3x+2=5x-3$;$ x^2+5x +3=0$;$ e^x =x^e$(e is a constant). Univariables are further divided into many categories depending upon the degree of the variable. Some most common are: 1. Linear Univariables: Equations having the maximum power (degree) of the variable 1.$ ax+b=c$is a general example of linear equations in one variable, where a, b and c are arbitrary constants. 2. Quadratic Equations: Also known as Square Equations, are ones in which the maximum power of the variable is 2.$ ax^2+bx+c=0$is a general example of quadratic equations, where a,b,c are constants. 3. Cubic Equations: Equations of third degree (maximum power=3) are called Cubic. A cubic equation is of type$ ax^3+bx^2+cx+d=0$; where a,b,c,d are constants. 4. Quartic Equations: Equations of fourth degree are Quartic. A quartic equation is of type$ ax^4+bx^3+cx^2+dx+e=0$. Similarly, equation of an n-th degree can be defined if the variable of the equation has maximum power n. Multivariable Equations: Some equations have more than one variables, as$ ax^2+2hxy+by^2=0$etc. Such equations are termed as Multivariable Equations. Depending on the number of variables present in the equations, multivariable equations can be classified as: 1. Bi-variable Equations – Equations having exactly two variables are called bi-variables.$ x+y=5$;$ x^2+y^2=4$;$ r^2+\theta^2=k^2$, where k is constant; etc are equations with two variables. Bivariable equations can also be divided into many categories, as same as univariables were. A.Linear Bivariable Equations: Power of a variable or sum of powers of product of two variables does not exceed 1. For example:$ ax+by=c$is a linear but$ axy=b$is not. B. Second Order Bivariable Equations: Power of a variable or sum of powers of product of two variables does not exceed 2. For example:$ axy=b$,$ ax^2+by^2+cxy+dx+ey+f=0$are of second order. Similarly you can easily define n-th order Bivariable equations. 2. Tri-variable Equations: Equations having exactly three variables are called tri-variable equations.$ x+y+z=5$;$ x^2+y^2-z^2=4$;$ r^3+\theta^3+\phi^3=k^3$, where k is constant; etc are trivariables. (Further classification of Trivariables are not necessary, but I hope that you can divide them into more categories as we did above.) similarly, you can easily define any n-variable equation as an equation in which the number of variables is n. Out of these equations, we shall discuss only Linear Univariable Equations here (actually we are discussing them). //// We have already discussed them above, for particular example. Here we’ll discuss them for general cases. As told earlier, a general example of linear univariable equation is$ ax+b=c$. We can adjust it by transferring constants to one side and keeping variable to other.$ ax+b = c$or,$ ax = c-b$or,$ x = \frac{c-b}{a}$this is the required solution. Example: Solve$ 3x+5=0$. We have,$ 3x+5=0$or,$ 3x = 0-5 =-5$or,$ x = \frac{-5}{3}\$ ////
### Posted by Gaurav Tiwari
A designer by profession, a mathematician by education but a Blogger by hobby. With an experience of over seven years with WordPress, PHP and CSS3, Gaurav is capable of doing almost anything related to these. Beyond that, He is a mathematics graduate & a civil service aspirant.
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# PATHSUMS ,October 2020 cookoff , Unofficial Editorial
Approach:-
Just assign value = 2 , for all even level nodes and value = 1 to all odd level nodes
Proof :
Let consider the path between two nodes whose length is even (>=2)
so what will be the pattern for value -
2 1 2 1 2 1
OR
1 2 1 2 1 2
means if path length is X then “2” occurs X/2 times and “1” also occurs X/2 times
total sum of this path values = (X/2)*2 + (X/2)*1 = (X + X/2) which is not completely divisible by X [path length] (as (n/2)%n !=0 it will be n/2 take n =6 then n/2 = 3 => 3%6 !=0 )
Now let consider the path between two nodes whose length is odd (>=3)
so what will be the pattern for value -
2 1 2 1 2
OR
1 2 1 2 1
means if path length is X then “2” occurs (X+1)/2 times and “1” occurs X/2 times
or “2” occurs X/2 times and “1” occurs (X+1)/2 times and in both case it will not completely divisible by X [ path length ] .
``````int vis[MAX];
lli depth[MAX];
void clear(lli n)
{
loop(i,n+1)
{
vis[i]=0;
depth[i]=0;
}
}
void bfs(lli src)
{
vis[src]=1;
queue<lli>q;
q.push(src);
depth[src]=0;
while(!q.empty())
{
lli u = q.front();q.pop();
{
if(!vis[xt])
{
vis[xt]=1;
q.push(xt);
depth[xt] = depth[u]+1;
}
}
}
}
int main(){
fastio
lli t=1;
cin>>t;
while(t--) {
lli n;
cin>>n;
loop(i,n-1)
{
lli x,y;
cin>>x>>y;
}
bfs(1);
lli a[n+1]={0};
FOR(i,1,n+1)
{
if(depth[i]%2)
{
a[i]=1;
}
else
a[i]=2;
}
FOR(i,1,n+1)
cout<<a[i]<<" ";
cout<<endl;
clear(n);
}
return 0;
}
``````
4 Likes
To be precise we do a level order traversal on the tree and assign 1 for the odd level and 2 for even (or) assign 2 for the odd level and 1 for even, so here he used bfs to get the level and assign the values accordingly.
1 Like
Just do BFS and find depth of each node after that assign 2 and 1 (or 1 and 2) for alternate level nodes.
1 Like
instead of finding depth each time, u can pass an array as an argument to ur bfs function and assign the values while you are pushing it in the queue right?
Check my code , I did only single time BFS and then it will calculate Depth of each node.
oh…k bruh
I didn’t understand, if there is a specific pattern to store it in the array and print.
can we directly print
1 2 1 2 1 2…n times ?
No , u don’t know whether the node number 2 is on odd level or even level , according to u , node number 2 is always at odd level.
I did simple one dfs
1 Like
nope you get it wrong, the pattern is you need to assign the opposite parity to the children’s value
1 Like
Great Doggo .
How you had observed that pattern I had tried but not able to find out
Take a paper and draw a graph for the first test case. Now assign 1 to each node on the odd level and 2 to each node on even level or vice-versa. Try out every path until you are convinced.
just hit and try bro , and also after observing i proved in my copy and coded it . I m unable to solve 4’th one as not observe the specific pattern in 4’th .
A simple BFS won’t work here. You need to do DFS as the assigning of node values depend upon the depth of nodes from the root.
In my case, BFS failed so I went for DFS.
i brute forced by assigning values lol
yupp bro, I tried so hard for 4th one but no pattern,
1 Like
Check my code .
OK bro I am thinking in same way but that 1,2,1 idea had not strike in mind i am thinking in term of odd even
1 Like
1-2 idea is also odd even
1 Like
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What are the second partial derivatives , z (x,y) and z (y,x) if z = x^2*y + 2x*e^1/y ?
# What are the second partial derivatives , z (x,y) and z (y,x) if z = x^2*y + 2x*e^1/y ?
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This one is tough. No doubt about it. If one were to assess affirmative action in a vacuum and without any sort of context, I do believe that an argument could be made that it goes against the…
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# You Can't Predict Small Counts
January 17, 2015
A small restaurant is interested in predicting how many customers will come in on a given night. This is valuable information to know ahead of time, for example, so that the restaurant can figure out whether to ask employees to work extra shifts. Unfortunately, under reasonable conditions no amount of data will permit even the most talented data scientist or statistician to make particularly good predictions.
## Poisson Model
The reason is that there is too much fundamental variance associated with the nightly counts. As an illustrative approximation, we can model the process as follows: suppose the restaurant is in a city with n people. Each night, every person in the city decides whether or not to come in that night by independently flipping a p-coin. This model leads to a binomial number of people coming in each night: $$Y = \sum_{i=1}^n X_i,$$ where $Y$ is the total number of people coming in that night and the $X_i$'s are independent Bernoulli(p)'s.
Now, n is probably pretty large, (there are a lot of people in the city), and p is probably pretty small, (any individual is relatively unlikely to attend on a given night), so that this binomial distribution is approximately Poisson. An important feature of the Poisson distribution is that it has equal mean and variance.
So, for example, suppose that on a given day the expected number of people coming in that night is about 50. Then the standard deviation of that count would be about 7 (actually sqrt(50)). Since the Poisson distribution with a large mean is approximately normal, if we go out two standard deviations on both sides we'll get an approximate 95% prediction interval: 36 to 64 people.
In other words, with approximately 95% probability somewhere between 36 and 64 people will show up. That is not a particularly impressive guess.
In practice, of course, we don't know the true expected number of people who'll show up. So we gather a bunch of historical restaurant attendance data and train a state of the art machine learning model to estimate that attendance from the day of week, day of month, day of year, weather that day, attendance the previous days, etc. But in the best case, all this allows us to do is predict the true expected number of customers.
Suppose, for example, that with all that fanciness we predict that the true expected number of people who'll show up is 48.0123, instead of our ballpark guess of 50. Since sqrt(50) is about the same as sqrt(48.0123), the error bars will still be of about the same width, (in fact, slightly less wide). So instead of our old interval, (from 36 to 64), we instead estimate the interval from 34 to 62. For many applications, isn't a very exciting difference.
(There are some applications where it is pretty useful, however: If a sushi restaurant has a policy of buying fish that morning so that they only run out 2.5% of the time, then using machine learning would allow them to buy for 62 people instead of 64 people, saving them a little money. The fundamental variance in the nightly arrivals, however, means that even with the best machine learning the restaurant will still be wasting quite a lot of fish on a typical day).
## Overdispersed Poisson
There are a few interesting violations of the above Poisson model. One is that people do not choose whether to come to the restaurant independently. In fact, people often decide in small groups where to eat dinner, and then all go out together.
This leads to what statisticians call "overdispersion", where the variance in the number of people arriving is even higher than it would be under the Poisson distribution. A more reasonable model in this circumstance is that a Poisson number of groups come, and each group has some iid distribution over sizes. Letting $N$ be Poisson, and $Z_i$ be the size of the ith group, this model is $$Y = \sum_{i=1}^N Z_i$$ (In the special case where $Z_i = 1$, then $Y = N$ and this is the Poisson model above).
It's not too hard to derive that the expected number of customers is the mean number of groups times the mean group size, and that the variance in the number of customers is the mean number of groups times the expected squared group size. That is, $$E\left[Y\right] = E\left[N\right] E\left[Z_1\right]$$ and $$Var\left[Y\right] = E\left[N\right] E\left[Z_1^2\right].$$
This leads to significantly higher variance than in the above naive model. Instead of the variance equaling the expectation, it's higher by a factor of $$\frac{E\left[Z_1^2\right]}{E\left[Z_1\right]}.$$
For example, suppose that each night an average of 25 groups come in. Suppose that a third of the groups have one person, a third have two people, and a third have three people. Then the average group has two people in it, so that an average of 50 (=25*2) people come in every night. The expected squared group size, however, is 14/3 = 4.67. This makes the standard deviation of the number of people who come in each night 10.8 (=sqrt(4.67*25)), substantially greater than 7.1, the standard deviation we'd have if each group had just one person.
The approximate 95% prediction interval then widens to anywhere from 28 to 72 people, making the prediction even less exciting. And this window will be even wider if you have a decent fraction of groups larger than 3 people. No matter how good your machine learning is, under this reasonable model you won't be able to make predictions with a mean square error of better than 10.8 people.
## Individual-Level Predictions
Another interesting violation of the Poisson model is in the assumption that p is the same and small for all individuals or groups in the city. This assumption is clearly violated, for example, if the restaurant takes reservations. In that case, there are some people in the city whose probability of coming that night is near to 1. Unlike in the overdispersed Poisson model above, this violation has the effect of lowering the variance relative to the naive Poisson model, rather than increasing it.
The models above also implicitly assume temporal independence; that each night people make decisions to come independent of their previous decisions. However, if the restaurant has a population of "regulars" who come in most Friday nights, for example, then this assumption is violated and the variance in arrivals will be lower on Fridays than it otherwise would be.
More generally, you can prove the following theorem: Suppose $$Y = \sum_{i=1}^n Z_i X_i,$$ where $n$ is the total number of potential groups, $Z_i$ (the size of the ith group) are iid, $X_i$ (whether the group comes) is a Bernoulli with success parameter $p_i$, and all the $Z_i$ and $X_i$ are independent. Then among all choices of $p_i$ with $E\left[Y\right] = \mu$, the choice of the $p_i$ that maximizes $Var\left[Y\right]$ has all $p_i = p$.
This is a long-winded way of saying that if some particular groups are more likely to come than others, the variance in the number of arrivals will be lower than it otherwise would be. If you have data that allows you to estimate which particular groups are going to come, you'll be able to do better than the "Poisson bound".
## Fundamental Variance
Regardless of the true model, the thing to keep in mind is that for count data the fundamental variance associated with the counts can be quite high. This is particularly true if the counts are small or if you have overdispersion. If this is the case, it may be impossible for you to make good predictions, no matter smart you are or how many observations you collect.
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# Geometry
Total word count: 796
Pages: 3
### Calculate the Price
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275 words
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Parallel lines Two lines that lie in the same plane and do not intersect. Perpendicular lines Two lines that intersect to form a right angle. Skew lines Two lines that do not lie in the same plane and do not intersect Parallel planes Two planes that do not intersect Transversal A line that intersects two or more coplanar lines at different points Corresponding angles Two angles formed by two lines and a transversal, and occupy corresponding positions Alternate interior angles Two angles formed by two lines and a transversal and lie between the two lines on the opposite sides of the transversal Alternate exterior angles Two angles formed by two lines and a transversal and lie outside the two lines on the opposite sides of the transversal Same-side interior angles Two angles formed by two lines and a transversal and lie between the two lines on the same side of the transversal Converse The statement formed by switching the hypothesis and the conclusion in an if-then statement Hypothesis The "if" part of an if-then statement Conclusion The "then" part of an if-then statement All right angles are ____________ congruent If two lines are perpendicular, then they intersect to form four _______ ________. right angles If two lines intersect to form adjacent congruent angles, then the lines are _____________________. perpendicular If two sides of adjacent acute angles are perpendicular, then the angles are ___________________. complementary If two parallel lines are cut by a transversal, then ______________ interior angles are congruent. alternate If two parallel lines are cut by a transversal, then _____________ exterior angles are congruent. alternate If two parallel lines are cut by a transversal, then _______________ interior angles are supplementary. same-side If two lines are cut by a transversal so that alternate interior angles are congruent, then the lines are ___________________. parallel If two lines are cut by a transversal so that alternate exterior angles are congruent, then the lines are _____________________. parallel If two lines are cut by a transversal so that same-side interior angles are _________________, then the lines are parallel. supplementary If two lines are parallel to the same line, then they are _________ to each other. parallel In a plane, if two lines are perpendicular to the same line, then they are ___________ to each other. parallel Midpoint The point on a segment that divides the segment into two equal parts. Segment bisector Something that intersects a segment at it’s midpoint. Bisect To divide into two congruent parts. Angle bisector A ray that divides an angle into two equal angles. Complementary angles Two angles whose measures have a sum of 90 degrees. Supplementary angles Two angles whose measure has a sum of 180 degrees. Adjacent angles Two angles with a common vertex and side but no common interior points. Theorem A true statement that follows from other true statements. Vertical angles Two nonadjacent angles formed by two intersecting lines. Linear pair Two adjacent angles whose noncommon sides are on the same line (supplementary). If-then statements A statement with a hypothesis (if) and conclusion (then). Deductive reasoning Using facts, definitions, accepted properties to make a logical argument. Congruent complements theorem If two angles are complementary to the same angle, then they are congruent. Congruent supplements theorem If two angles are supplements to the same angle, then they are congruent. Vertical angles theorem Vertical angles are congruent. Reflexive Property A=A Symmetric Property If A=B, then B=A Transitive Property 2.1 Congruent Complements Theorem If two angles are complements to the same angle, then they are congruent. 2.2 Congruent Supplements Theorem If two angles are supplements to the same angle, then they are congruent. 2.3 Vertical Angles Theorem When two lines intersect, they form two sets of vertical angels. Vertical angles are congruent.
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Check Price
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A Guide to Using Checks
# See How to Write Dollars and Cents on a Check
Sometimes it’s the small things that get you: You might be familiar with checks, but you get stuck writing out the amount. Writing a check with cents is especially tricky, but with a little bit of practice, you’ll soon be able to do it without thinking.
• ### 01 Sample Check With Dollars and Cents
For example, assume you need to write a check for eight dollars and fifteen cents (that’s \$8.15). There are two steps:
1. Write the amount using numbers (see the red number one in the picture above).
2. Write the amount using words (see the red number two in the image above).
First, write the amount in numeric form in the dollar box, located on the right side of your check next to the dollar sign (“\$”). Start by writing the number of dollars (“8”) followed by a decimal point or period (“.”), and then the number of cents (“15”). Ultimately, you’ll have “8.15” in the dollar box. For more examples and practice questions, scroll down.
Next, to write out the check’s amount in words, the two steps are similar:
1. Write out the dollar amount.
2. Write the word “and.”
3. Write out the number of cents.
The tricky part is putting the number of cents into fraction format. To do so, write the number of cents, then write a slash (“/”), and then write the number 100. Technically, this is the fractional amount of whole dollars.
Using our \$8.15 example, write the following:
1. “Eight dollars”
2. “and”
3. “15/100”
Write everything together on one line so that it reads “Eight dollars and 15/100.” For a detailed example of how to write a check, see a step-by-step tutorial that uses the same amount.
Now that you have the basic idea, let’s look at the example in more detail.
No “cents”: You might notice that the word “cents” doesn’t appear anywhere—you don’t need to use it when writing a check. It is sufficient to simply put the number of cents into the format above. If you want, you can certainly write “fifteen cents,” but it’s easier and faster to use the fraction format. Plus, your check probably has the word “Dollars” at the end of the line, so it would not make sense.
The word “and”: Include the word “and” just before you write how many cents the check is for (or just after you write out the full dollar amount). You are writing a check for dollars and cents. If you like, you can use an ampersand (“&”) or plus sign (“+”) instead. It is best not to use the word “and” elsewhere when you write out the amount. For example, the following example is incorrect, and the word “and” should be removed: “One hundred and five dollars.”
### Percentages
It might help to think in terms of percentages: The word percent comes from a Latin term that roughly translates to “per 100.” That’s why cents are called “cents”—each one is one percent of a dollar. Another way of looking at it is to consider that each cent is one one-hundredth of a dollar. When you write a check, you note how many dollars the check is for, including whole dollars as well as partial dollars—or cents.
• ### 02 More Examples
To solidify the concept and develop the habit, try several different dollar amounts.
Example: Write ten dollars and 99 cents on a check.
1. 10.99
2. Ten and 99/100
Example: Write eleven dollars and five cents on a check.
1. 11.05
2. Eleven and 5/100
Example: write a check for five dollars.
1. 5.00. Note the double zero—you should always have two digits to the right of the decimal.
2. Five and 00/100. Here, you can use one or two zeros, but two is safer.
Example: Write a seventy-five-cent check.
1. 0.75
2. Zero dollars and 75/100
You might have noticed that the last example was for less than one dollar. To write a check for less than a full dollar, use a zero to show that there aren’t any dollars. After that, include the number of cents just like all of the other examples. You can also write “No dollars and….” if you prefer.
The five dollar example can also be confusing. Just write a zero (or double zero) when there isn’t any other number to use. Some people would write that amount out as “Five dollars only,” which is also fine.
• ### 03 Tip: Don't Write Checks
Want to make your life even easier? Use fewer checks—or at least have your bank write your checks for you.
Online bill payment allows you to set up automatic payments or just pay when you feel like it. Your bank will pay electronically if possible, or print a check and mail it. This service is typically free with most checking accounts, and you can send payments to businesses and individuals.
Debit cards can be used at merchants and online retailers. Just like a check, your debit card pulls funds from your checking account. For everyday spending, it may be safer to use credit cards to reduce the chance of errors and fraud hitting your checking account.
P2P payment services help you send money to friends and family, often for free. Those services draw money from your checking account electronically, so you’ll automatically have a record of every transaction.
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In triangle ABC, PQ drawn parallel to BC meets the side AB at P and the side AC at Q. The lines CP and BQ intersect each other at S. SR drawn parallel to AB meets BC at R. Prove that BR RC = AQ AC
Theorems: A line drawn parallel to a side of a triangle divides the other two sides proportionally. If two triangles are equiangular then the corresponding sides are proportional. If the three sides of a triangle are proportional to the three sides of another triangle, then the two triangles are equiangular.
This answer has been revised as a result of comments and corrections and is now believed to be correct.
Point T has been inserted as the intersection of RS extended to meet PQ.
Similar triangles ABC and APQ: AQ/AC=PQ/BC.
To make it easier to see the similar triangles PQS and BCS, the points P', Q' and T' have been inserted such that P'Q' is the same length as PQ, making triangles PQS and P'Q'S congruent. It's then easier to appreciate the ratios of the sides of the similar triangles, because ST=S'T', PS=P'S, etc.
Similar triangles PQS and BCS: PQ/BC=QS/BS=PS/CS (angles QPS=BCS, PQS=CBS, PSQ=BSC, alternate angles between parallel lines). With the introduction of the extra points, we can also write P'Q'/BC=Q'S/BS=P'S/CS.
BR=PT in parallelogram BRTP and PT=P'T'. AQ/AC=PQ/BC=(PT+QT)/(BR+CR)=(BR+QT)/(BR+CR).
P'T'/CR=Q'T'/BR so PT/CR=QT/BR, because ST'/RS=ST/RS is a common ratio in triangles P'ST' and CRS, and Q'ST' and BRS.
But BR=PT so BR/CR=QT/BR, QT=BR^2/CR and AQ/AC=PQ/BC=(BR+BR^2/CR)/(BR+CR)=BR(CR+BR)/(CR(BR+CR))=BR/CR. Hence AQ/AC=BR/CR, in other words BR/RC=AQ/AC.
answered Sep 8, 2016 by Top Rated User (425,260 points)
reshown Sep 8, 2016 by Rod
Sorry. There's mistake in the question. Here is the right one.
In triangle ABC, PQ drawn parallel to BC meets the side AB at P and the side AC at Q. The lines CP and BQ intersect each other at S. SR drawn parallel to AB meets BC at R. Prove that BR/RC = AQ/AC
And Mr.Rod there is mistake in your picture. SR not parallel to AC. It's parallel to AB.
Thank you for that and for pointing out my error. I'll start again.
I think this the best answer for this math. Thank you so much Mr.Rod.
Much appreciated, math93. Got there in the end!
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# Thread: Differential Equations in Clinical Medicine
1. ## Differential Equations in Clinical Medicine
Group Project B, Chapter 2, Page 80 in Fundamentals of Differential Equations 8E.
Does anyone have Fundamentals of Differential Equations 8E?
I need help with Group Project B. It is very.. lengthy.
The book can be found on slideshare, or
Ecuaciones diferenciales 4ta edición - r. nagle, e. saff & a. snide?
2. ## Re: Differential Equations in Clinical Medicine
If you understand the material and time management is the only problem, then I suggest you start working on this now. By the way, the provided page number is not for Group project B for the given link.
3. ## Re: Differential Equations in Clinical Medicine
Who said anything about time management? I have a specific question about solving the linear equation in part (b), re equation (3). I'm looking for someone who has time to go to slideshare.net and look up the right book, who has time to read the whole problem, and has the knowledge to spare and share. Nice quotes
4. ## Re: Differential Equations in Clinical Medicine
The solution of a first order linear differential equation of the form:
$\displaystyle \alpha \frac{dy}{dx} + f(x) y = r(x) \implies \frac{dy}{dx} + \frac{f(x)}{\alpha} y = \frac{r(x)}{\alpha} , \,\, r(x) \not\equiv 0$
where alpha is a constants, then we have:
$\displaystyle y= \frac{1}{\mu (x)} \left ( \int \mu (x) \cdot r(x) dx + C \right ) , \, \, \mu (x)= e^{ \int \frac{f(x)}{\alpha} dx}$
I took a glance at your problem, and it is just using different variables.
Ps. I went and looked up the problem because I am currently studying DE and wanted more practice, but someone else may not have. In the future please post all questions in the thread itself (show some initiative if you want help).
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# NFA TO DFA.
## Presentation on theme: "NFA TO DFA."— Presentation transcript:
NFA TO DFA
Non Deterministic Features of NFA
There are three main cases of non- determinism in NFAs: Transition to a state without consuming any input. Multiple transitions on the same input symbol. No transition on an input symbol. To convert NFAs to DFAs we need to get rid of non-determinism from NFAs.
Subset Construction Method
Using Subset construction method to convert NFA to DFA involves the following steps: For every state in the NFA, determine all reachable states for every input symbol. The set of reachable states constitute a single state in the converted DFA (Each state in the DFA corresponds to a subset of states in the NFA). Find reachable states for each new DFA state, until no more new states can be found.
Subset Construction Method
Fig1. NFA without λ-transitions
Subset Construction Method
Fig1. NFA without λ-transitions 3 a b a a 2 a b a,b 1 5 a a,b 4 b
Subset Construction Method
Step1 Fig1. NFA without λ-transitions 3 a b a a 2 a b Construct a transition table showing all reachable states for every state for every input signal. a,b 1 5 a a,b 4 b
Subset Construction Method
Fig1. NFA without λ-transitions Fig2. Transition table 3 a b a a 2 a b a,b 1 5 a a,b 4 b
Subset Construction Method
Fig2. Transition table Fig1. NFA without λ-transitions 3 q δ(q,a) δ(q,b) 1 {1,2,3,4,5} {4,5} 2 {3} {5} 3 {2} 4 {4} 5 a b a a 2 a b a,b 1 5 a a,b 4 b
Subset Construction Method
Transition from state q with input a Fig2. Transition table Transition from state q with input b Fig1. NFA without λ-transitions 3 q δ(q,a) δ(q,b) 1 {1,2,3,4,5} {4,5} 2 {3} {5} 3 {2} 4 {4} 5 a b Starts here a a 2 a b a,b 1 5 a a,b 4 b
Subset Construction Method
Fig2. Transition table q δ(q,a) δ(q,b) 1 {1,2,3,4,5} {4,5} 2 {3} {5} 3 {2} 4 {4} 5 Step2 The set of states resulting from every transition function constitutes a new state. Calculate all reachable states for every such state for every input signal.
∅ q δ(q,a) δ(q,b) q δ(q,a) δ(q,b) 4 5 Fig2. Transition table 1 {4,5} 2
Fig3. Subset Construction table q δ(q,a) δ(q,b) 1 {1,2,3,4,5} {4,5} 2 {3} {5} 3 {2} 4 {4} 5 Starts with Initial state q δ(q,a) δ(q,b) 1 {1,2,3,4,5} {4,5}
∅ q δ(q,a) δ(q,b) q δ(q,a) δ(q,b) 4 5 Fig2. Transition table 1 {4,5} 2
Fig3. Subset Construction table q δ(q,a) δ(q,b) 1 {1,2,3,4,5} {4,5} 2 {3} {5} 3 {2} 4 {4} 5 Starts with Initial state q δ(q,a) δ(q,b) 1 {1,2,3,4,5} {4,5}
∅ q δ(q,a) δ(q,b) q δ(q,a) δ(q,b) Step3 4 5 Fig2. Transition table 1
Fig3. Subset Construction table q δ(q,a) δ(q,b) 1 {1,2,3,4,5} {4,5} 2 {3} {5} 3 {2} 4 {4} 5 Starts with Initial state q δ(q,a) δ(q,b) 1 {1,2,3,4,5} {4,5} Step3 Repeat this process(step2) until no more new states are reachable.
∅ q δ(q,a) δ(q,b) q δ(q,a) δ(q,b) 4 5 Fig2. Transition table 1 {4,5} 2
Fig3. Subset Construction table q δ(q,a) δ(q,b) 1 {1,2,3,4,5} {4,5} 2 {3} {5} 3 {2} 4 {4} 5 q δ(q,a) δ(q,b) 1 {1,2,3,4,5} {4,5} {2,4,5}
∅ q δ(q,a) δ(q,b) q δ(q,a) δ(q,b) 4 5 5 4 Fig2. Transition table 1
Fig3. Subset Construction table q δ(q,a) δ(q,b) 1 {1,2,3,4,5} {4,5} 2 {3} {5} 3 {2} 4 {4} 5 q δ(q,a) δ(q,b) 1 {1,2,3,4,5} {4,5} {2,4,5} 5 4
∅ q δ(q,a) δ(q,b) q δ(q,a) δ(q,b) 4 5 5 4 Fig2. Transition table 1
Fig3. Subset Construction table q δ(q,a) δ(q,b) 1 {1,2,3,4,5} {4,5} 2 {3} {5} 3 {2} 4 {4} 5 q δ(q,a) δ(q,b) 1 {1,2,3,4,5} {4,5} {2,4,5} 5 4 {3,5}
∅ ∅ q δ(q,a) δ(q,b) q δ(q,a) δ(q,b) 4 5 5 4 Fig2. Transition table 1
Fig3. Subset Construction table q δ(q,a) δ(q,b) 1 {1,2,3,4,5} {4,5} 2 {3} {5} 3 {2} 4 {4} 5 q δ(q,a) δ(q,b) 1 {1,2,3,4,5} {4,5} {2,4,5} 5 4 {3,5}
∅ ∅ q δ(q,a) δ(q,b) q δ(q,a) δ(q,b) 4 5 5 4 Fig2. Transition table 1
Fig3. Subset Construction table q δ(q,a) δ(q,b) 1 {1,2,3,4,5} {4,5} 2 {3} {5} 3 {2} 4 {4} 5 q δ(q,a) δ(q,b) 1 {1,2,3,4,5} {4,5} {2,4,5} 5 4 {3,5} We already got 4 and 5. So we don’t add them again.
∅ ∅ q δ(q,a) δ(q,b) q δ(q,a) δ(q,b) 4 5 5 4 Fig2. Transition table 1
Fig3. Subset Construction table q δ(q,a) δ(q,b) 1 {1,2,3,4,5} {4,5} 2 {3} {5} 3 {2} 4 {4} 5 q δ(q,a) δ(q,b) 1 {1,2,3,4,5} {4,5} {2,4,5} 5 4 {3,5} 2
∅ ∅ q δ(q,a) δ(q,b) q δ(q,a) δ(q,b) 4 5 5 4 Fig2. Transition table 1
Fig3. Subset Construction table q δ(q,a) δ(q,b) 1 {1,2,3,4,5} {4,5} 2 {3} {5} 3 {2} 4 {4} 5 q δ(q,a) δ(q,b) 1 {1,2,3,4,5} {4,5} {2,4,5} 5 4 {3,5} 2
∅ ∅ q δ(q,a) δ(q,b) q δ(q,a) δ(q,b) 4 5 5 4 Fig2. Transition table 1
Fig3. Subset Construction table q δ(q,a) δ(q,b) 1 {1,2,3,4,5} {4,5} 2 {3} {5} 3 {2} 4 {4} 5 q δ(q,a) δ(q,b) 1 {1,2,3,4,5} {4,5} {2,4,5} 5 4 {3,5} 2 3
∅ ∅ q δ(q,a) δ(q,b) q δ(q,a) δ(q,b) 4 5 5 4 Fig2. Transition table 1
Fig3. Subset Construction table q δ(q,a) δ(q,b) 1 {1,2,3,4,5} {4,5} 2 {3} {5} 3 {2} 4 {4} 5 q δ(q,a) δ(q,b) 1 {1,2,3,4,5} {4,5} {2,4,5} 5 4 {3,5} 2 3 Stops here as there are no more reachable states
Fig4. Resulting FA after applying Subset Construction to fig1
Fig3. Subset Construction table a q δ(q,a) δ(q,b) 1 {1,2,3,4,5} {4,5} {2,4,5} 5 4 {3,5} 2 3 b a 12345 245 35 a a,b a b b a 1 3 b a,b b a 2 a 45 5 b b a 4 b
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# Thread: Factorising problem? with two "squared" brackets
1. ## Factorising problem? with two "squared" brackets
So here is my math problem - I think it was supposed to be just straightforward factorising, but since it isn't a generic quadratic equation (or maybe it is haha, I can't tell if it is, in this form!) I can't work out how to factorise right down to the answer the book says. Here is the problem.
Factorise the expression
(x - 3)2 - (y - 2)2
The answer is apparently: (x + y - 5)(x - y - 1)
but the only thing I could work out how to do was: (x - 3)(x - 3) - (y - 2)(y - 2)
Hint: use the difference of squares formula: $a^2-b^2=(a+b)(a-b)$
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Students of Grade 3 can get a strong foundation by referring to the Eureka Math Book. It was developed by highly professional mathematics educators and the solutions prepared by them are in a concise manner for easy grasping. To achieve high scores in Grade 3, students need to solve all questions and exercises included in Eureka Grade 3 Book.
Engage NY Eureka Math 3rd Grade Module 3 Lesson 12 Answer Key
Teachers and students can find this Eureka Book Answer Key for Grade 3 more helpful in raising students’ scores and supporting teachers to educate the students. Learning activities are the best option to educate elementary school kids and make them understand the basic mathematical concepts like addition, subtraction, multiplication, division, etc. Grade 3 elementary school students can find these fun-learning exercises for all math concepts through the Eureka Book.
Eureka Math Grade 3 Module 3 Lesson 12 Pattern Sheet Answer Key
Multiply
8 x 1 = 8, 8 x 2 = 16, 8 x 3 = 24, 8 x 5 = 40, 8 x 6 = 48, 8 x 7 = 56, 8 x 8 = 64, 8 x 9 = 72, 8 x 10 = 80, 8 x 5 = 40, 8 x 6 = 48, 8 x 5 = 40, 8 x 7 = 56, 8 x 5 = 40, 8 x 6 = 48, 8 x 5 = 40, 8 x 7 = 56, 8 x 5 = 40, 8 x 8 = 64, 8 x 5 = 40, 8 x 9 = 72, 8 x 5 = 40, 8 x 10 = 80, 8 x 6 = 48, 8 x 5 = 40, 8 x 6 = 48, 8 x 7 = 56, 8 x 6 = 48, 8 x 8 = 64, 8 x 6 = 48, 8 x 9 = 72, 8 x 6 = 48, 8 x 7 = 56, 8 x 6 = 48, 8 x 7 = 56, 8 x 8 = 64, 8 x 7 = 56, 8 x 9 = 72, 8 x 7 = 56, 8 x 8 = 64, 8 x 6 48, 8 x 8 = 64, 8 x 7 = 56, 8 x 8 = 64, 8 x 9 = 72, 8 x 9 = 72, 8 x 6 = 48, 8 x 9 = 72, 8 x 7 = 56, 8 x 9 = 72, 8 x 8 = 64, 8 x 9 = 72, 8 x 8 = 64, 8 x 6 = 48, 8 x 9 = 72, 8 x 7 = 56, 8 x 9 = 72, 8 x 6 = 48, 8 x 8 = 64, 8 x 9 = 72, 8 x 7 = 56, 8 x 6 = 48, 8 x 8 = 64.
Explanation:
In the above-given question,
given that,
the table of 8 and multiply.
8 x 1 = 8, 8 x 2 = 16, 8 x 3 = 24, 8 x 5 = 40, 8 x 6 = 48, 8 x 7 = 56, 8 x 8 = 64, 8 x 9 = 72, 8 x 10 = 80, 8 x 5 = 40, 8 x 6 = 48, 8 x 5 = 40, 8 x 7 = 56, 8 x 5 = 40, 8 x 6 = 48, 8 x 5 = 40, 8 x 7 = 56, 8 x 5 = 40, 8 x 8 = 64, 8 x 5 = 40, 8 x 9 = 72, 8 x 5 = 40, 8 x 10 = 80, 8 x 6 = 48, 8 x 5 = 40, 8 x 6 = 48, 8 x 7 = 56, 8 x 6 = 48, 8 x 8 = 64, 8 x 6 = 48, 8 x 9 = 72, 8 x 6 = 48, 8 x 7 = 56, 8 x 6 = 48, 8 x 7 = 56, 8 x 8 = 64, 8 x 7 = 56, 8 x 9 = 72, 8 x 7 = 56, 8 x 8 = 64, 8 x 6 48, 8 x 8 = 64, 8 x 7 = 56, 8 x 8 = 64, 8 x 9 = 72, 8 x 9 = 72, 8 x 6 = 48, 8 x 9 = 72, 8 x 7 = 56, 8 x 9 = 72, 8 x 8 = 64, 8 x 9 = 72, 8 x 8 = 64, 8 x 6 = 48, 8 x 9 = 72, 8 x 7 = 56, 8 x 9 = 72, 8 x 6 = 48, 8 x 8 = 64, 8 x 9 = 72, 8 x 7 = 56, 8 x 6 = 48, 8 x 8 = 64.
Eureka Math Grade 3 Module 3 Lesson 12 Problem Set Answer Key
Question 1.
Each has a value of 9. Find the value of each row. Then, add the rows to find the total.
a. 6 × 9 = ___54____
6 × 9 = (5 + 1) × 9
= (5 × 9) + (1 × 9)
= 45 + ___9__
= __54___
6 x 9 = 54.
Explanation:
In the above-given question,
given that,
6 x 9 = ( 5 + 1 ) x 9.
( 5 x 9 ) + ( 1 x 9 )
45 + 9
54.
b. 7 × 9 = ____63___
7 × 9 = (5 + __2__) × 9
= (5 × 9) + (__2__ × 9)
= 45 + __18___
= ___63__
7 x 9 = 63.
Explanation:
In the above-given question,
given that,
7 x 9 = ( 5 + 2 ) x 9.
( 5 x 9 ) + ( 2 x 9 )
45 + 18.
63.
c. 8 × 9 = ___72____
8 × 9 = (5 + ___3__) × 9
= (5 × 9) + (___3__ × __9___)
= 45 + __27___
= __72___
8 x 9 = 72.
Explanation:
In the above-given question,
given that,
8 x 9 = ( 5 + 3 ) x 9.
( 5 x 9 ) + ( 3 x 9 )
45 + 27.
72.
d. 9 × 9 = ____81___
9 × 9 = (5 + ___4__) × 9
= (5 × 9) + (__4___ × ___9__)
= 45 + __36___
= __81___
9 x 9 = 81.
Explanation:
In the above-given question,
given that,
9 x 9 = ( 5 + 4 ) x 9.
( 5 x 9 ) + ( 4 x 9 )
45 + 36.
81.
Question 2.
Find the total value of the shaded blocks.
a. 9 × 6 = 54
9 sixes = 10 sixes – 1 six
=__60___ – 6
= __54___
9 x 6 = 54.
Explanation:
In the above-given equation,
given that,
9 sixes = 10 sixes – 1 six.
60 – 6.
54.
b. 9 × 7 = 63
9 sevens = 10 sevens – 1 seven
= __70___ – 7
= __63___
9 x 7 = 63.
Explanation:
In the above-given equation,
given that,
9 sevens = 10 sevens – 1 seven.
70 – 7.
63.
c. 9 × 8 = 72.
9 eights = 10 eights – 1 eight
= __80___ – 8
= __72___
9 x 8 = 72.
Explanation:
In the above-given equation,
given that,
9 eights = 10 eights – 1 eight.
80 – 8.
72.
d. 9 × 9 = 81.
9 nines = 10 nines – 1 nine
= __90___ – __9___
= __81___
9 x 9 = 81.
Explanation:
In the above-given equation,
given that,
9 nines = 10 nines – 1 nine.
90 – 9.
81.
Question 3.
Matt buys a pack of postage stamps. He counts 9 rows of 4 stamps. He thinks of 10 fours to find the total number of stamps. Show the strategy that Matt might have used to find the total number of stamps.
9 x 4 = 36.
Explanation:
In the above-given equation,
given that,
9 fours = 10 fourss – 1 four.
40 – 4.
36.
Question 4.
Match
3 x 9 = 27, 45 / 9 = 5.
9 x 9 = 81, 9 / 9 = 1.
8 x 9 = 72, 90 / 9 = 10.
9 x 4 = 36, 72 / 9 = 8.
2 x 9 = 18, 54 / 9 = 6.
Explanation:
In the above-given question,
given that,
the multiplication and division of 9.
3 x 9 = 27, 45 / 9 = 5.
9 x 9 = 81, 9 / 9 = 1.
8 x 9 = 72, 90 / 9 = 10.
9 x 4 = 36, 72 / 9 = 8.
2 x 9 = 18, 54 / 9 = 6.
Eureka Math Grade 3 Module 3 Lesson 12 Exit Ticket Answer Key
Question 1.
Each has a value of 9. Complete the equations to find the total value of the tower of blocks.
___5__ × 9 = (5 + __1__ ) × 9
= (5 × ___9__ ) + ( __1___ × __9___ )
= 45 + __9___
= __54___
6 x 9 = 54.
Explanation:
In the above-given question,
given that,
5 x 9 = ( 5 + 1 ) x 9.
( 5 x 9 ) + ( 1 x 9 )
45 + 9.
54.
Question 2.
Hector solves 9 × 8 by subtracting 1 eight from 10 eights. Draw a model, and explain Hector’s strategy.
9 x 8 = 72.
Explanation:
In the above-given question,
given that,
9 x 8 = 72.
10 eights – 1 eight.
80 – 8.
72.
Question 1.
Find the value of each row. Then, add the rows to find the total.
a. Each has a value of 6.
9 × 6 = ___54_____
9 × 6 = (5 + 4) × 6
= (5 × 6) + (4 × 6)
= 30 + __24___
= __54___
6 x 9 = 54.
Explanation:
In the above-given question,
given that,
6 x 9 = ( 5 + 4 ) x 6.
( 5 x 6 ) + ( 4 x 6 )
30 + 24.
54.
b. Each has a value of 7.
9 × 7 = ___63_____
9 × 7 = (5 + __4___) × 7
= (5 × 7) + (__4___ × 7)
= 35 + __28___
= ___63__
9 x 7 = 63.
Explanation:
In the above-given question,
given that,
9 x 7 = ( 5 + 4 ) x 7.
( 5 x 7 ) + ( 4 x 9 )
35 + 28.
63.
c. Each has a value of 8.
9 × 8 = ____72____
9 × 8 = (5 + __4___) × 8
= (5 × 8) + (__4___ × ___8__)
= 40 + ___32__
= __72___
9 x 8 = 72.
Explanation:
In the above-given question,
given that,
9 x 8 = ( 5 + 4 ) x 8.
( 5 x 8 ) + ( 4 x 8 )
40 + 32.
72.
d. Each has a value of 9.
9 × 9 = ____81____
9 × 9 = (5 + _____) × 9
= (5 × 9) + (__4___ × __9___)
= 45 + ___36__
= __81___
9 x 9 = 81.
Explanation:
In the above-given question,
given that,
9 x 9 = ( 5 + 4 ) x 9.
( 5 x 9 ) + ( 4 x 9 )
45 + 36.
81.
Question 2.
Match
50 – 5 = 45, 9 x 5 = 45.
60 – 6 = 54, 9 x 6 = 54.
70 – 7 = 63, 9 x 7 = 63.
80 – 8 = 72, 9 x 8 = 72.
90 – 9 = 81, 9 x 9 = 81.
40 – 4 = 36, 9 x 4 = 36.
Explanation:
In the above-given question,
given that,
the multiplication and division of 9 is given.
50 – 5 = 45, 9 x 5 = 45.
60 – 6 = 54, 9 x 6 = 54.
70 – 7 = 63, 9 x 7 = 63.
80 – 8 = 72, 9 x 8 = 72.
90 – 9 = 81, 9 x 9 = 81.
40 – 4 = 36, 9 x 4 = 36.
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# Attractive force between Venus and Sun
1. Oct 29, 2012
I just need someone to double check my answer and confirm they get the same answer as I do. Thanks.
Question: Find the attractive force between Vneus and the sun. The distance between their centers is 1.08x10^11.
Equation: F = Gm1m2/d^2
G= 6.67x10^-11 N.m^2/kg^2
m1 (mass of sun)= (1.98x10^30kg)
m2 (mass of venus) = (4.83x10^24kg)
d= (1.08x10^11)
= (6.67x10^-11 N.m^2/kg^2)(1.98x10^30kg)(4.83x10^24kg)/(1.08x10^11)^2
=5.4687 x 10^22
Please let me know if this is the correct answer asap. Thanks!
2. Oct 29, 2012
### Staff: Mentor
3. Oct 29, 2012
### PeterO
You have just omitted the next digits rather than round off to the 5 figures you have chosen to specify.
And as per the previous post - you need some units.
4. Oct 29, 2012
Thanks folks, appreciate it!
5. Oct 29, 2012
### Staff: Mentor
WolframAlpha gets a slightly different result: 5.54*10^22 N
It uses a mass of venus which is larger by 1% and a mass of sun which is larger by .5%.
6. Oct 29, 2012
### PeterO
So you should have been saying 5.47 (correctly rounded) not 5.46 (merely truncated)
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# Angle Vector Problem
• goaliejoe35
In summary: I think i understand now!In summary, the sum of the four vectors in unit-vector notation is (-3.18m)i + (4.73m)j, with a magnitude of 5.70 m and an angle of -56.09 degrees (or 123.91 degrees when adjusted for quadrant). The confusion may have been due to the calculator giving an angle in the opposite direction, which can be resolved by adding 180 degrees to the answer.
#### goaliejoe35
What is the sum of the following four vectors in (a) unit-vector notation, and as (b) a magnitude and (c) an angle? Positive angles are counterclockwise from the positive direction of the x axis; negative angles are clockwise.
http://edugen.wiley.com/edugen/courses/crs1650/art/qb/qu/c03/eq03_84.gif
a) (-3.18m)i + (4.73m)j
b) 5.70 m
c) -56.09 degrees
Part C was marked wrong and I don't understand why... i did tan inverse of (4.73/-3.18) to get my answer. Could someone tell me what I did wrong?
Hi goaliejoe35,
If you draw a set of x- and y- axes on a paper and draw the vector that you gave in part a, what quadrant is it in? What quadrant is the angle the you put as the answer to part c in?
the vector from part a is in quadrant 2 right? and the angle i gave is in quadrant 4?
(-3.18m)i + (4.73m)j
and then think about the vector in the opposite direction:
(3.18m)i + (-4.73m)j
Use inverse tangent on both of them:
on yours: arctan( 4.73 / -3.18 ) = arctan( -1.48742)
on the other: arctan ( -4.73 / 3.18 ) = arctan(-1.48742)
and that's the problem. Most calculators can't tell the difference between the two cases, so the answer it gives is either the true answer that you're looking for, or it's in the opposite direction from what you want (180 degrees away).
So in your case, you would need to recognize that your vector is in the second quadrant, but the calculator is giving you an angle in the fourth quadrant, so you need to add 180 degrees to the calculator answer to get the real answer.
Ok awesome! Thanks for all your help!
## 1. What is an angle vector problem?
An angle vector problem is a mathematical problem that involves finding the magnitude and direction of a vector in terms of its components. It is commonly used in physics and engineering to solve problems involving forces, velocities, and displacements.
## 2. How do you solve an angle vector problem?
To solve an angle vector problem, you can use trigonometric functions such as sine, cosine, and tangent to find the magnitude and direction of the vector. You can also use vector addition and subtraction to solve more complex problems.
## 3. What are the components of a vector?
A vector has two main components: magnitude and direction. The magnitude is the length of the vector, and the direction is the angle at which the vector is pointing. These components can be represented by numbers or variables.
## 4. Can angle vector problems be solved graphically?
Yes, angle vector problems can be solved graphically using vector diagrams. By drawing the vectors and their components on a coordinate system, you can determine the magnitude and direction of the resulting vector visually.
## 5. What are some real-world applications of angle vector problems?
Angle vector problems are used in many fields, including physics, engineering, and navigation. For example, they can be used to calculate the velocity and acceleration of an object, determine the forces acting on a structure, or plot the course of a plane or ship.
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Using functions to separate a compact set from a closed set in a completely regular space
Suppose $X$ is completely regular, $K\subset X$ compact and $C\subset X$ closed such that $K\cap C$ empty.
• Prove that there exists $f_x\in C(X,[0,1])$ such that $f_x(x)=1$ and $f_x=1$ on $C$ for $x\in K$ (this follows from my point of view from the $T_{3,5}$ property)
• Construct with help of the $f_x$ a cover of $K$ (can we use the inverse of $f_x$ vor every $x$ in $1$?)
• Use compactness to find a $g\in C(X,[0,\infty))$ such that $g=0$ on $C$ and $g>\frac{1}{2}$ on $K$ (can we use Urysohn?)
• Modify $g$ to get an $f\in C(X,[0,1])$ such that $f=1$ on $K$ and $f=0$ on $C$
I think it is not so difficult but I don't see it. Can someone help me? Thanks
-
In the first bullet point you want $f_x=0$ on $C$. – Brian M. Scott Jan 30 '13 at 21:13
Yes, you're right, the first bullet point follows from $T_{3.5}$ applied to the point $x \notin C$ and the closed set $C$. There is a typo: you want $f_x(x) = 1$ and $f_x = 0$ on $C$ (not $f_x = 1$ on $C$).
The sets $U_x = \{y \in X : f_x(y) \gt 1/2\}$, $x \in K$ form an open cover of $K$.
Take a finite subcover $U_{x_1},\dots,U_{x_n}$ and set $g = f_{x_1} + \dots + f_{x_n}$. Then $g \gt 1/2$ on $K$ and $g = 0$ on $C$.
Now set $f(x) = 2\min\{g(x),1/2\}$.
-
The first remark is true: the existence of the $f_x$ follows from applying complete regularity to $x$ and $C$, which can be done as all $x \in K$ are not in $C$. (nitpick: we also use the axiom of choice, to simultaneously choose such $f_x$, but topology without choice gets hairy pretty fast.)
Use $U_x = (f_x)^{-1}[ [0, \frac{1}{2} ) ]$ as open set that contains $x$.
There are finitely many $U_{x_1},...,U_{x_n}$ that also cover $K$.
Use $g' = \min_{i=1 \ldots n} f_{x_i}$, which is continuous as a minimum of finitely many continuous functions (and the minimum is well defined as we map into $[0,1]$).
All $f_x$ are constantly 1 on $C$, and hence so is $g'$ and if $x \in K$, $x$ is in some $U_{x_i}$, so $g'(x) \le f_{x_i}(x) < \frac{1}{2}$, so $g = 1 - g'$ is as required for the third point.
Now try the last modification yourself...
-
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# How To Find Arc Length
## Online Tutoring Is The Easiest, Most Cost-Effective Way For Students To Get The Help They Need Whenever They Need It.
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We know Arc is a segment of a circle. In other words we can say Arc is a curve. We Know Arc length is the distance along the curved line. We can say length of an arc is the part of the circumference. We can call the angle which is made by the end points of the arc with centre as central angle.
We know the formula to find the arc length that is
Arc length = (Ø/ 360o) x 2πr
Here, r is the radius and
Ø is the central angle.
Example 1: Find the arc length if the radius of the arc is 8 cm and its central angle is 30o.
Solution: We can write
r = radius of the arc = 8 cm
ø = central angle = 30o
The formula to find arc length = (Ø/ 360o) x 2πr
We can plug the r and Ø values in the above formula. Then we have
arc length = (30o/360o) x 2 (22/7) x 8
= (1/ 12) x 16 x (22/7)
= 4.19 cm
Example 2: Find the arc length of a circle with diameter 12 cm and the central angle 180o.
Solution: from the problem, we can write
r = radius of the arc = (1/2 ) diameter = (1/2) x 12 cm = 6 cm
Ø = central angle = 180o
The formula to find arc length = (Ø/ 360o) x 2πr
We can plug the r and Ø values in the above formula. Then we have
arc length = (180o/360o) x 2 (22/7) x 6= 18.85 cm
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# Math Snap
## Solve $x^{\wedge} 2-2^{\star} x+y^{\wedge} 2-4 y+5=0$
#### STEP 1
Assumptions
1. The given equation is $x^2 - 2x + y^2 - 4y + 5 = 0$.
2. We need to solve this equation to find the values of $x$ and $y$.
#### STEP 2
First, we will complete the square for the $x$-terms in the equation $x^2 - 2x$.
#### STEP 3
To complete the square for $x^2 - 2x$, we add and subtract $\left(\frac{-2}{2}\right)^2 = 1$.
$x^2 - 2x = (x - 1)^2 - 1$
#### STEP 4
Next, we will complete the square for the $y$-terms in the equation $y^2 - 4y$.
#### STEP 5
To complete the square for $y^2 - 4y$, we add and subtract $\left(\frac{-4}{2}\right)^2 = 4$.
$y^2 - 4y = (y - 2)^2 - 4$
#### STEP 6
Substitute the completed squares back into the original equation.
$(x - 1)^2 - 1 + (y - 2)^2 - 4 + 5 = 0$
#### STEP 7
Simplify the equation.
$(x - 1)^2 - 1 + (y - 2)^2 - 4 + 5 = 0$
$(x - 1)^2 + (y - 2)^2 - 1 - 4 + 5 = 0$
$(x - 1)^2 + (y - 2)^2 = 0$
#### STEP 8
Since the sum of two squares equals zero, each square must individually be zero.
$(x - 1)^2 = 0$
$(y - 2)^2 = 0$
#### STEP 9
Solve for $x$ and $y$.
$x - 1 = 0$
$y - 2 = 0$
##### SOLUTION
Thus, we find:
$x = 1$
$y = 2$
The solution to the equation $x^2 - 2x + y^2 - 4y + 5 = 0$ is $x = 1$ and $y = 2$.
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|
Associated Topics || Dr. Math Home || Search Dr. Math
### Interesting Number Sequence Pattern
```Date: 11/01/2004 at 11:28:14
From: Kamsin
Subject: Number series
The sequence of digits 1,2,3,4,0,9,6,9,4,8,7,... is constructed in the
following way: every digit starting from the fifth is the last digit
of the sum of the previous four digits.
a) Do the digits 2,0,0,4 appear in the sequence in that order?
b) Do the initial digits 1,2,3,4 appear again in the sequence in that
order?
I haven't come across such a series before. I really don't know how to
continue.
```
```
Date: 11/01/2004 at 15:20:29
From: Doctor Vogler
Subject: Re: Number series
Hi Kamsin,
Thanks for writing to Dr. Math. Do you know what modular arithmetic
is? You can refer to
Mod, Modulus, Modular Arithmetic
http://mathforum.org/library/drmath/view/62930.html
Thinking of this in terms of modular arithmetic will probably be
helpful. Then you have: Each term is the sum of the previous four,
mod 10. By the Chinese Remainder Theorem, we can break this into two
problems: Answer each question mod 2 and mod 5. That will make
things easier.
In case none of that made sense, I will try to avoid talk of modular
arithmetic. But if you wonder how I came up with these ideas, then
you should start by learning modular arithmetic.
Now arithmetic mod 2 is simply a matter of odds or evens. So look at
the pattern of odds and evens in your sequence:
odd, even, odd, even, even, odd, even, odd, even, even, ...
You should notice a repeat after every five terms. Can you prove that
this pattern will continue indefinitely? Now answer part (a).
Part (b) is a little harder, but it only requires you to establish
that this pattern has to repeat. So I'll give you the general ideas
and hope that you can fill in the details and understand what is going
on. First of all, you should notice that if you know any four
consecutive numbers in the sequence, then you can start listing off
all of the numbers that follow it, and you can start listing off all
of the numbers that preceded it.
Let's practice. Suppose that
5, 8, 1, 4
appears somewhere in the sequence (which it may or may not). If it
does, then what do the next three or four numbers have to be? And
what does the number before the 5 have to be? What about the two
numbers before that?
Okay, so if the initial sequence
1, 2, 3, 4
ever appears later in the sequence (after the first time), then that
means that the sequence repeats, right? Well, what if it doesn't
repeat? What can happen?
Well, there are only 10^4 possible sequences of four numbers. So
let's think about the first 10^4 + 4 numbers, and every string of four
of them makes a total of 10^4 + 1 different strings of four numbers.
Do you know the pigeonhole principal? It tells us that some string of
four numbers has to occur twice (at least).
But that means that it repeats! Let's suppose this string (and we
don't know what numbers it has) starts the first time at position n
and the second time at position n+m. Then the m numbers (including
the four from the first string) from the beginning of the first
occurrance to the beginning of the second occurrance MUST be the next
m numbers as well! Because the next m numbers are determined by the
next four. So that means it has to repeat, every m numbers.
But now let's go backwards. The m numbers before the first occurrance
must be the same as the m numbers after it, so that same pattern
repeats. That means that every string of four numbers that appears in
the sequence at all must repeat every m numbers!
Is that right? Why can't we start going along and then get stuck
somewhere? For example, why can't we have lots of random numbers
coming along until we suddenly find four zeros in a row? If that
ever happens, then it will just keep giving us zeros. So why can't
that happen? (Hint: What was the last nonzero digit?)
Does all of this discussion make sense to you? So what is the answer
to the second question?
back and show me what you have been able to do, and I will try to
offer further suggestions.
- Doctor Vogler, The Math Forum
http://mathforum.org/dr.math/
```
```
Date: 11/02/2004 at 12:43:15
From: Kamsin
Subject: Thank you (Number series)
Dear Dr. Vogler,
Thank you very much for the explanation to this problem which I've
been trying to solve for weeks. This is a great site for a maths
enthusiast!
```
Associated Topics:
High School Sequences, Series
Search the Dr. Math Library:
Find items containing (put spaces between keywords): Click only once for faster results: [ Choose "whole words" when searching for a word like age.] all keywords, in any order at least one, that exact phrase parts of words whole words
Submit your own question to Dr. Math
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## How to Solve Systems of Algebraic Equations Containing Two Variables
Last Updated: July 30, 2023 Fact Checked
This article was reviewed by Grace Imson, MA . Grace Imson is a math teacher with over 40 years of teaching experience. Grace is currently a math instructor at the City College of San Francisco and was previously in the Math Department at Saint Louis University. She has taught math at the elementary, middle, high school, and college levels. She has an MA in Education, specializing in Administration and Supervision from Saint Louis University. There are 8 references cited in this article, which can be found at the bottom of the page. This article has been fact-checked, ensuring the accuracy of any cited facts and confirming the authority of its sources. This article has been viewed 1,027,843 times.
In a "system of equations," you are asked to solve two or more equations at the same time. When these have two different variables in them, such as x and y, or a and b, it can be tricky at first glance to see how to solve them. [1] X Research source Fortunately, once you know what to do, all you need is basic algebra skills (and sometimes some knowledge of fractions) to solve the problem. If you are a visual learner or if your teacher requires it, learn how to graph the equations as well. Graphing can be useful to "see what's going on" or to check your work, but it can be slower than the other methods, and doesn't work well for all systems of equations.
## Using the Substitution Method
• This method often uses fractions later on. You can try the elimination method below instead if you don't like fractions.
• 4x = 8 - 2y
• (4x)/4 = (8/4) - (2y/4)
• You know that x = 2 - ½y .
• Your second equation, that you haven't yet altered, is 5x + 3y = 9 .
• In the second equation, replace x with "2 - ½y": 5(2 - ½y) + 3y = 9 .
• 5(2 - ½y) + 3y = 9
• 10 – (5/2)y + 3y = 9
• 10 – (5/2)y + (6/2)y = 9 (If you don't understand this step, learn how to add fractions . This is often, but not always, necessary for this method.)
• 10 + ½y = 9
• You know that y = -2
• One of the original equations is 4x + 2y = 8 . (You can use either equation for this step.)
• Plug in -2 instead of y: 4x + 2(-2) = 8 .
• If you end up with an equation that has no variables and isn't true (for instance, 3 = 5), the problem has no solution . (If you graphed both of the equations, you'd see they were parallel and never intersect.)
• If you end up with an equation without variables that is true (such as 3 = 3), the problem has infinite solutions . The two equations are exactly equal to each other. (If you graphed the two equations, you'd see they were the same line.)
## Using the Elimination Method
• You have the system of equations 3x - y = 3 and -x + 2y = 4 .
• Let's change the first equation so that the y variable will cancel out. (You can choose x instead, and you'll get the same answer in the end.)
• The - y on the first equation needs to cancel with the + 2y in the second equation. We can make this happen by multiplying - y by 2.
• Multiply both sides of the first equation by 2, like this: 2(3x - y)=2(3) , so 6x - 2y = 6 . Now the - 2y will cancel out with the +2y in the second equation.
• Your equations are 6x - 2y = 6 and -x + 2y = 4 .
• Combine the left sides: 6x - 2y - x + 2y = ?
• Combine the right sides: 6x - 2y - x + 2y = 6 + 4 .
• You have 6x - 2y - x + 2y = 6 + 4 .
• Group the x and y variables together: 6x - x - 2y + 2y = 6 + 4 .
• Simplify: 5x = 10
• Solve for x: (5x)/5 = 10/5 , so x = 2 .
• You know that x = 2 , and one of your original equations is 3x - y = 3 .
• Plug in 2 instead of x: 3(2) - y = 3 .
• Solve for y in the equation: 6 - y = 3
• 6 - y + y = 3 + y , so 6 = 3 + y
• If your combined equation has no variables and is not true (like 2 = 7), there is no solution that will work on both equations. (If you graph both equations, you'll see they're parallel and never cross.)
• If your combined equation has no variables and is true (like 0 = 0), there are infinite solutions . The two equations are actually identical. (If you graph them, you'll see that they're the same line.)
## Graphing the Equations
• The basic idea is to graph both equations, and find the point where they intersect. The x and y values at this point will give us the value of x and the value of y in the system of equations.
• Your first equation is 2x + y = 5 . Change this to y = -2x + 5 .
• Your second equation is -3x + 6y = 0 . Change this to 6y = 3x + 0 , then simplify to y = ½x + 0 .
• If both equations are identical , the entire line will be an "intersection". Write infinite solutions .
• If you don't have graph paper, use a ruler to make sure the numbers are spaced precisely apart.
• If you are using large numbers or decimals, you may need to scale your graph differently. (For example, 10, 20, 30 or 0.1, 0.2, 0.3 instead of 1, 2, 3).
• In our examples from earlier, one line ( y = -2x + 5 ) intercepts the y-axis at 5 . The other ( y = ½x + 0 ) intercepts at 0 . (These are points (0,5) and (0,0) on the graph.)
• Use different colored pens or pencils if possible for the two lines.
• In our example, the line y = -2x + 5 has a slope of -2 . At x = 1, the line moves down 2 from the point at x = 0. Draw the line segment between (0,5) and (1,3).
• The line y = ½x + 0 has a slope of ½ . At x = 1, the line moves up ½ from the point at x=0. Draw the line segment between (0,0) and (1,½).
• If the lines have the same slope , the lines will never intersect, so there is no answer to the system of equations. Write no solution .
• If the lines are moving toward each other, keep plotting points in that direction.
• If the lines are moving away from each other, move back and plot points in the other direction, starting at x = -1.
• If the lines are nowhere near each other, try jumping ahead and plotting more distant points, such as at x = 10.
## Video . By using this service, some information may be shared with YouTube.
• You can check your work by plugging the answers back into the original equations. If the equations end up true (for instance, 3 = 3), your answer is correct. Thanks Helpful 3 Not Helpful 1
• In the elimination method, you will sometimes have to multiply one equation by a negative number in order to get a variable to cancel out. Thanks Helpful 1 Not Helpful 1
• These methods cannot be used if there is a variable raised to an exponent, such as x 2 . For more information on equations of this type, look up a guide to factoring quadratics with two variables. [11] X Research source Thanks Helpful 0 Not Helpful 0
## You Might Also Like
• ↑ https://www.mathsisfun.com/definitions/system-of-equations.html
• ↑ https://calcworkshop.com/systems-equations/substitution-method/
• ↑ https://www.cuemath.com/algebra/substitution-method/
• ↑ https://tutorial.math.lamar.edu/Classes/Alg/SystemsTwoVrble.aspx
• ↑ http://www.purplemath.com/modules/systlin2.htm
• ↑ http://www.virtualnerd.com/algebra-2/linear-systems/graphing/solve-by-graphing/equations-solution-by-graphing
To solve systems of algebraic equations containing two variables, start by moving the variables to different sides of the equation. Then, divide both sides of the equation by one of the variables to solve for that variable. Next, take that number and plug it into the formula to solve for the other variable. Finally, take your answer and plug it into the original equation to solve for the other variable. To learn how to solve systems of algebraic equations using the elimination method, scroll down! Did this summary help you? Yes No
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## Course: Algebra 1 > Unit 4
• Two-variable linear equations intro
## Solutions to 2-variable equations
• Worked example: solutions to 2-variable equations
• Completing solutions to 2-variable equations
• Complete solutions to 2-variable equations
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## How Do You Solve Two Equations with Two Variables?
Trying to solve two equations each with the same two unknown variables? Take one of the equations and solve it for one of the variables. Then plug that into the other equation and solve for the variable. Plug that value into either equation to get the value for the other variable. This tutorial will take you through this process of substitution step-by-step!
• substitution
• solve by substitution
• apply distributive property
• linear equation
• 2 equations 2 unknowns
## Background Tutorials
Evaluating expressions.
## What is a Variable?
You can't do algebra without working with variables, but variables can be confusing. If you've ever wondered what variables are, then this tutorial is for you!
## How Do You Use the Distributive Property to Simplify an Expression?
In this tutorial you'll see how to apply the distributive property. Remember that this is important when you are trying to simplify an expression and get rid of parentheses!
## How Do You Simplify an Expression?
Simplifying an algebraic expression is a fundamental part of solving math problems. Get some practice putting an expression in simplest form by following along with this tutorial.
## How Do You Solve a Two-Step Equation?
Solving an equation for a variable? Perform the order of operations in reverse! Check it out in this tutorial.
## How Do You Check Your Answer When You Have Two Equations?
Imagine you have two equations with two variables that you're trying to solve for, and someone hands you the answer. How do you know that the answer is right? After watching this tutorial you'll see exactly what it takes to check that the answer you have is correct for BOTH equations!
## Further Exploration
Solving systems using substitution.
## How Do You Solve a Word Problem Using Two Equations?
Sometimes word problems describe a system of equations, two equations each with two unknowns. Solving word problems like this one aren't so bad if you know what to do. Check it out with this tutorial!
## What Are the Ways You Can Solve a System of Linear Equations?
Knowing the definition of a system of equations is great, but you should also know how to solve them! This tutorial introduces you to the graphing method, substitution method, and elimination method for solving a system of equations. Take a look and learn them all!
## Solving Equations With Two Variables
Related Pages Solving Equations More Lessons for GRE Math More Algebra Lessons
This is part of a series of lessons for the quantitative reasoning section of the GRE revised General Test. In these lessons, we will learn:
• Linear Equations in Two Variables
• Solving Simultaneous Equations or Systems of Equations
• Using the Substitution Method
• Using the Elimination Method
## Linear Equation In Two Variables
A linear equation in two variables, x and y, can be written in the form ax + by = c where x and y are real numbers and a and b are not both zero.
For example, 3x + 2y = 8 is a linear equation in two variables.
A solution of such an equation is an ordered pair of numbers (x, y) that makes the equation true when the values of x and y are substituted into the equation.
For example, both (2, 1) and (0, 4) are solutions of the equation but (2, 0) is not a solution. A linear equation in two variables has infinitely many solutions.
The following video shows how to complete ordered pairs to make a solution to linear equations.
## Systems of Equations or Simultaneous Equations
The following diagram shows examples of how to solve systems of equations using substitution or elimination.. Scroll down the page for more examples and solutions on how to solve systems of equations or simultaneous equations..
If another linear equation in the same variables is given, it is usually possible to find a unique solution of both equations. Two equations with the same variables are called a system of equations , and the equations in the system are called simultaneous equations . To solve a system of two equations means to find an ordered pair of numbers that satisfies both equations in the system.
There are two basic methods for solving systems of linear equations, by substitution or by elimination.
## Substitution Method
In the substitution method, one equation is manipulated to express one variable in terms of the other. Then the expression is substituted in the other equation.
For example, to solve the system of equations 3x + 2y = 2 y + 8 = 3x
Isolate the variable y in the equation y + 8 = 3x to get y = 3x – 8.
Then, substitute 3x – 8 for y into the equation 3x + 2y = 2. 3x + 2 (3x – 8) = 2 3x + 6x – 16 = 2 9x – 16 = 2 9x = 18
Substitute x = 2 into y = 3x – 8.to get the value for y y = 3 (2) – 8 y = 6 – 8 = – 2
Answer: x = 2 and y = –2
How to solve simultaneous equations using substitution?
## Elimination Method
In the elimination method, the object is to make the coefficients of one variable the same in both equations so that one variable can be eliminated either by adding the equations together or by subtracting one from the other.
Consider the following example: 2x + 3y = –2 4x – 3y = 14
In this example the coefficients of y are already opposites (+3 and –3). Just add the two equations to eliminate y.
To get the value of y, we need to substitute x = 2 into the equation 2x + 3y = –2 2(2) + 3y = –2 4 + 3y = –2 3y = –6 y = –2
How to solve simultaneous equations using the substitution method and elimination (or combination) method
Example of the GRE Quantitative Comparison question that involves simultaneous equations
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• Study Guides
• Number Problems with Two Variables
• Preliminaries
• Quiz: Preliminaries
• Properties of Basic Mathematical Operations
• Quiz: Properties of Basic Mathematical Operations
• Multiplying and Dividing Using Zero
• Quiz: Multiplying and Dividing Using Zero
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• Quiz: Powers and Exponents
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• Quiz: Square Roots and Cube Roots
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• Quiz: Grouping Symbols
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• Quiz: Divisibility Rules
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• Quiz: Signed Numbers (Positive Numbers and Negative Numbers)
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• Quiz: Variables and Algebraic Expressions
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• Quiz: Solving Systems of Equations (Simultaneous Equations)
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• Quiz: Word Problems
Here are some examples for solving number problems with two variables.
The sum of two numbers is 15. The difference of the same two numbers is 7. What are the two numbers?
First, circle what you're looking for— the two numbers. Let x stand for the larger number and y stand for the second number. Now, set up two equations.
The sum of the two numbers is 15.
The difference is 7.
Now, solve by adding the two equations.
Now, plugging into the first equation gives
The numbers are 11 and 4.
The sum of twice one number and three times another number is 23 and their product is 20. Find the numbers.
First, circle what you must find— the numbers . Let x stand for the number that is being multiplied by 2 and y stand for the number being multiplied by 3.
Now set up two equations.
The sum of twice a number and three times another number is 23.
2 x + 3 y = 23
Their product is 20.
x ( y ) = 20
Rearranging the first equation gives
3 y = 23 – 2 x
Dividing each side of the equation by 3 gives
Now, substituting the first equation into the second gives
Multiplying each side of the equation by 3 gives
23 x – 2 x 2 = 60
Rewriting this equation in standard quadratic form gives
2 x 2 – 23 x + 60 = 0
Solving this quadratic equation using factoring gives
(2 x – 15)( x – 4) = 0
Setting each factor equal to 0 and solving gives
With each x value we can find its corresponding y value.
Therefore, this problem has two sets of solutions.
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## Module 13: Systems of Equations and Inequalities
Systems of linear equations: two variables, learning outcomes.
• Solve systems of equations by graphing, substitution, and addition.
• Identify inconsistent systems of equations containing two variables.
• Express the solution of a system of dependent equations containing two variables using standard notations.
A skateboard manufacturer introduces a new line of boards. The manufacturer tracks its costs, which is the amount it spends to produce the boards, and its revenue, which is the amount it earns through sales of its boards. How can the company determine if it is making a profit with its new line? How many skateboards must be produced and sold before a profit is possible? In this section we will consider linear equations with two variables to answer these and similar questions.
(credit: Thomas Sørenes)
## Introduction to Solutions of Systems
In order to investigate situations such as that of the skateboard manufacturer, we need to recognize that we are dealing with more than one variable and likely more than one equation. A system of linear equations consists of two or more linear equations made up of two or more variables such that all equations in the system are considered simultaneously. To find the unique solution to a system of linear equations, we must find a numerical value for each variable in the system that will satisfy all equations in the system at the same time. Some linear systems may not have a solution and others may have an infinite number of solutions. In order for a linear system to have a unique solution, there must be at least as many equations as there are variables. Even so, this does not guarantee a unique solution.
In this section, we will look at systems of linear equations in two variables, which consist of two equations that contain two different variables. For example, consider the following system of linear equations in two variables.
\begin{align}2x+y&=15\\[1mm] 3x-y&=5\end{align}
The solution to a system of linear equations in two variables is any ordered pair that satisfies each equation independently. In this example, the ordered pair $(4,7)$ is the solution to the system of linear equations. We can verify the solution by substituting the values into each equation to see if the ordered pair satisfies both equations. Shortly we will investigate methods of finding such a solution if it exists.
\begin{align}2\left(4\right)+\left(7\right)&=15 &&\text{True} \\[1mm] 3\left(4\right)-\left(7\right)&=5 &&\text{True} \end{align}
In addition to considering the number of equations and variables, we can categorize systems of linear equations by the number of solutions. A consistent system of equations has at least one solution. A consistent system is considered to be an independent system if it has a single solution, such as the example we just explored. The two lines have different slopes and intersect at one point in the plane. A consistent system is considered to be a dependent system if the equations have the same slope and the same y -intercepts. In other words, the lines coincide so the equations represent the same line. Every point on the line represents a coordinate pair that satisfies the system. Thus, there are an infinite number of solutions.
Another type of system of linear equations is an inconsistent system , which is one in which the equations represent two parallel lines. The lines have the same slope and different y- intercepts. There are no points common to both lines; hence, there is no solution to the system.
## A General Note: Types of Linear Systems
There are three types of systems of linear equations in two variables, and three types of solutions.
• An independent system has exactly one solution pair $\left(x,y\right)$. The point where the two lines intersect is the only solution.
• An inconsistent system has no solution. Notice that the two lines are parallel and will never intersect.
• A dependent system has infinitely many solutions. The lines are coincident. They are the same line, so every coordinate pair on the line is a solution to both equations.
Below is a comparison of graphical representations of each type of system.
## How To: Given a system of linear equations and an ordered pair, determine whether the ordered pair is a solution.
• Substitute the ordered pair into each equation in the system.
• Determine whether true statements result from the substitution in both equations; if so, the ordered pair is a solution.
## Example: Determining Whether an Ordered Pair Is a Solution to a System of Equations
Determine whether the ordered pair $\left(5,1\right)$ is a solution to the given system of equations.
\begin{align}x+3y&=8\\ 2x-9&=y \end{align}
Substitute the ordered pair $\left(5,1\right)$ into both equations.
\begin{align}\left(5\right)+3\left(1\right)&=8 \\[1mm] 8&=8 &&\text{True} \\[3mm] 2\left(5\right)-9&=\left(1\right) \\[1mm] 1&=1 &&\text{True} \end{align}
The ordered pair $\left(5,1\right)$ satisfies both equations, so it is the solution to the system.
## Analysis of the Solution
We can see the solution clearly by plotting the graph of each equation. Since the solution is an ordered pair that satisfies both equations, it is a point on both of the lines and thus the point of intersection of the two lines.
Determine whether the ordered pair $\left(8,5\right)$ is a solution to the following system.
\begin{align}5x-4y&=20\\ 2x+1&=3y\end{align}
Not a solution.
## Solving Systems of Equations by Graphing
There are multiple methods of solving systems of linear equations. For a system of linear equations in two variables, we can determine both the type of system and the solution by graphing the system of equations on the same set of axes.
## Example: Solving a System of Equations in Two Variables by Graphing
Solve the following system of equations by graphing. Identify the type of system.
\begin{align}2x+y&=-8\\ x-y&=-1\end{align}
Solve the first equation for $y$.
\begin{align}2x+y&=-8\\ y&=-2x-8\end{align}
Solve the second equation for $y$.
\begin{align}x-y&=-1\\ y&=x+1\end{align}
Graph both equations on the same set of axes:
The lines appear to intersect at the point $\left(-3,-2\right)$. We can check to make sure that this is the solution to the system by substituting the ordered pair into both equations.
\begin{align}2\left(-3\right)+\left(-2\right)&=-8 \\[1mm] -8=-8 &&\text{True} \\[3mm] \left(-3\right)-\left(-2\right)&=-1 \\[1mm] -1&=-1 &&\text{True} \end{align}
The solution to the system is the ordered pair $\left(-3,-2\right)$, so the system is independent.
Solve the following system of equations by graphing.
$\begin{gathered}2x - 5y=-25 \\ -4x+5y=35 \end{gathered}$
The solution to the system is the ordered pair $\left(-5,3\right)$.
## Can graphing be used if the system is inconsistent or dependent?
Yes, in both cases we can still graph the system to determine the type of system and solution. If the two lines are parallel, the system has no solution and is inconsistent. If the two lines are identical, the system has infinite solutions and is a dependent system.
Plot the three different systems with an online graphing tool. Categorize each solution as either consistent or inconsistent. If the system is consistent determine whether it is dependent or independent. You may find it easier to plot each system individually, then clear out your entries before you plot the next. 1) $5x-3y = -19$ $x=2y-1$
2) $4x+y=11$ $-2y=-25+8x$
3) $y = -3x+6$ $-\frac{1}{3}y+2=x$
• One solution – consistent, independent
• No solutions, inconsistent, neither dependent nor independent
• Many solutions – consistent, dependent
## Solving Systems of Equations by Substitution
Solving a linear system in two variables by graphing works well when the solution consists of integer values, but if our solution contains decimals or fractions, it is not the most precise method. We will consider two more methods of solving a system of linear equations that are more precise than graphing. One such method is solving a system of equations by the substitution method , in which we solve one of the equations for one variable and then substitute the result into the second equation to solve for the second variable. Recall that we can solve for only one variable at a time, which is the reason the substitution method is both valuable and practical.
## How To: Given a system of two equations in two variables, solve using the substitution method.
• Solve one of the two equations for one of the variables in terms of the other.
• Substitute the expression for this variable into the second equation, then solve for the remaining variable.
• Substitute that solution into either of the original equations to find the value of the first variable. If possible, write the solution as an ordered pair.
• Check the solution in both equations.
## Example: Solving a System of Equations in Two Variables by Substitution
Solve the following system of equations by substitution.
\begin{align}-x+y&=-5 \\ 2x-5y&=1 \end{align}
First, we will solve the first equation for $y$.
\begin{align}-x+y&=-5 \\ y&=x - 5 \end{align}
Now we can substitute the expression $x - 5$ for $y$ in the second equation.
\begin{align}2x - 5y&=1 \\ 2x - 5\left(x - 5\right)&=1 \\ 2x - 5x+25&=1 \\ -3x&=-24 \\ x&=8 \end{align}
Now, we substitute $x=8$ into the first equation and solve for $y$.
\begin{align}-\left(8\right)+y&=-5 \\ y&=3 \end{align}
Our solution is $\left(8,3\right)$.
Check the solution by substituting $\left(8,3\right)$ into both equations.
\begin{align}-x+y&=-5 \\ -\left(8\right)+\left(3\right)&=-5 && \text{True} \\[3mm] 2x - 5y&=1 \\ 2\left(8\right)-5\left(3\right)&=1 && \text{True} \end{align}
\begin{align}x&=y+3 \\ 4&=3x - 2y \end{align}
$\left(-2,-5\right)$
## Can the substitution method be used to solve any linear system in two variables?
Yes, but the method works best if one of the equations contains a coefficient of 1 or –1 so that we do not have to deal with fractions.
The following video is ~10 minutes long and provides a mini-lesson on using the substitution method to solve a system of linear equations. We present three different examples, and also use a graphing tool to help summarize the solution for each example.
## Solving Systems of Equations in Two Variables by the Addition Method
A third method of solving systems of linear equations is the addition method, this method is also called the elimination method . In this method, we add two terms with the same variable, but opposite coefficients, so that the sum is zero. Of course, not all systems are set up with the two terms of one variable having opposite coefficients. Often we must adjust one or both of the equations by multiplication so that one variable will be eliminated by addition.
## How To: Given a system of equations, solve using the addition method.
• Write both equations with x – and y -variables on the left side of the equal sign and constants on the right.
• Write one equation above the other, lining up corresponding variables. If one of the variables in the top equation has the opposite coefficient of the same variable in the bottom equation, add the equations together, eliminating one variable. If not, use multiplication by a nonzero number so that one of the variables in the top equation has the opposite coefficient of the same variable in the bottom equation, then add the equations to eliminate the variable.
• Solve the resulting equation for the remaining variable.
• Substitute that value into one of the original equations and solve for the second variable.
• Check the solution by substituting the values into the other equation.
## Example: Solving a System by the Addition Method
Solve the given system of equations by addition.
\begin{align}x+2y&=-1 \\ -x+y&=3 \end{align}
Both equations are already set equal to a constant. Notice that the coefficient of $x$ in the second equation, –1, is the opposite of the coefficient of $x$ in the first equation, 1. We can add the two equations to eliminate $x$ without needing to multiply by a constant.
\begin{align} x+2y&=-1 \\ -x+y&=3 \\ \hline 3y&=2\end{align}
Now that we have eliminated $x$, we can solve the resulting equation for $y$.
\begin{align}3y&=2 \\ y&=\dfrac{2}{3} \end{align}
Then, we substitute this value for $y$ into one of the original equations and solve for $x$.
\begin{align}-x+y&=3 \\ -x+\frac{2}{3}&=3 \\ -x&=3-\frac{2}{3} \\ -x&=\frac{7}{3} \\ x&=-\frac{7}{3} \end{align}
The solution to this system is $\left(-\frac{7}{3},\frac{2}{3}\right)$.
Check the solution in the first equation.
\begin{align}x+2y&=-1 \\ \left(-\frac{7}{3}\right)+2\left(\frac{2}{3}\right)&= \\ -\frac{7}{3}+\frac{4}{3}&= \\ \-\frac{3}{3}&= \\ -1&=-1&& \text{True} \end{align}
We gain an important perspective on systems of equations by looking at the graphical representation. See the graph below to find that the equations intersect at the solution. We do not need to ask whether there may be a second solution because observing the graph confirms that the system has exactly one solution.
## Example: Using the Addition Method When Multiplication of One Equation Is Required
Solve the given system of equations by the addition method .
\begin{align}3x+5y&=-11 \\ x - 2y&=11 \end{align}
Adding these equations as presented will not eliminate a variable. However, we see that the first equation has $3x$ in it and the second equation has $x$. So if we multiply the second equation by $-3,\text{}$ the x -terms will add to zero.
\begin{align}x - 2y&=11 \\ -3\left(x - 2y\right)&=-3\left(11\right) && \text{Multiply both sides by }-3 \\ -3x+6y&=-33 && \text{Use the distributive property}. \end{align}
\begin{align}3x+5y&=−11 \\ −3x+6y&=−33 \\ \hline 11y&=−44 \\ y&=−4 \end{align}
For the last step, we substitute $y=-4$ into one of the original equations and solve for $x$.
\begin{align}3x+5y&=-11\\ 3x+5\left(-4\right)&=-11\\ 3x - 20&=-11\\ 3x&=9\\ x&=3\end{align}
Our solution is the ordered pair $\left(3,-4\right)$. Check the solution in the original second equation.
\begin{align}x - 2y&=11 \\ \left(3\right)-2\left(-4\right)&=3+8 \\ &=11 && \text{True} \end{align}
Solve the system of equations by addition.
\begin{align}2x - 7y&=2\\ 3x+y&=-20\end{align}
$\left(-6,-2\right)$
## Example: Using the Addition Method When Multiplication of Both Equations Is Required
Solve the given system of equations in two variables by addition.
\begin{align}2x+3y&=-16 \\ 5x - 10y&=30\end{align}
One equation has $2x$ and the other has $5x$. The least common multiple is $10x$ so we will have to multiply both equations by a constant in order to eliminate one variable. Let’s eliminate $x$ by multiplying the first equation by $-5$ and the second equation by $2$.
\begin{align} -5\left(2x+3y\right)&=-5\left(-16\right) \\ -10x - 15y&=80 \\[3mm] 2\left(5x - 10y\right)&=2\left(30\right) \\ 10x - 20y&=60 \end{align}
Then, we add the two equations together.
\begin{align} −10x−15y&=80 \\ 10x−20y&=60 \\ \hline −35y&=140 \\ y&=−4 \end{align}
Substitute $y=-4$ into the original first equation.
\begin{align}2x+3\left(-4\right)&=-16\\ 2x - 12&=-16\\ 2x&=-4\\ x&=-2\end{align}
The solution is $\left(-2,-4\right)$. Check it in the other equation.
\begin{align} 5x - 10y&=30\\ 5\left(-2\right)-10\left(-4\right)&=30\\ -10+40&=30\\ 30&=30\end{align}
## Example: Using the Addition Method in Systems of Equations Containing Fractions
\begin{align}\frac{x}{3}+\frac{y}{6}&=3 \\[1mm] \frac{x}{2}-\frac{y}{4}&=1 \end{align}
First clear each equation of fractions by multiplying both sides of the equation by the least common denominator.
\begin{align}6\left(\frac{x}{3}+\frac{y}{6}\right)&=6\left(3\right) \\[1mm] 2x+y&=18 \\[3mm] 4\left(\frac{x}{2}-\frac{y}{4}\right)&=4\left(1\right) \\[1mm] 2x-y&=4 \end{align}
Now multiply the second equation by $-1$ so that we can eliminate x .
\begin{align}-1\left(2x-y\right)&=-1\left(4\right) \\[1mm] -2x+y&=-4 \end{align}
Add the two equations to eliminate x and solve the resulting equation for y .
\begin{align} 2x+y&=18 \\ −2x+y&=−4 \\ \hline 2y&=14 \\ y&=7 \end{align}
Substitute $y=7$ into the first equation.
\begin{align}2x+\left(7\right)&=18 \\ 2x&=11 \\ x&=\frac{11}{2} \\ &=7.5 \end{align}
The solution is $\left(\frac{11}{2},7\right)$. Check it in the other equation.
\begin{align}\frac{x}{2}-\frac{y}{4}&=1\\[1mm] \frac{\frac{11}{2}}{2}-\frac{7}{4}&=1\\[1mm] \frac{11}{4}-\frac{7}{4}&=1\\[1mm] \frac{4}{4}&=1\end{align}
\begin{align}2x+3y&=8\\ 3x+5y&=10\end{align}
$\left(10,-4\right)$
in the following video we present more examples of how to use the addition (elimination) method to solve a system of two linear equations.
## Classify Solutions to Systems
Now that we have several methods for solving systems of equations, we can use the methods to identify inconsistent systems. Recall that an inconsistent system consists of parallel lines that have the same slope but different $y$ -intercepts. They will never intersect. When searching for a solution to an inconsistent system, we will come up with a false statement, such as $12=0$.
## Example: Solving an Inconsistent System of Equations
Solve the following system of equations.
$\begin{gathered}&x=9 - 2y \\ &x+2y=13 \end{gathered}$
We can approach this problem in two ways. Because one equation is already solved for $x$, the most obvious step is to use substitution.
\begin{align}x+2y&=13 \\ \left(9 - 2y\right)+2y&=13 \\ 9+0y&=13 \\ 9&=13 \end{align}
Clearly, this statement is a contradiction because $9\ne 13$. Therefore, the system has no solution.
The second approach would be to first manipulate the equations so that they are both in slope-intercept form. We manipulate the first equation as follows.
$\begin{gathered}x=9 - 2y \\ 2y=-x+9 \\ y=-\frac{1}{2}x+\frac{9}{2} \end{gathered}$
We then convert the second equation expressed to slope-intercept form.
$\begin{gathered}x+2y=13 \\ 2y=-x+13 \\ y=-\frac{1}{2}x+\frac{13}{2} \end{gathered}$
Comparing the equations, we see that they have the same slope but different y -intercepts. Therefore, the lines are parallel and do not intersect.
$\begin{gathered}y=-\frac{1}{2}x+\frac{9}{2} \\ y=-\frac{1}{2}x+\frac{13}{2} \end{gathered}$
Writing the equations in slope-intercept form confirms that the system is inconsistent because all lines will intersect eventually unless they are parallel. Parallel lines will never intersect; thus, the two lines have no points in common. The graphs of the equations in this example are shown below.
Solve the following system of equations in two variables.
$\begin{gathered}2y - 2x=2\\ 2y - 2x=6\end{gathered}$
No solution. It is an inconsistent system.
## Expressing the Solution of a System of Dependent Equations Containing Two Variables
Recall that a dependent system of equations in two variables is a system in which the two equations represent the same line. Dependent systems have an infinite number of solutions because all of the points on one line are also on the other line. After using substitution or addition, the resulting equation will be an identity, such as $0=0$.
## Example: Finding a Solution to a Dependent System of Linear Equations
Find a solution to the system of equations using the addition method .
$\begin{gathered}x+3y=2\\ 3x+9y=6\end{gathered}$
With the addition method, we want to eliminate one of the variables by adding the equations. In this case, let’s focus on eliminating $x$. If we multiply both sides of the first equation by $-3$, then we will be able to eliminate the $x$ -variable.
\begin{align}x+3y&=2 \\ \left(-3\right)\left(x+3y\right)&=\left(-3\right)\left(2\right) \\ -3x - 9y&=-6 \end{align}
\begin{align} −3x−9y&=−6 \\ +3x+9y&=6 \\ \hline 0&=0 \end{align}
We can see that there will be an infinite number of solutions that satisfy both equations.
If we rewrote both equations in the slope-intercept form, we might know what the solution would look like before adding. Let’s look at what happens when we convert the system to slope-intercept form.
\begin{align}\begin{gathered}x+3y=2 \\ 3y=-x+2 \\ y=-\frac{1}{3}x+\frac{2}{3} \end{gathered} \hspace{2cm} \begin{gathered} 3x+9y=6 \\9y=-3x+6 \\ y=-\frac{3}{9}x+\frac{6}{9} \\ y=-\frac{1}{3}x+\frac{2}{3} \end{gathered}\end{align}
Look at the graph below. Notice the results are the same. The general solution to the system is $\left(x, -\frac{1}{3}x+\frac{2}{3}\right)$.
## Writing the general solution
In the previous example, we presented an analysis of the solution to the following system of equations:
After a little algebra, we found that these two equations were exactly the same. We then wrote the general solution as $\left(x, -\frac{1}{3}x+\frac{2}{3}\right)$. Why would we write the solution this way? In some ways, this representation tells us a lot. It tells us that x can be anything, x is x . It also tells us that y is going to depend on x , just like when we write a function rule. In this case, depending on what you put in for x , y will be defined in terms of x as $-\frac{1}{3}x+\frac{2}{3}$.
In other words, there are infinitely many ( x , y ) pairs that will satisfy this system of equations, and they all fall on the line $f(x)-\frac{1}{3}x+\frac{2}{3}$.
$\begin{gathered}y - 2x=5 \\ -3y+6x=-15 \end{gathered}$
The system is dependent so there are infinitely many solutions of the form $\left(x,2x+5\right)$.
## Using Systems of Equations to Investigate Profits
Using what we have learned about systems of equations, we can return to the skateboard manufacturing problem at the beginning of the section. The skateboard manufacturer’s revenue function is the function used to calculate the amount of money that comes into the business. It can be represented by the equation $R=xp$, where $x=$ quantity and $p=$ price. The revenue function is shown in orange in the graph below.
The cost function is the function used to calculate the costs of doing business. It includes fixed costs, such as rent and salaries, and variable costs, such as utilities. The cost function is shown in blue in the graph below. The x -axis represents quantity in hundreds of units. The y -axis represents either cost or revenue in hundreds of dollars.
The point at which the two lines intersect is called the break-even point . We can see from the graph that if 700 units are produced, the cost is $3,300 and the revenue is also$3,300. In other words, the company breaks even if they produce and sell 700 units. They neither make money nor lose money.
The shaded region to the right of the break-even point represents quantities for which the company makes a profit. The shaded region to the left represents quantities for which the company suffers a loss. The profit function is the revenue function minus the cost function, written as $P\left(x\right)=R\left(x\right)-C\left(x\right)$. Clearly, knowing the quantity for which the cost equals the revenue is of great importance to businesses.
## Example: Finding the Break-Even Point and the Profit Function Using Substitution
Given the cost function $C\left(x\right)=0.85x+35{,}000$ and the revenue function $R\left(x\right)=1.55x$, find the break-even point and the profit function.
Write the system of equations using $y$ to replace function notation.
\begin{align} y&=0.85x+35{,}000 \\ y&=1.55x \end{align}
Substitute the expression $0.85x+35{,}000$ from the first equation into the second equation and solve for $x$.
$\begin{gathered}0.85x+35{,}000=1.55x\\ 35{,}000=0.7x\\ 50{,}000=x\end{gathered}$
Then, we substitute $x=50{,}000$ into either the cost function or the revenue function.
$1.55\left(50{,}000\right)=77{,}500$
The break-even point is $\left(50{,}000,77{,}500\right)$.
The profit function is found using the formula $P\left(x\right)=R\left(x\right)-C\left(x\right)$.
\begin{align}P\left(x\right)&=1.55x-\left(0.85x+35{,}000\right) \\ &=0.7x - 35{,}000 \end{align}
The profit function is $P\left(x\right)=0.7x - 35{,}000$.
The cost to produce 50,000 units is $77,500, and the revenue from the sales of 50,000 units is also$77,500. To make a profit, the business must produce and sell more than 50,000 units.
We see from the graph below that the profit function has a negative value until $x=50{,}000$, when the graph crosses the x -axis. Then, the graph emerges into positive y -values and continues on this path as the profit function is a straight line. This illustrates that the break-even point for businesses occurs when the profit function is 0. The area to the left of the break-even point represents operating at a loss.
## Writing a System of Linear Equations Given a Situation
It is rare to be given equations that neatly model behaviors that you encounter in business, rather, you will probably be faced with a situation for which you know key information as in the example above. Below, we summarize three key factors that will help guide you in translating a situation into a system.
## How To: Given a situation that represents a system of linear equations, write the system of equations and identify the solution.
• Identify the input and output of each linear model.
• Identify the slope and y -intercept of each linear model.
• Find the solution by setting the two linear functions equal to another and solving for x , or find the point of intersection on a graph.
Now let’s practice putting these key factors to work. In the next example, we determine how many different types of tickets are sold given information about the total revenue and amount of tickets sold to an event.
## Example: Writing and Solving a System of Equations in Two Variables
The cost of a ticket to the circus is $25.00 for children and$50.00 for adults. On a certain day, attendance at the circus is 2,000 and the total gate revenue is $70,000. How many children and how many adults bought tickets? Let c = the number of children and a = the number of adults in attendance. The total number of people is 2,000. We can use this to write an equation for the number of people at the circus that day. $c+a=2{,}000$ The revenue from all children can be found by multiplying$25.00 by the number of children, $25c$. The revenue from all adults can be found by multiplying $50.00 by the number of adults, $50a$. The total revenue is$70,000. We can use this to write an equation for the revenue.
$25c+50a=70{,}000$
We now have a system of linear equations in two variables.
$\begin{gathered}c+a=2,000\\ 25c+50a=70{,}000\end{gathered}$
In the first equation, the coefficient of both variables is 1. We can quickly solve the first equation for either $c$ or $a$. We will solve for $a$.
$\begin{gathered}c+a=2{,}000\\ a=2{,}000-c\end{gathered}$
Substitute the expression $2{,}000-c$ in the second equation for $a$ and solve for $c$.
\begin{align} 25c+50\left(2{,}000-c\right)&=70{,}000 \\ 25c+100{,}000 - 50c&=70{,}000 \\ -25c&=-30{,}000 \\ c&=1{,}200 \end{align}
Substitute $c=1{,}200$ into the first equation to solve for $a$.
\begin{align}1{,}200+a&=2{,}000 \\ a&=800 \end{align}
We find that 1,200 children and 800 adults bought tickets to the circus that day.
Meal tickets at the circus cost $4.00 for children and$12.00 for adults. If 1,650 meal tickets were bought for a total of $14,200, how many children and how many adults bought meal tickets? 700 children, 950 adults Sometimes, a system of equations can inform a decision. In our next example, we help answer the question, “Which truck rental company will give the best value?” ## Example: Building a System of Linear Models to Choose a Truck Rental Company Jamal is choosing between two truck-rental companies. The first, Keep on Trucking, Inc., charges an up-front fee of$20, then 59 cents a mile. The second, Move It Your Way, charges an up-front fee of \$16, then 63 cents a mile. [1] When will Keep on Trucking, Inc. be the better choice for Jamal?
The two important quantities in this problem are the cost and the number of miles driven. Because we have two companies to consider, we will define two functions.
A linear function is of the form $f\left(x\right)=mx+b$. Using the rates of change and initial charges, we can write the equations
\begin{align}K\left(d\right)=0.59d+20\\ M\left(d\right)=0.63d+16\end{align}
Using these equations, we can determine when Keep on Trucking, Inc., will be the better choice. Because all we have to make that decision from is the costs, we are looking for when Move It Your Way, will cost less, or when $K\left(d\right)<M\left(d\right)$. The solution pathway will lead us to find the equations for the two functions, find the intersection, and then see where the $K\left(d\right)$ function is smaller.
These graphs are sketched above, with K ( d ) in blue.
To find the intersection, we set the equations equal and solve:
\begin{align}K\left(d\right)&=M\left(d\right) \\ 0.59d+20&=0.63d+16 \\ 4&=0.04d \\ 100&=d \\ d&=100 \end{align}
This tells us that the cost from the two companies will be the same if 100 miles are driven. Either by looking at the graph, or noting that $K\left(d\right)$ is growing at a slower rate, we can conclude that Keep on Trucking, Inc. will be the cheaper price when more than 100 miles are driven, that is $d>100$.
The applications for systems seems almost endless, but we will just show one more. In the next example, we determine the amount 80% methane solution to add to a 50% solution to give a final solution of 60%.
## Example: Solve a Chemical Mixture Problem
A chemist has 70 mL of a 50% methane solution. How much of a 80% solution must she add so the final solution is 60% methane?
We will use the following table to help us solve this mixture problem:
We start with 70 mL of solution, and the unknown amount can be x . The part is the percentages, or concentration of solution 0.5 for start, 0.8 for add.
Add amount column to get final amount. The part for this amount is 0.6 because we want the final solution to be 60% methane.
Multiply amount by part to get total. be sure to distribute on the last row:$(70 + x)0.6$.
If we add the start and add entries in the Total column, we get the final equation that represents the total amount and it’s concentration.
\begin{align}35+0.8x& = 42+0.6x \\ 0.2x&=7 \\ \frac{0.2}{0.2}x&=\frac{7}{0.2} \\ x&=35 \end{align}
35mL of 80% solution must be added to 70mL of 50% solution to get a 60% solution of Methane.
The same process can be used if the starting and final amount have a price attached to them, rather than a percentage.
## Key Concepts
• A system of linear equations consists of two or more equations made up of two or more variables such that all equations in the system are considered simultaneously.
• The solution to a system of linear equations in two variables is any ordered pair that satisfies each equation independently.
• Systems of equations are classified as independent with one solution, dependent with an infinite number of solutions, or inconsistent with no solution.
• One method of solving a system of linear equations in two variables is by graphing. In this method, we graph the equations on the same set of axes.
• Another method of solving a system of linear equations is by substitution. In this method, we solve for one variable in one equation and substitute the result into the second equation.
• A third method of solving a system of linear equations is by addition, in which we can eliminate a variable by adding opposite coefficients of corresponding variables.
• It is often necessary to multiply one or both equations by a constant to facilitate elimination of a variable when adding the two equations together.
• Either method of solving a system of equations results in a false statement for inconsistent systems because they are made up of parallel lines that never intersect.
• The solution to a system of dependent equations will always be true because both equations describe the same line.
• Systems of equations can be used to solve real-world problems that involve more than one variable, such as those relating to revenue, cost, and profit.
addition method an algebraic technique used to solve systems of linear equations in which the equations are added in a way that eliminates one variable, allowing the resulting equation to be solved for the remaining variable; substitution is then used to solve for the first variable
break-even point the point at which a cost function intersects a revenue function; where profit is zero
consistent system a system for which there is a single solution to all equations in the system and it is an independent system, or if there are an infinite number of solutions and it is a dependent system
cost function the function used to calculate the costs of doing business; it usually has two parts, fixed costs and variable costs
dependent system a system of linear equations in which the two equations represent the same line; there are an infinite number of solutions to a dependent system
inconsistent system a system of linear equations with no common solution because they represent parallel lines, which have no point or line in common
independent system a system of linear equations with exactly one solution pair $\left(x,y\right)$
profit function the profit function is written as $P\left(x\right)=R\left(x\right)-C\left(x\right)$, revenue minus cost
revenue function the function that is used to calculate revenue, simply written as $R=xp$, where $x=$ quantity and $p=$ price
substitution method an algebraic technique used to solve systems of linear equations in which one of the two equations is solved for one variable and then substituted into the second equation to solve for the second variable
system of linear equations a set of two or more equations in two or more variables that must be considered simultaneously.
• Rates retrieved Aug 2, 2010 from http://www.budgettruck.com and http://www.uhaul.com/ ↵
• Solving Systems of Equations using Elimination. Authored by : James Sousa (Mathispower4u.com). Located at : https://youtu.be/ova8GSmPV4o . License : CC BY: Attribution
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Systems of Linear Equations
## Solve Systems of Linear Equations with Two Variables
Learning objectives.
By the end of this section, you will be able to:
• Determine whether an ordered pair is a solution of a system of equations
• Solve a system of linear equations by graphing
• Solve a system of equations by substitution
• Solve a system of equations by elimination
• Choose the most convenient method to solve a system of linear equations
Before you get started, take this readiness quiz.
Determine Whether an Ordered Pair is a Solution of a System of Equations
In Solving Linear Equations , we learned how to solve linear equations with one variable. Now we will work with two or more linear equations grouped together, which is known as a system of linear equations .
When two or more linear equations are grouped together, they form a system of linear equations .
In this section, we will focus our work on systems of two linear equations in two unknowns. We will solve larger systems of equations later in this chapter.
An example of a system of two linear equations is shown below. We use a brace to show the two equations are grouped together to form a system of equations.
To determine if an ordered pair is a solution to a system of two equations, we substitute the values of the variables into each equation. If the ordered pair makes both equations true, it is a solution to the system.
Solve a System of Linear Equations by Graphing
In this section, we will use three methods to solve a system of linear equations. The first method we’ll use is graphing.
The graph of a linear equation is a line. Each point on the line is a solution to the equation. For a system of two equations, we will graph two lines. Then we can see all the points that are solutions to each equation. And, by finding what the lines have in common, we’ll find the solution to the system.
Most linear equations in one variable have one solution, but we saw that some equations, called contradictions, have no solutions and for other equations, called identities, all numbers are solutions.
Similarly, when we solve a system of two linear equations represented by a graph of two lines in the same plane, there are three possible cases, as shown.
Each time we demonstrate a new method, we will use it on the same system of linear equations. At the end of the section you’ll decide which method was the most convenient way to solve this system.
The steps to use to solve a system of linear equations by graphing are shown here.
• Graph the first equation.
• Graph the second equation on the same rectangular coordinate system.
• Determine whether the lines intersect, are parallel, or are the same line.
If the lines intersect, identify the point of intersection. This is the solution to the system.
If the lines are parallel, the system has no solution.
• If the lines are the same, the system has an infinite number of solutions.
• Check the solution in both equations.
In the next example, we’ll first re-write the equations into slope–intercept form as this will make it easy for us to quickly graph the lines.
In all the systems of linear equations so far, the lines intersected and the solution was one point. In the next two examples, we’ll look at a system of equations that has no solution and at a system of equations that has an infinite number of solutions.
no solution
Sometimes the equations in a system represent the same line. Since every point on the line makes both equations true, there are infinitely many ordered pairs that make both equations true. There are infinitely many solutions to the system.
If you write the second equation in slope-intercept form, you may recognize that the equations have the same slope and same y -intercept.
infinitely many solutions
When we graphed the second line in the last example, we drew it right over the first line. We say the two lines are coincident . Coincident lines have the same slope and same y- intercept.
Coincident lines have the same slope and same y- intercept.
The systems of equations in (Figure) and (Figure) each had two intersecting lines. Each system had one solution.
In (Figure) , the equations gave coincident lines, and so the system had infinitely many solutions.
The systems in those three examples had at least one solution. A system of equations that has at least one solution is called a consistent system.
A system with parallel lines, like (Figure) , has no solution. We call a system of equations like this inconsistent. It has no solution.
A consistent system of equations is a system of equations with at least one solution.
An inconsistent system of equations is a system of equations with no solution.
We also categorize the equations in a system of equations by calling the equations independent or dependent . If two equations are independent, they each have their own set of solutions. Intersecting lines and parallel lines are independent.
If two equations are dependent, all the solutions of one equation are also solutions of the other equation. When we graph two dependent equations, we get coincident lines.
Let’s sum this up by looking at the graphs of the three types of systems. See below and (Figure) .
Without graphing, determine the number of solutions and then classify the system of equations.
ⓐ We will compare the slopes and intercepts of the two lines.
A system of equations whose graphs are parallel lines has no solution and is inconsistent and independent.
ⓑ We will compare the slope and intercepts of the two lines.
A system of equations whose graphs are intersect has 1 solution and is consistent and independent.
ⓐ no solution, inconsistent, independent ⓑ one solution, consistent, independent
Solve a System of Equations by Substitution
We will now solve systems of linear equations by the substitution method.
We will use the same system we used first for graphing.
We will first solve one of the equations for either x or y . We can choose either equation and solve for either variable—but we’ll try to make a choice that will keep the work easy.
Then we substitute that expression into the other equation. The result is an equation with just one variable—and we know how to solve those!
After we find the value of one variable, we will substitute that value into one of the original equations and solve for the other variable. Finally, we check our solution and make sure it makes both equations true.
• Solve one of the equations for either variable.
• Substitute the expression from Step 1 into the other equation.
• Solve the resulting equation.
• Substitute the solution in Step 3 into either of the original equations to find the other variable.
• Write the solution as an ordered pair.
• Check that the ordered pair is a solution to both original equations.
Be very careful with the signs in the next example.
We need to solve one equation for one variable. We will solve the first equation for y .
Solve a System of Equations by Elimination
We have solved systems of linear equations by graphing and by substitution. Graphing works well when the variable coefficients are small and the solution has integer values. Substitution works well when we can easily solve one equation for one of the variables and not have too many fractions in the resulting expression.
The third method of solving systems of linear equations is called the Elimination Method. When we solved a system by substitution, we started with two equations and two variables and reduced it to one equation with one variable. This is what we’ll do with the elimination method, too, but we’ll have a different way to get there.
The Elimination Method is based on the Addition Property of Equality. The Addition Property of Equality says that when you add the same quantity to both sides of an equation, you still have equality. We will extend the Addition Property of Equality to say that when you add equal quantities to both sides of an equation, the results are equal.
For any expressions a, b, c, and d .
To solve a system of equations by elimination, we start with both equations in standard form. Then we decide which variable will be easiest to eliminate. How do we decide? We want to have the coefficients of one variable be opposites, so that we can add the equations together and eliminate that variable.
Notice how that works when we add these two equations together:
The y ’s add to zero and we have one equation with one variable.
Let’s try another one:
This time we don’t see a variable that can be immediately eliminated if we add the equations.
Then rewrite the system of equations.
Now we see that the coefficients of the x terms are opposites, so x will be eliminated when we add these two equations.
Once we get an equation with just one variable, we solve it. Then we substitute that value into one of the original equations to solve for the remaining variable. And, as always, we check our answer to make sure it is a solution to both of the original equations.
Now we’ll see how to use elimination to solve the same system of equations we solved by graphing and by substitution.
The steps are listed here for easy reference.
• Write both equations in standard form. If any coefficients are fractions, clear them.
Decide which variable you will eliminate.
• Multiply one or both equations so that the coefficients of that variable are opposites.
• Add the equations resulting from Step 2 to eliminate one variable.
• Solve for the remaining variable.
• Substitute the solution from Step 4 into one of the original equations. Then solve for the other variable.
Now we’ll do an example where we need to multiply both equations by constants in order to make the coefficients of one variable opposites.
In this example, we cannot multiply just one equation by any constant to get opposite coefficients. So we will strategically multiply both equations by different constants to get the opposites.
When the system of equations contains fractions, we will first clear the fractions by multiplying each equation by the LCD of all the fractions in the equation.
In this example, both equations have fractions. Our first step will be to multiply each equation by the LCD of all the fractions in the equation to clear the fractions.
When we solved the system by graphing, we saw that not all systems of linear equations have a single ordered pair as a solution. When the two equations were really the same line, there were infinitely many solutions. We called that a consistent system. When the two equations described parallel lines, there was no solution. We called that an inconsistent system.
The same is true using substitution or elimination. If the equation at the end of substitution or elimination is a true statement, we have a consistent but dependent system and the system of equations has infinitely many solutions. If the equation at the end of substitution or elimination is a false statement, we have an inconsistent system and the system of equations has no solution.
This is a true statement. The equations are consistent but dependent. Their graphs would be the same line. The system has infinitely many solutions.
After we cleared the fractions in the second equation, did you notice that the two equations were the same? That means we have coincident lines.
Choose the Most Convenient Method to Solve a System of Linear Equations
When you solve a system of linear equations in in an application, you will not be told which method to use. You will need to make that decision yourself. So you’ll want to choose the method that is easiest to do and minimizes your chance of making mistakes.
For each system of linear equations, decide whether it would be more convenient to solve it by substitution or elimination. Explain your answer.
Since both equations are in standard form, using elimination will be most convenient.
Since one equation is already solved for y , using substitution will be most convenient.
For each system of linear equations decide whether it would be more convenient to solve it by substitution or elimination. Explain your answer.
ⓐ Since both equations are in standard form, using elimination will be most convenient. ⓑ Since one equation is already solved for x , using substitution will be most convenient.
ⓐ Since one equation is already solved for y , using substitution will be most convenient. ⓑ Since both equations are in standard form, using elimination will be most convenient.
## Practice Makes Perfect
In the following exercises, determine if the following points are solutions to the given system of equations.
In the following exercises, solve the following systems of equations by graphing.
infinite solutions
1 point, consistent and independent
infinite solutions, consistent, dependent
In the following exercises, solve the systems of equations by substitution.
In the following exercises, solve the systems of equations by elimination.
infinitely many
In the following exercises, decide whether it would be more convenient to solve the system of equations by substitution or elimination.
ⓐ substitution ⓑ elimination
ⓐ elimination ⓑ substituion
## Writing Exercises
In a system of linear equations, the two equations have the same intercepts. Describe the possible solutions to the system.
ⓐ by graphing ⓑ by substitution
After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.
If most of your checks were:
…confidently. Congratulations! You have achieved the objectives in this section. Reflect on the study skills you used so that you can continue to use them. What did you do to become confident of your ability to do these things? Be specific.
…with some help. This must be addressed quickly because topics you do not master become potholes in your road to success. In math every topic builds upon previous work. It is important to make sure you have a strong foundation before you move on. Who can you ask for help? Your fellow classmates and instructor are good resources. Is there a place on campus where math tutors are available? Can your study skills be improved?
…no – I don’t get it! This is a warning sign and you must not ignore it. You should get help right away or you will quickly be overwhelmed. See your instructor as soon as you can to discuss your situation. Together you can come up with a plan to get you the help you need.
#### IMAGES
1. 01
2. Number Problems with Two Variables
3. How to Solve Systems of Equations with two variables
4. How to solve two equations with three variables
5. Solving Word Problems with Two Variables
6. How To Solve Math Problems With Two Variables
#### VIDEO
1. DAY 3/75 HARD CHALLENGE. LINEAR EQUATION IN TWO VARIABLES (PART-2) CLASS 10 (SSC BOARD)
2. DAY 5/75 HARD CHALLENGE . LINEAR EQUATION IN TWO VARIABLES (PART-1) SSC BOARD
3. DAY 1/75 HARD CHALLENGE . LINEAR EQUATION IN TWO VARIABLES (PART-1) SSC BOARD
4. Solve linear equations in two variables in just 10sec #maths #shorts #khantrickster
5. DAY 7/75 HARD CHALLENGE . Linear Equations In Two Variables (MATHS-1) Class 10 . Maharashtra Board
6. How to Solve for a Specific Variable
1. Solve for Y in Terms of X?
According to the University of Regina, another way to express solving for y in terms of x is solving an equation for y. The solution is not a numerical value; instead, it is an expression equal to y involving the variable x. An example prob...
2. What Are the Four Steps for Solving an Equation?
The four steps for solving an equation include the combination of like terms, the isolation of terms containing variables, the isolation of the variable and the substitution of the answer into the original equation to check the answer.
3. What Are the Six Steps of Problem Solving?
The six steps of problem solving involve problem definition, problem analysis, developing possible solutions, selecting a solution, implementing the solution and evaluating the outcome. Problem solving models are used to address issues that...
4. 4 Ways to Solve Systems of Algebraic Equations Containing Two
Then, divide both sides of the equation by one of the variables to solve for that variable. Next, take that number and plug it into the formula
5. Solutions to 2-variable equations (video)
They help optimize systems, analyze data, and solve problems across various domains. Comment Button navigates to signup page. (4 votes). Upvote. Button
6. Two-Variable Equations
A two-variable equation is solved by plugging in x = 0 and solving for y and then plugging in y = 0 and solving
7. 01
Get more lessons and courses at http://www.MathTutorDVD.com. Learn how to solve equations in algebra that have more than one variable in
8. How Do You Solve Two Equations with Two Variables?
This tutorial will take you through this process of substitution step-by-step! Keywords: problem; equation; substitution; solve by substitution; simplify
9. Solving Equations With Two Variables
where x and y are real numbers and a and b are not both zero. For example, 3x + 2y = 8 is a linear equation in two variables.
10. How do you solve an equation with two variables?
An equation with two variables requires a second, independent equation to solve . Independent means it is not just a multiple of the first, such
11. Solve Equations with Multiple Variables
MathIsSimple #Grade8 #Variables #Equations #Algebra Solve Equations with Multiple Variables ... problems from Middle School to Algebra on Math is
12. Number Problems with Two Variables
Now, solve by adding the two equations. Now, plugging into the first equation gives. The numbers are 11 and 4. Example 2. The
13. Systems of Linear Equations: Two Variables
Systems of equations can be used to solve real-world problems
14. Solve Systems of Linear Equations with Two Variables
If you missed this problem, review (Figure). Determine Whether an Ordered Pair is a Solution of a System of Equations. In Solving Linear Equations, we learned
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# Newton's laws vs energy for solving a problem
I have a problem I solved using kinematics/Newton's 2nd law.
It gives the mass of a walker as 55kg. It then says she starts from rest and walks 20m is 7s. It wants to know the horizontal force acting on her.
From kinematics for constant acceleration, I know $$\vec{r}=\frac{a}{2}t^2\hat{i}$$. Plugging in the known time and the known distance I solved for the acceleration and then I could get the force by multiplying the acceleration by the walker's mass. So I got the problem right... but then I got to wondering: Was there a way to do this problem using energy? I have in mind $$\vec{F}\cdot\Delta\vec{r}=\Delta K$$. I tried but I don't know the final velocity (from the given information).
Edit: I realized after looking at some of the feedback that I do know the final velocity (because the linear dependance of velocity on time means the average velocity must be half the final velocity). Therefore, you can see below, that I have posted the answer I was hoping to write back when I wished I knew the final velocity.
• This problem is extremely unclear (not your fault). What is a walker? Is it a person or a thing? Is there friction? If we're talking about a human walking, then that sounds like common-core, because the kinesiology of walking is not amenable to simple analysis, which is why physics problems general discuss masses on frictionless surfaces. – JEB Nov 24 '18 at 1:14
• When I walk, I don’t accelerate uniformly and go faster and faster. – G. Smith Nov 24 '18 at 1:16
• @JEB please take the force of static friction of ground on walker to be the only relevant force; and treat the walker as a point mass. – okcapp Nov 24 '18 at 1:17
Assuming constant acceleration from rest the velocity against time graph looks like this:
Knowing the displacement $$s$$, which is the area under the graph, and the time $$t$$ one can link these two quantities either to the acceleration $$a$$ using $$s = \frac 12 \,at\,t = \frac 12at^2$$ (compare with the constant acceleration kinematic equation $$s = ut + \frac 12 at^2$$ with the initial velocity $$u = 0$$) or the final velocity $$v$$ using $$s = \frac12 \,v\,t$$ (compare with the constant acceleration kinematic equation $$s = \frac 12 \frac{(u+v)}{t}$$ with the initial velocity $$u=0$$).
One can then use either Newton's second law $$F = ma$$ or the work-energy theorem $$Fs = \frac 12 m v^2$$ to find the force $$F$$.
OK, so a female point mass $$m$$ accelerates from $$v=0$$ at constant acceleration and covers distance $$r$$ in time $$t$$, so using:
$$d = \frac 1 2 a t^2$$
we get
$$a = 2d/t^2$$
so that:
$$F = ma = 2md/t^2$$.
The question is, can this problem be solved using energy? Let's try:
We have to tilt it and use an equivalent gravitational field $$a$$, in which an at rest mass falls $$d$$ in time $$t$$, which mean the potential energy:
$$U = mad$$
is converted into kinetic energy:
$$K = ?$$.
Now what? Well, we know the average velocity is:
$$\bar v = d/t$$
and we know the final velocity is twice the average velocity, so:
$$v = 2\bar v = 2d/t$$
so that the kinetic energy is:
$$K = \frac 1 2 m v^2 = 2md^2/t^2$$
an of course:
$$K = U$$
so that:
$$2md^2/t^2 = mad$$
or:
$$a = 2d/t^2$$
Now at this point we could use $$F=ma$$ and get the right answer, but we're not using Newton's Laws. We're going to use:
$$F = \frac{\partial U}{\partial d}$$
so plugging $$a$$ in to the expression for $$U$$:
$$U = mad = m(2d/t^2)d = 2md^2/t^2$$
so
$$\partial U/\partial d = 2md/t^2 = F$$
which is correct. So the answer to your question is "yes", you can use energy.
Using the work-kinetic energy theorem like you stated is a good start. As you said, that method requires knowing the final velocity. So, just use the basic kinematic relation,
$$v_{f}^{2} = v_{i}^{2} + 2a\Delta x = 2a\Delta x$$
where $$\Delta x$$ is the displacement which is given in the problem statement. I think it's kinda straight forward from here:
$$W = \Delta K$$
$$F \Delta x = \frac{1}{2}m v_{f}^{2} = \frac{1}{2}m (2a \Delta x) = ma \Delta x$$
$$F = ma$$
So indeed, Newton's second law is recovered, and you would just use the relation that you provided to find the acceleration. In this problem, using energy involves a bit more work than what you did originally, but it's still a workable path :)
• I'm a little confused by this. Your kinematic equation is exactly the same as your second equation, so you've somehow used the same equation twice to recover Newton's second law. I"m not sure what exactly you did. There is a typo in the last equation, by the way. – garyp Nov 24 '18 at 1:45
• Thank you for pointing out the typo, it's been fixed. And to clarify, I agree that the kinematic equation I provided can be algebraically manipulated into the work-kinetic energy theorem, but if you want to use energy to solve the OP's problem then you need the final velocity, and if you want to use the work-kinetic energy theorem then it's a rather circular method of solving. One can instead use an artificial potential energy as JEB did in his solution, but I just wanted to point out the circularity of the OP's supposition. – N. Steinle Nov 24 '18 at 1:53
So from energy conservation $$F.s = mv^2/2$$ ;$$F.s=ma^2t^2/2$$ ; $$F.s=\frac{ 2m(at^2/2)^2}{t^2}$$ ;F.s=$$\frac{2m \times (at^2/2)^2}{t^2}= 2ms^2/t^2$$ ; note that $$v = at$$ and $$s=at^2/2$$ s= displacement v= velocity. I get the force as $$F= 2 \times m \times s/t^2$$ so i conclude the result can be also obtained by energy conservation.
It occurred to me that since $$\vec{v}=at\hat{i}$$, it is clear that $$v_{final}=2v_{average}$$. Well, since $$v_{average}=\frac{|\Delta\vec{x}|}{t_{total}}$$, we know that $$v_{final}=2v_{average}=\frac{2|\Delta\vec{x}|}{t_{total}}$$. This means that $$\vec{F}\cdot\Delta\vec{x}=\Delta K=\frac{1}{2}mv_{final}^2$$ can be solved for $$|\vec{F}|$$ using the known mass, the known distance, and $$\vec{v}_{final}=\frac{2|\Delta\vec{x}|}{t_{total}}$$: $$|\vec{F}|=\biggl(\frac{1}{|\Delta\vec{x}|}\biggr)\biggl(\frac{1}{2}\biggr)m\biggl(\frac{2|\Delta\vec{x}|}{t_{total}}\biggr)^2$$ Note that $$\vec{F}$$ and $$\Delta \vec{x}$$ both only have components in the positive $$\hat{i}$$ direction, so I took for granted that: $$\vec{F}\cdot\Delta\vec{x}=|\vec{F}||\Delta\vec{x}|$$
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Home » Stones to Kilograms » 9 and a Half Stone to Kg – 9 and a Half Stone in Kilograms
9 and a Half Stone to Kg – 9 and a Half Stone in Kilograms
9 and a half stone to kg means the same as changing 9 and a half stone to kilograms 🙂
9 and a Half Stone to Kilograms Converter
+
=
Reset
Right below you can find all you want to know about the conversion:
What is 9 and a Half Stone in Kilos?
The answer to the question is:
9 1/2 st = 60.32779 kg
9 1/2 stone = 60.32779 kilograms
• A stone (st) is an English and British imperial unit of mass equal to 14 pounds
• A kilogram (kg) is the base unit of mass equal to 2.2046226218 pounds
9 and a Half Stone Equal How Many Kilos as a Fraction?
Here is the result as a fraction:
9 and a half stone in kg ≈ 60 10536/32143 kilograms
Next, we explain the math:
How to Convert 9 and a Half Stone to Kilograms?
You can get the result in two different ways:
1. 9 1/2 st × 6.35029318 kg1 st = 60.32779 kg, because there are 6.35029318 kilos per stone
2. or
3. 9 1/2 st 0.1574730444 st × 1 kg = 60.32779 kg, because there are 0.1574730444 stone weight per kilo
It is recommended using our app above; way easier than a formula!
Similar calculations include, for example:
For further information regarding the units in this post check our articles located in the header menu.
– Article written by Mark
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## What simple interest rate will susan
Simple Interest. When you save money in a bank account, the bank pays you interest.. When you borrow money from a bank, you must pay both the loan and the interest to the bank.. The Principal is the amount either saved or borrowed.. Simple Interest is also called Flat Interest.. Simple Interest Rules are: Simple Interest (\$) = Principal (\$) × Interest Rate (as a decimal) × Number of Years (time)
What simple interest rate will Susan need to secure to make \$2,500 in interest on a \$10,000 principal over 5 years? Find answers now! No. 1 Questions & Answers Place. Simple interest is calculated only on the initial amount (principal) that you invested. Example: Suppose you give \\$100 to a bank which pays you 5% simple interest at the end of every year. After one year you will have \\$105, and after two years you will have \\$110. Simple Interest. When you save money in a bank account, the bank pays you interest.. When you borrow money from a bank, you must pay both the loan and the interest to the bank.. The Principal is the amount either saved or borrowed.. Simple Interest is also called Flat Interest.. Simple Interest Rules are: Simple Interest (\$) = Principal (\$) × Interest Rate (as a decimal) × Number of Years (time) Simple interest is money earned on a savings account or an investment. It can also be money you pay for borrowing money. Simple Interest = Principal × Annual Interest Rate × Time (\$) (\$) (% per yr) (Years) I = Prt Work with a partner. You put \$100 in a savings account. The account earns 6% simple interest per year. Principal, rate of simple interest, and amount problems Our mission is to provide a free, world-class education to anyone, anywhere. Khan Academy is a 501(c)(3) nonprofit organization.
## Simple Interest Worksheet - Free download as PDF File (.pdf), Text File (.txt) or read Now suppose you want to know what interest rate would be needed to earn \$7250 Susan borrows \$8650 to buy a used car and is charged 4.5% interest.
17 Jan 2020 Here, we'll say that you borrowed \$1,000 for 2 years with a simple interest rate of 5%. In this case, you will pay back a total of \$1,100. This is The general interest Investment Writing Update is my monthly compilation of useful ISI's Straszheim: China's interest rate hike is “tapping the brakes” November 2010 Research study: How financial services firms will make money in the future November 2007; Is your investment commentary too simple?, October 2007 PENALTY (WHETHER EARNED OR NOT) based on simple interest rate in effect for the account on early withdrawal date. Penalty will be waived under certain Susan Edmunds14:10, Sep 03 2019. Facebook; Twitter; Whats App But, simple or compounding, the interest rate is too high. The risk with payday lending is
### Compound Interest Rate Calculation - This calculator simply takes the final amount you The loan can be a mortgage, car loan, or any other simple interest
SOLUTION: What simple interest rate will Susan need to secure to make \$2,500 in interest on a \$10,000 principal over 5 years?
### Best Answer: Simple interest at a rate of 5% on \$10,000 would earn \$500 per year or \$2,500 over 5 years.
The general interest Investment Writing Update is my monthly compilation of useful ISI's Straszheim: China's interest rate hike is “tapping the brakes” November 2010 Research study: How financial services firms will make money in the future November 2007; Is your investment commentary too simple?, October 2007 PENALTY (WHETHER EARNED OR NOT) based on simple interest rate in effect for the account on early withdrawal date. Penalty will be waived under certain Susan Edmunds14:10, Sep 03 2019. Facebook; Twitter; Whats App But, simple or compounding, the interest rate is too high. The risk with payday lending is Contact Susan M DeMello at (703) 395-9129 to learn about Chase refinancing or new home mortgage What is better: a fixed or an adjustable interest rate? But buyers who are short on cash and can qualify for a higher interest rate may find a no-point loan will significantly cut their closing costs. The most common form of interest is "fee simple" or "fee," which is the highest type of interest an When you chose financing with Susan Schein Automotive, you will not only receive one of the best interest rates from one of our approved lenders, you will also What simple interest rate will Susan need to secure to make \$2,500 in interest on a \$10,000 principal over 5 years? -----I = Prt 2500 = 10000*r*5 r = 2500/50000 = 0.05 = 5% ===== Cheers, Stan H.
## SOLUTION: What simple interest rate will Susan need to secure to make \$2,500 in interest on a \$10,000 principal over 5 years?
She has to make \$500 per year on a \$10,000 principal. Divide 500 by 10000, then express the decimal fraction as a percentage. John My calculator said it, I 14 Nov 2013 William G. asked • 10/08/13 . What simple interest rate will Susan need to secure to make \$2,500 in interest on a \$10,000 principal over 5 years? . What simple (b) How much will she have altogether in her bank account after 3 years? Answer : (a) Simple Interest = Principal × Interest Rate × Number of Years = P × i × n 29 Feb 2020 Exercise 6.66: Find the simple interest earned after 2 years on \$700 at an interest rate of 4%. In the next example, we will use the
replacing the plus sign with a minus sign, we change the formula to simple interest Bank B: 12.44%. Bank C: Effective interest rate will only differ from nominal interest rate over 2 years. What will Susan's mother's monthly instalments be? 7 Sep 2019 Until the interest rates rise again, the consumer benefits and can use the Of course, the simple opportunity for a loan invites you to financially 12 Dec 2008 What will be the price of a \$20,000 car in 5 years if the inflation rate is 6%?. At an interest rate of 8% compounded continuously, how many years The Susan G. Komen® Cash Rewards credit card lets you earn cash rewards while helping to support a cure for Before you leave, please take a moment to answer these 4 simple questions. 1. After the intro APR offer ends, a Variable APR that's currently 15.49% to 25.49% will apply. Interest Rates & Fees Summary † 1 Nov 2012 Summary review of simple interest and exploration of continuously compounding interest. How long will it take \$2000 to grow to \$25,000 at a 5% interest rate? Susan has had a saving account for a few years now.
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## How to
When solving these problems, learning by doing is much better than learning by reading. I encourage to you read only as far in the solution as you need, then trying to solve the problem. If you get stuck, try reading a little further. And of course, let me know if you find a better solution!
## Wednesday, February 29, 2012
### Write OR and AND from NAND.
Problem: You are told that you only have one logical operator, !&& which is NAND. Write or()and and().
bool or(bool a, bool b) {...}
bool and(bool a, bool b) {...}
Solution:
Let's start by making sure that we understand NAND well. NAND takes the AND function and then negates it, returning the opposite of the standard AND function. It's worth drawing a table (called a truth table) which shows the output of NAND
The truth table below shows NAND and its values. (As you can see, it's the opposite of an AND truth table)
Table 1: NAND truth table
INPUT OUTPUT A B A NAND B 0 0 1 0 1 1 1 0 1 1 1 0
Let's start now by comparing this with the OR function. Or is true when either or both of the values are true. This gives us the truth table of:
Table 2: OR truth table
INPUT OUTPUT A B A OR B 0 0 0 0 1 1 1 0 1 1 1 1
Now, let's think about how we could write the or() function using only the !&& operator. Well let's think about some ideas here. If we simply NAND the input together, we will get the outputs of table 1. This is close and works for the middle two case, but we need to flip the two edges to get the correct output. We could try NAND-ing the values together again, but this doesn't get us anywhere.
This is a good time to take a step back and start over. Clearly, our 'just get going approach' isn't going anywhere.
Now, let's look at what happens if we NAND a value with itself. Well, if the value is 1, 1 NAND 1 is 0. And, if the value is 0, 0 NAND 0 is 1. So, we can essentially, flip or negate a value.
Now, look at the truth tables again. See if this can be useful.
In fact, it can be useful, as if we flight each input, and then NAND the output, we will only return false if both inputs are originally 0, which is what we want to create a proper or.
Now, coded up, this is:
Instead, let's go back to our original case and maybe we don't want to NAND the numbers together immediately. Let's look at the truth table 2 again, and we can see that so long as there is at least one 1, we want to output 1. In the truth table in Table 1, we only get a 0 when we have two 1s.
This is a critical insight, and we can apply it if we consider what happens when we NAND a value with 1 (or true).
bool or(bool a, bool b) {
return ((a !&& a) !&& (b !&& b));
}
Now, we can probably already see how to make the AND gate if we were paying attention. We saw that we can flip (or negate a value) by NANDING with itself, as we did with both a and b to make or(). Hence, NAND is simply Not-And, so we can apply the not or negate operator again by NANDING the outpub with itself.
As code, this is:
bool or(bool a, bool b) {
return ((a! && b) !&& (a !&& b));
}
It's worth doing the truth table for both or and and to convince yourself, and your interviewer, that these are correct in all four cases of a and b.
This is a good example of where it's useful to try examples and gain insights (e.g., NANDing a and b, NAND-ing a value with itself to see that it flips) and also be willing to go back a step if you're not making progress. Also, drawing pictures and truth tables is very helpful.
It's worth nothing that this isn't as contrived a problem as it seems. In digital logic, NAND gates (and NOR gates) are known as universal gates because you can make all other logical operations (OR, AND, NOT, XOR, NAND, AND) from any one of these. So, often, these operations are quite useful. If you would like further practice, try doing the same problem but with !|| as your only operator instead.
/* Please let me know if you find any bugs or if you have an alternative solution. Thank you, Noah*/
## Tuesday, February 28, 2012
### What goes with if?
Problem: Write what goes inside of the if statement to produce, "Hello World.\n"
void printHelloWorld () {
if(...) {
printf("Hello ");
} else {
printf("World.\n");
}
}
Solution: Let's start by trying some of the obvious conditions that could go inside the if statement. The most obvious are true and false.
Clearly, neither of these works. If we use true, the output is only "Hello". And, if we do false, the output is only "World.\n".
So this is tricky, which it would almost have to be to be an interview question.
One thing that we know about an if statement is that 0 evaluates to false, and everything else to true. Can we use this fact? Well, we could put in something that could be contingent, but ultimately, the if statement would evaluate to true of false, so this doesn't help us to have both the clause after if and the clause after else print.
But, that's not to say that it won't help us. We can have something print in the if statement. For example, we can have a function in the if statement, which would both print and return a boolean value. If we do this, we can then print part of the if/else statement.
And, since the evaluation part comes first, we would want to print the first part of the "Hello World\n." statement and then trigger the second half (We can't do it backwards by printing World\n. last).
And, we'll need a function that always returns false, but that is easy too.
Now, we can code up just such a function as:
bool printHelloAndReturnFalse () {
printf("Hello ");
return (false);
}
Now, let's test our function and make sure if works. When we call PrintHelloWorld(), the function will call what we wrote above and print, "Hello ". The false will trigger the else clause and print, "World.\n". This is exactly what we want.
Now, this is a good solution, but try not to stop here. Other than a programming exercise, can you think of a time when you may have to do something like this? (Even if you're not asked this, it's good to think about it.)
Well, we may need an evaluator function for certain packages. And, we may want the evaluator function to do more than just evaluate (e.g., print to a log, etc.). In this case, we would want to make sure that an evaluator did more than just return a boolean value, and it's a case of where we would write such a function.
This is a good example of a problem where it's important to try (and quickly realize that they don't work) obvious answers, then think about the challenges to the question and how it can be overcome. Write a solution (pretty easy) -- and then stretch your answer to show your interviewer that you're always looking for an even better solution.
/*Please let me know if there are bugs or if you find a better solution. Thank you. Noah */
## Monday, February 27, 2012
### 3 integers that sum to zero
Problem: Given an unsorted array of integers, return true if there is a 3-element subset, of distinct elements, that sums to zero
bool threeSumToZero(int a[]){
...
}
Soution: Well, there seems to be an obvious answer here. We can iterate through the array and for each element, we iterate and then iterate again to check all three element combinations. And, so long as we're careful and make sure that we use distinct elements, we can get this right.
The code for this is:
bool threeSumToZero(int a[]){
int i, j, k;
for(i=0; i < a.length, i++) {
for (j=0; j < a.length; j++) {
for (k=0; k < a.length; k++) {
//check for distinct elements
if (i !=j && i != k && j!= k) {
if ((a[i] + a[j] + a[k]) == 0 )
return (true);
}
}
}
}
return (false);
}
Well, this works beautifully and is a very simple solution. There are a few optimizations possible (like starting j at 1 and k at 2) and combining the two if statements, but basically this is pretty sound.
What is the algorithmic run time for this?
Well, since for each element, we check each element and then check each element for that, we have an O(n) with another O(n) for each n and then another O(n) for each of those. This results in a run time of O(n^3).
We could stop here, but O(n^3) is pretty slow. So, let's think about a potential optimization. You may see an optimization, but if you don't, let's try an example:
Pic 1 shows an array that will sum to zero. How do we know this. Well, we can see that -3 + 2 + 1 = 0.
Now, how did we inuit this?
What you probably did, was check the sums, and then see if there was a value which was opposite that (e.g., negative if the sum was positive, and positive if it was negative). You didn't have to actually check the sum of each element, only the membership of the opposite of the sum.
This hints at the solution to make it faster. Since we don't have to check all sums of triplets; we can just check the sum of every pair, and then see if there is an opposite. And, since look up can be a constant time operation after we build a data structure with O(n) time, we can:
1. Build a constant time lookup, like a hashtable, an O(n) operation
2. Determine the sum of all pairs, an O(n^2) operation, and
3. Check if we have the opposite of the sum in a hash table, an O(1) operation
This will give us an O(n^2) algorithm, but at the tradeoff of needing a data structure.
The code for this is:
bool threeSumToZero(int a[]){
int i, j;
//assume we have hashtable functions:
hash *h;
h = buildHashTable(a);
for(i=0; i < a.length, i++) {
for (j=0; j < a.length; j++) {
//check for distinct elements
if (i !=j) {
if (lookup(h, -(a[i] + a[j])))
return (true);
}
}
}
//you could free the hashtable here if needed
return (false);
}
Beautiful. This is an example of an obvious solution being nice, but inefficient. It's good to always get to a working solution fast. Then, by doing an example and intuiting what we were doing, it's possible to build a much more efficient algorithm. And, the coding isn't too difficult; it's mostly the algorithm here.
/*Please let me know if you find any bugs or have comments. Thanks, Noah */
## Friday, February 24, 2012
### Let's Make a Deal
Problem: (Thank you to Mr. Schwartz in 9th grade for this). In the classic TV game show, "Let's Make a Deal," you win a prize if you can choose which of three doors the big prize is behind. After choosing the door, the host opens one of the other doors which has a booby prize (like a bag of sand). You are then asked whether you want to switch to the one remaining closed door, or stick with your original choice. Should you switch?
Solution:
Let's start with a drawing of the situation. Pic 1 shows the basics where there are three closed doors, two with booby prizes and one with the big prize behind it. We are asked to choose any door first. Let's say we choose door 2.
Then, the host opens another door, let's say door 1, revealing a booby prize (e.g., a bag of sand). We are then asked if we want to switch -- from door 2 to door 3. We know that one of the remaining closed doors has the big prize, and one another booby prize This situation is shown in pic 2.
Well, should we switch?
The obvious answer is that it just doesn't matter. We know that there is a prize behind one of two doors, and we have a 50% shot. So, switch or don't switch; we should be indifferent and we will have a 50% chance of winning regardless of strategy.
This is a fairly straight-forward answer and is tempting. However, before we accept it, let's think about the situation a little more and if we're using all available information.
Certainly, if we were presented with two doors and told to choose 1, we could make an overwhelming case for the strategy not mattering -- similar to the strategy not mattering if you call heads or tails on a fair coin toss.
However, we were given lots more information here with a longer set up, one choice, some information revealed, and then asked to make another choice.
Unless your an information/logic expert, a good strategy is to try some examples to see what happens when we choose different doors (with big prize, without) and what happens when we switch/don't switch to see if it helps us figure out the optimal strategy:
Strategy 1: Never switch
A- Choose big prize door out of three (chance 1/3)
- Don't switch when given choice
- Win
- Chance: 1/3
B-Choose booby prize door (chance 2/3)
- Don't switch when given choice
- Lose
- Chance 2/3
Strategy 2: Always switch
A-Choose big prize door out of three (chance 1/3)
- Switch when given choice
- Lose
- Chance 1/3
B-Choose booby prize door out of three (chance 2/3)
- Switch when given choice
- Win
- Chance 2/3
To win:
- If we never switch, we have a 1/3 chance of winning; we need to choose the correct door initially
- If we always switch, we have a 2/3 chance of winning; we need to choose the incorrect door initially
Hence, we should always switch to improve our odds of winning.
This answer is correct, but there's something that can be unsatisfying about it, because intuitively, staring at two doors -- which is our situation, it still feels like we should be able to choose either and have a 50% chance.
So, why isn't this the case?
Well, we were given significant information by the door that was opened. Since the door had to be either: 1) the other booby prize (if we chose the first), or 2) could be either other door, we are given information because it cannot simply be a random door that was opened. This information can be used to improve our odds by switching.
If you're interested there are lots of simulations run showing the benefits of switching vs. not switching.
This is a good example of the first, intuitive answer of 50% not being correct, which is usually the case in brainteasers. When stumped, try doing examples, and when possible, the exhaustive answer set, as that can help bring insights into the correct solution.
/* Please let me know if you find bugs or have alternative solutions. Thank you, Noah */
## Thursday, February 23, 2012
### Data structures and DAWGS
Problem: (Thank you to Stanford’s CS 106X class). Discuss a data structure, under 64k bytes, which can in near constant time determine whether a string is a valid word in the English language, or is invalid.
Solution:
Let’s start by asking a few more assumptions about this problem. First, do we have to build the data structure? No – we just need to figure out what sort of data structure would be able to do this.
Well, let’s start out with one of the parts of the questions – ‘near constant time.’ Well, there are very few common data structures which can achieve this. Some data structures that fail immediately are binary search trees (log n), sorted arrays (log n), linked lists (n or log n).
One common data structure that is nearly constant time is a hash table. And, at first this seems like a great data structure. We could hash based upon the string, end up in a hash bucket, and so long as we keep the hash buckets relatively small, we could get a near constant time data structure. Well, this is true. This would work as being nearly constant time. Pic 1 shows such a hash table.
Now, let’s see if this would work on the space constraint. Let’s make an assumption that there are 50k words in English (especially with variations like sing, sang, sung, singing, singer, sings, etc.). This seems about right. Maybe it’s higher, but let’s see if we can create a hash table with a dataset of this size in the space that we have.
Let’s also assume that we use ascii and the average word is 5 letters (seems low, but let’s go with it). So, the average word is 6 bytes long, assuming that it’s null terminated. Storing all of these words, not include the over head of the hash table, we would have a data structure of 6 * 50k = 300k. This is much larger than we want, and probably it would be larger as our estimate of words is probably low. With the overhead of the hash table, we would have an even bigger amount.
So, a hash table can’t work as its data store will be too large.
So, this creates a problem that any data structure would have – we cannot uniquely store each word and meet the space requirements.
Well, where to go from here?
Well, let’s think about what we know about the English language to look for optimizations. We know that the words are not random character strings, but words with many common parts.
Again, a good example is: sing, sang, sung, sings, singer, singing. Here, we have six distinct words. In a hash table, we would have six distinct entries and we know that this approach will not work.
However, looking at the words, we can see that they have both common prefix and common suffix. Four of the words have the common prefix ‘sing’ and four of them end with ‘ng’ and 3 with ‘ing.’
Let’s think about what we could do here. Well, this screams for a tree approach where individual characters are nodes. We previously rejected a tree because of the time required, but let’s go with this for now. We can make ach node only branches when it has a new letter.
By doing this, we have a tree with s as the root and then the structure as shown in pic 2.
We could put a bit on each bit indicating whether stopping at the node is a valid word or not.
This tree would be fairly quick to search to see if a word were in it, but the structure isn’t quite good enough for searching. For example, we would have to search all of the lower nodes to see if a descendant node with the character we’re looking for.
One quick optimization is to create nodes with all 26 characters as pointers and a ‘true’ bit. We can then easily go down the tree and see if we have a valid word or not. And, this is almost constant time as we never have to look up more elements the number of characters in the word. This type of structure is shown in pic 3, and is called a Directed Acyclical Word Graph (DAWG).
If we think of the size of this structure, it probably reduces the size of the words by a factor of 10 given all of the common prefixes in words (probably a greater reduction as over time, most prefixes are pretty common, but seems pretty close). This will get us close to the size we’re looking for, but let’s think about how we can do better.
So far, we have done a good job constructing a data structure that optimizes around prefixes, but if we look at the example of the words like, ‘sing’, we see that we also have lots of common suffixes. As simple optimizations, we could imagine uniting common prefixes. For example, many words in English end in ‘ing’; we could have all of these words point to a common ‘ing’ ending.
This would be a further optimization, but would obviously require balancing and make the function which inserts words longer.
This would give us a much optimized and fast data structure.
This is a good example where it’s good to try the first thought of what could work (e.g., a hash table), but then move on and look at optimizations to get to the correct answer.
/* Please let me know if you find bugs or know alternative solutions. Thanks, Noah */
## Wednesday, February 22, 2012
### Elevator Algorithm
Problem: Describe an algorithm that you could use if programming an elevator.
Solution: No custom elevator knowledge is assumed here. Let’s start by drawing a picture to allow us to make sure that we understand the problem.
Pic 1 is a simple picture of an elevator and some floors it can stop on.
Let’s start by talking about the characteristics of the problem, which we will have to optimize:
• There are a certain number of floors on which the elevator can stop
• It takes a certain amount of time for the elevator doors to open and close
• It takes time for the elevator to accelerate, stop and has a max speed
• People can signal the elevator and indicate whether they are going up or down
• The elevator has a maximum capacity
Now, lets’ talk about some of the basic elements of an algorithm for an elevator:
• Elevators should minimize waiting times for people who are signaling the elevator
• Once an elevator starts going up and someone indicates a higher floor on it, the evator should not go down
• Same as 2, but for down
• Elevators should minimize transit times for people in the elevator
Well, already, this is looking much more complicated. We have multiple characteristics and rules. Let’s now put this together, and start talking about what we know about elevators:
• When someone is in the elevator, it is going up or down.
• It will go up and stop at every floor where either: 1) someone in the elevator has selected, or 2) someone outside the elevator has pressed the up button
• It will go down and stop at every floor where: 1) someone in the elevator has selected, or 2) someone outside the elevator has pressed the button
• It will always stop at the bottom floor when going down if someone has pressed the up button
• It will always stop at the top floor when going up if someone has pressed the down button
These are the basics. Now, let’s try an example. Assume that someone walks into a building and presses the up button on the bottom floor – what happens?
Well, assuming no one is in the elevator, it will come down from the floor it is on. So, the person will have to wait while the elevator comes down.
This example highlights another issue. Where does the elevator wait when it is not in use?
One first thought would be to wait in the middle. This would seem to optimize between people going up and people going down. We can probably assume too that there are relatively few trips between floors – though even this may not be true if for example a company has offices on more than one floor.
But, let’s keep it simple. But, think about someone coming in to go up. When do people do this? And, when do people go down primarily?
This would depend a little on the type of building. In an office building, people will often come in in the morning and leave in the evening. In an apartment building, it would be the opposite.
One optimization could be to move the elevator closer to where people are expected at different times of day. This could reduce wait times dramatically.
We have most of what we need to write out an algorithm. Let’s do it now:
First, let’s bring in all the rules that are expected:
1 When someone is in the elevator, it is going up or down.
a. It will go up and stop at every floor where either: 1) someone in the elevator has selected, or 2) someone outside the elevator has pressed the up button
b. It will go down and stop at every floor where: 1) someone in the elevator has selected, or 2) someone outside the elevator has pressed the button
2 It will always stop at the bottom floor when going down if someone has pressed the up button
3 It will always stop at the top floor when going up if someone has pressed the down button
And, let’s add, assuming we have an office building:
• The elevator will wait at the bottom floor when not in use between 7am-10am.
• The elevator will wait at the top floor when not in use between 3pm – 7pm.
• At other times, the elevator will wait in the middle floor
Well, we have a good algorithm here.
Most important in this problem, is to 1) not ignore the obvious such as the rules of how an elevator works and what it is trying to optimize, and 2) always seek to ‘stretch’ your answer. In this case, we stretched with an optimization problem.
/* Please let me know if you find bugs or have alternate solutions. Thx, Noah */
## Tuesday, February 21, 2012
### Implement division without using the division or multiplication operation
Problem: Implement division without using the division or multiplication operation. Round to the nearest whole number
int divide(int numerator, int denominator) {
...
}
Numerator= 108
Denominator = 10
We know that the answer to this is 10.8, which rounds up to 11. But, let’s think about how we got that. We know that 10 can be multiplied by 10 to get 100. This is equivalent to adding 10, ten times.
Similarly, we can subtract 10, ten times from 100 to get zero.
This starts to intuit an algorithm, where we see how many times we can subtract and that is most of the answer.
But, how do we deal with what is left over? In this case it is 8. We’re supposed to round up or down, so we can compare 8 to 10. We’re not allowed to divide, but we are allowed to add an compare. So, we can double 8 by adding it to itself. If the result is greater than 10, we round up.
Putting this into code, we get:
for (i=0, answer=0, curValue=numerator; i < numerator; i+=denominator) {
curValue -= denominator;
}
// round up if necessary
if ( curValue + curvalue <= denominator) answer++;
}
This is almost right, but it has at least problem. Can you see it? If not, try a negative number example.
What do we do if we get a negative number? In this case, our algorithm fails as one number will go up or down indefinitely.
To solve this, we need to use some if statements to make sure that we are always iterating in the right direction. We also can’t divide by zero and need to check that case too. We can solve this with three if statements a shown:
int divide(int numerator, int denominator) {
//check for 0 denominator
if (denominator == 0) return ERROR_VALUE;
for (i=0, answer=0, curValue=numerator; (numerator > 0 && i < numerator) || (numerator < 0 && i > numerator);) {
//if both numerator and denominator are positive or negative:
if (numberator >= 0 && denominator > 0) {
curValue -= denominator;
i+=denominator;
} else {
//one is negative
curValue+=denominator;
i-= denominator;
}
}
// round up if needed
if (curValue + curvalue <= denominator) answer++;
}
This is an example of a problem which starts out simple, and gets much more complicated. It’s good to do some examples now with varying positive and negative numerators and where you round down and up. There’s actually still a bug in this solution, but I’ll leave it to you to discover it.
/*Please let me know if you know alternate solutions or find bugs other than the one left in there. Thank you. Noah */
## Monday, February 20, 2012
### Number of Drops (Updated/Correct)
This is the updated/correct solution. Thank you to several community members for pointing this out.
Problem: You are given two identical cell phones and told to test them to see what is the highest floor of a hundred story building that they can be dropped from before they break. What is the strategy for the minimum number of drops required and what is that number?
Solution:
Let's start by asking some clarifying questions to ensure that we don't go off down the wrong path. - First, are all the floors the same height? Yes
- Do we have to account for that the cell phone may land differently, in a different spot with a different hardness, or in different atmospheric conditions (like wind)? No. Assume that the only variable that will influence whether a cell phone breaks or not is the floor it is dropped from
- Can we retest a broken cell phone for any additional information? No, once a cell phone is broken it can no longer be used usefully
These are good questions to ask. They show your interviewer that you're thoughtful, do not necessarily assume conditions not stated and do not want to try to solve a problem without fully understanding it.
Now that we have a better view of the problem, let's try to think of a potential solution to this problem.
Well, we could start at the bottom floor and go up one floor and drop the phone. When the phone broke, we would know that we were at the top. In the worst case, this would take up to 100 floors and 100 drops. O(n).
It's good to get a working solution, but this is clearly non-ideal. For starters, we didn't use our second phone.
A simple optimization that we could imagine when starting this is to skip by 2s: test on floor 2, 4, 6, 8,...
If the phone breaks, we can then test the phone on one floor lower. Then, the answer is one lower, else if it survives, the answer is that floor.
This results in a max number of drops of potentially 51 if the top floor is the answer. Still O(n), but much better.
At this point, think more about the optimization. We skipped floors, and then went back and tested some we had skipped once we found a range of a not-broken floor and a broken floor.
Previously, we only skipped one floor, but we could skip more. Let's try for an example, skipping 3 floors: We start by testing on floors: 3, 6, 9, 12...
Well, the worst case here is floor 90, which would be 34 drops potentially. So, it seems that skipping more floors, then going back and testing one-by-one from the last good floor is a good strategy.
It appears that skipping more floors reduces the number of drops, so let's try skipping 50 floors. Here, the worst case is now 51 floors (floor 99). So, clearly, skipping too many is not always good.
So, there seems to be an optimal number of floors to skip. If you work from here using math or simply trying bigger and smaller numbers, you'll get to an answer of about 10 (11 and 9 yield similar answers).
This gives a worst case of 19, which would be if the 99th floor were the fail floor.
It's tempting to stop here, and frankly, I have egg on my face for previously publishing an answer where I did.
There is a better solution.
Let's try to figure it out by doing some examples where we establish the min/max range by skipping 10 floors each time. Try 9.
We would test:
10: fail
1: pass
2: pass
...
8: pass
9: fail.
This would be a total of 10 tests.
Essentially, if we skip 10 floors each time, the worst case is always the number ending in 9, and the worst case floors take one more test each time:
9: 10 tests
19: 11 tests
29: 12 tests
...
99: 19 tests
Well, think about one of the assumptions that we had earlier -- always skip the same number of floors to establish a min/max range. There is no proof that this is the right solution. In fact, it looks like by doing this, we always have essentially one more test as we go higher. However, what if we used the fact that we essentially have 'tests' we can do early in a bigger range as we have not used them up to establish the range. In fact, on each subsequent upping of the potential min/max range, we lose 1 test, to establish the range.
Now, what if we accepted a larger min/max initially and then narrowed it as we went up. This would get around the problem of always having to get worse as we go higher. This hints at an optimization to our initial strategy.
As an example, let's try skipping 17 floors and then reducing the range by one each time (one better than our current best result). Note we should try 17, not 18, so that we are never worst than 18. This way, the highest range of the min/max will always yield 18, but that is better than our current best result
We could establish min/max ranges of:
1...17
18...33
34..48
....
Well, this works and we can get better than 19.
Now, keep trying smaller min/max ranges to start out with initially. If you go too small, you can't get to the top.
If you keep trying, or use math, you'll eventually narrow in on skipping 13 floors initially, which gives you a worst case of 14. This is quite a bit better than our initial case of 19.
/* Thank you to the community for help getting this right. Apologies for my previous incorrect answer. Noah */
### Largest value in an unordered, unablanced tree
Problem: Write a function with determines the largest value in unordered, unbalanced binary tree.
int largestValue(struct node * tree) {
...
}
Solution
It’s worth mentioning that if the tree were ordered, this would be trivial – going to the far right. However, we don’t have it so easy. So, let’s start by drawing out an example to make sure that we understand the problem. Pic 1 is a good example of a binary tree which is unbalanced binary tree.
Looking at pic 1, see if you can find a way to intuit the largest element. Nothing really helps here as the longest element is simply the element which is bigger than all ones.
With the only insight into finding the biggest value that we need to find the element larger than all others, let’s look at how we could do this.
First, it’s tempting to do something that we know how to do. We could do a breadth first (or depth first) search – put all the elements in an array and then very quickly iterate through the array looking for the largest element.
This solution would work, and it would be O(n), as each element is inspected once, but it’s inelegant as it requires building a new data structure. So, let’s only come back to this if we need to.
Let’s look at the tree and look at a sub tree with only two nodes as in pic 2. Essentially, we would return the largest of the value of the node, its left hand descendant and the right hand descendant. We can use this build a recursive case (as trees usually do) and return the largest of: 1) the current value, 2) the largest value of the left hand sub-tree, or 3) the largest value of the right hand subtree.
The base case here is when we have a node with no descendants, as a null node has no value.
Using these insights, we can get the code as:
int largestValue(struct node * tree) {
int rh, lh;
lh = rh = MAX_NEGATIVE;
// Check for null in case our node is empty
if (tree == null) return MAX_NEGATIVE; // or error value
//Recursive base case -- no descendant nodes
if (tree->left == null && tree->right == null)
return tree->value;
//else, return the largest of the lh, rh or value
if (tree->left != null)
lh = largestValue(tree->left);
if (tree->right !=null)
rh = largestValue(tree->right);
//return rh if it's the biggest
if (rh >= lh && rh >= tree->value)
return rh;
//return lh if it's the biggest
if (lh >= rh && lh >= tree->value)
return lh;
//else, the current node's value is the biggest
return tree->value;
}
Let’s now test this function on our sample tree in pic 1. Yup, it works. It's worth going through the exercise, but you can see that it returns 30, as desired.
Algorithmically, this function is O(n) as we examine each node once, with recursive overhead. O(n) is probably the fastest that we can get as we need to look at each code. There are certainly optimizations that we could do so we wouldn’t have to use MAX_NEGATIVE, which would be more if statements to make sure that we don’t de-reference a null pointer, but this doesn’t seem hard and we can certainly do the work as we have the structure of the algorithm.
This is a pretty good tree problem and illustrates why it’s always good to draw a picture, think of a simple sub-tree and try to intuit a problem, especially with trees.
/* Please let me know if you have alternative solutions or find bugs. Thx, Noah */
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# Basic principles by southraze
VIEWS: 33 PAGES: 26
• pg 1
``` 224 C1 Basic principles of seismology
C1.1 General introduction to wave phenomena
A wave can be defined as a periodic disturbance that transmits energy through a medium,
without the permanent displacement of the medium. Also required that energy is converted
back and forward between two different types.
Before considering the details, it is useful to review some basic definitions. Consider the two
waves shown in the MATLAB movie waves.m
Frequency (f) : The number of cycles a given point moves through in 1 second. Frequency is
measured in Hertz (Hz). If the frequency is very low, then it is common to refer to the period (T)
of the signal in seconds. T = 1/f
Angular frequency (ω): Frequency is the number of rotations per second. The angular frequency
is the number of radians per second and given by ω=2πf
Wavelength (λ) : Distance in metres between two points of the wave having the same phase (e.g.
two crests or two troughs).
If the waves moves at a velocity, v, then v = f λ
● Can you check that this relationship is correct for the figure above?
● Note that points on the wave move up and down, they are not translated to the right.
● In this case, the velocity is independent of frequency. This type of wave behaviour is called
non-dispersive. If velocity varies with frequency, the wave is said to be dispersive
1
In seismology, we need to understand how waves will travel in the Earth. For example, how fast
will they go, which direction, how will amplitude vary with distance etc. In general this requires
the solution of some complicated differential equations.
In Geophysics 224 we will approach this subject through visualization. Wave propagation can be
considered in two ways, by considering either wavefronts or rays. These are complementary
Rays denote the direction in which the
wave travels.
Wavefronts are points on the wave with
the same phase (e.g. a line along the crest of
a wave is a wavefront).
Note that wavefronts and rays are at right
angles to each other.
One way to visualize wave propagation over time is through Huyghens Principle. This states
that all points on a wavefront can be considered secondary sources of wavelets. These secondary
wavelets propagate outwards and at a time later, the overall wavefront is the envelope of
secondary wavelets. Examples for a point source is shown below.
2
C1.2 Wave propagation in the Earth
Having considered some general aspects of wave propagation we now need to consider how
waves will propagate in Earth. Waves in the Earth can be divided into two main categories:
(a) Body waves travel through the bulk medium.
(b) Surface waves are confined to interfaces, primarily the Earth-Air interface.
C1.2.1 Body waves
Body waves in the Earth can be divided into two types:
P-waves : Particle motion is in the same direction as the wave propagation. They are
also called compressional or longitudinal waves
and
S-waves: Particle motion is at right angles to the wave propagation. Also called shear
waves or transverse waves)
The velocity of body waves can be calculated from the properties of the material, as outlined
below. More details are given by Kearey pages 22-23, 3rd Edition. The mathematics is simplest
for the geometry shown below for a column of rock with cross-sectional area (A).
3
If a force (F) is applied at the left end, this can be considered an applied longitudinal stress =
F/A. The rock deforms elastically in response to the applied stress. The deformation of the
ΔL
leftmost disk can be quantified as the longitudinal strain =
L
The behaviour of the medium is described by the longitudinal modulus,
ψ = longitudinal stress / longitudinal strain
which is a measure the stiffness of the rock (i.e. how much stress is needed to produce a given
strain).
This strain produces a force that will cause the shaded section of the rock to accelerate to the
right. This lowers the stress to the left, but increases it to the right. This causes the next section of
the rock to move and so on. Can show that a wave motion will move down the column at a
velocity
ψ
v=
ρ
where ρ is the density of the material. Note that the stiffer the medium (larger ψ) the greater the
force on the shaded cylinder, thus acceleration is higher and wave velocity is greater. Similarly,
as density increases, the shaded section becomes heavier and it’s acceleration (and wave
velocity) for a given force will decrease.
In general, the calculation of velocity is more complicated as the deformation will involve both
compression and shearing. Two other modulii must be defined to fully understand these effects.
Bulk modulus, K = volume stress, P
volume strain, ΔV/V
and
Shear modulus, μ = shear stress, τ
shear strain, tan θ
4
4
It can be shown, that these modulii are related as ψ = K + μ
3
Thus the P-wave velocity can be written as
1
⎡K + 4 μ ⎤2
vP = ⎢ 3 ⎥
⎢ ρ ⎥
⎣ ⎦
and the S-wave velocity as
1
⎡μ ⎤2
vS = ⎢ ⎥
⎣ρ⎦
● Note that P-waves always travel faster than S-waves
● In a liquid μ=0 while K is always non-zero. Thus only P-waves can travel in a liquid, since
shear stresses cannot exist.
● These expressions for vP and vS do not depend on frequency, thus body waves (both P-waves
and S-waves) are non-dispersive.
● Note that as the rock cylinder is stretched, it will get longer and thinner. This effect can be
quantified through Poissons ratio. This is defined as:
σ= lateral strain
longitudinal strain
● Note that individual values of vP and vS depend on several modulii and density. This can make
it difficult to compare the velocities of similar rocks.
It can be shown that vp/vs = [2(1-σ)/(1-2σ)]½
For typical consolidated crustal rocks, σ ~ 0.25 and vp/vs ~ 1.7. An increase in vp/vs and/or
Poisson’s ratio can be indicative of the presence of fluids.
5
C1.2.2 Surface waves
These waves are localized at the Earth’s surface and can be divided into two types. The particle
motions are illustrated in Kearey Figure 3.4
Rayleigh Waves are like ripples on a lake. Particle motion is in a retrograde ellipse.
Love waves have a horizontal particle motion analogous to S-waves.
Note that:
●The velocity of a Rayleigh wave does not vary with frequency when travelling in a uniform
medium and it is slower than an S-wave. In a layered Earth the velocity of a Rayleigh wave
varies with frequency (it is dispersive) and can be used to infer velocity variation with depth.
●Love waves only exist if the Earth is layered and are always dispersive.
●Large earthquakes can generate surface waves that travel around the globe. They can be large in
amplitude and cause a lot of damage during earthquakes.
●In exploration seismology, ground roll is a Rayleigh wave that travels across the geophone
array.
6
C1.3 Typical seismic velocities for Earth materials
Typical values for P-wave velocities in km s-1 include:
Sand (dry) 0.2-1.0
Wet sand 1.5-2.0
Clay 1.0-2.5
Tertiary sandstone 2.0-2.5
Cambrian quartzite 5.5-6.0
Cretaceous chalk 2.0-2.5
Carboniferous limestone 5.0-5.5
Salt 4.5-5.0
Granite 5.5-6.0
Gabbro 6.5-7.0
Ultramafics 7.5-8.5
Air 0.3
Water 1.4-1.5
Ice 3.4
Petroleum 1.3-1.4
● why does vp apparently increase with density? e.g. for the sequence granite-ultramafics.
1
⎡K + 4 μ ⎤2
The equation v P = ⎢ 3 ⎥ suggests that vp should decrease as density increases.
⎢ ρ ⎥
⎣ ⎦
P-wave velocity of porous rocks
A rock sample comprises a matrix of rock
grains with pores in between. The fractional
volume occupied by the pore space is
termed the porosity (Φ). Pores are generally
filled with a fluid such as air, water, oil or
gas. If the matrix has a density ρm and the
pore fluid has a density ρf, then the overall
density of the rock will be
ρ = ρf Φ + (1- Φ) ρm
Thus if the velocities in the matrix and fluid
are vm and vf it is quite simple to show that
1 Φ (1 − Φ)
= +
v vf vm
This equation is sometime called the time-
average equation. An example showing
how porosity controls vP is in Kearey
Figure 3.6. Note that the pore fluid is water,
as evidenced by the density at 100%
porosity.
7
Example 1 : A rock has 30% porosity and the P-wave velocities in the rock matrix and pore fluid
are 1.5 and 2.9 km s-1 respectively. What is the overall seismic velocity?
Example 2 : Consider the gas reservoir shown below. The reservoir sand has 10% porosity. The
Material vP (km s-1)
Sandstone 4.27
Water 1.50
Gas 0.3
Shale 2.4
What is the average P-wave velocity in the water-saturated reservoir? What about in the gas
saturated reservoir?
Example 3 : In a similar reservoir, the P-wave velocity in the gas filled horizon was measured to
be 2.2 km s-1. What porosity does this imply?
Variation of seismic velocity with depth
With increasing depth, compaction increase the density of a rock through reduction of pore
space. The rigidity of the rock also increases with depth. The net effect is that velocity will
increase with depth, even if the lithology does not change.
The increase in velocity with burial depth is quite well defined, and deviations from the expected
velocity at a given depth are sometimes used to infer uplift histories in basins. This technique is
based on the observation that the compaction is essentially an irreversible process (Telford
Figure 4.21)
8
Rippability
In geotechnical studies, a knowledge of P-wave velocity can be used to determine if a
particular rock type can be ripped with a given tractor. This also gives a useful
compilation of the range of velocities for a given material. (Kearey Figure 5.25)
As a seismic wave travels through the Earth, several factors will change the direction and
amplitude of the waves. When detected at the surface, an understanding of these factors can tell
In sections 1.4 – 1.7 we will examine a number of these factors.
9
C1.4 Reflection coefficients at normal incidence
Consider a seismic wave that is travelling vertically downwards, as shown in the figure below. If
the wave reaches an interface, some of the energy will be transmitted, and some will be reflected.
Note that the waves travels from medium 1 into medium 2.
For each medium, the impedance is defined as the product of density (ρ) and seismic velocity (v),
Z = ρv
The reflection coefficient, R, is defined as
Ar v ρ −v ρ Z − Z1
=R= 2 2 1 1 = 2
Ai v 2 ρ 2 + v1 ρ1 Z 2 + Z 1
and the transmission coefficient, T, is defined as
At 2v1 ρ1 2Z1
=T = =
Ai v 2 ρ 2 + v1 ρ1 Z 2 + Z1
The reflection coefficient is a measure of the change in impedance across the interface. These
equations are called the Zoeppritz equations. If the wave is incident at an angle the become more
complicated.
Sometimes reflection co-efficients are expressed in terms of energy, not amplitude. The equations
for this situation are listed in Kearey, chapter 3.
10
Example 1
The MATLAB script reflect_v1.m generates a movie showing how a seismic pulse propagates in
the Earth. For simplicity, density is constant and only velocity varies with depth. In this example
there is an increase in seismic velocity (and impedance) at 500 m. Note that the reflection has the
same polarity as the down going pulse.
What are the values of R and T? Is the largest amplitude in the reflected or transmitted wave?
Example 2
Same geometry as Example 1, but a decrease in velocity (impedance) occurs at 500 m depth.
Note that the reflected pulse has the opposite (negative) polarity to the down going pulse. Is this
what the Zoeppritz equations predict?
What are the values of R and T? Is the largest amplitude in the reflected or transmitted wave?
11
Example 3
This has two interfaces, each associated with an increase in velocity. Both arrivals have a
positive polarity, showing that they originate at interfaces where the velocity (impedance)
increases.
What are the reflection co-efficients at each interface?
What other factor makes the amplitude of the second reflection smaller than the amplitude of the
first?
12
Example 4
This has a low velocity layer (LVZ). The first reflections has negative polarity show a decrease
in velocity, while the second has positive polarity, showing an increase in seismic velocity. Note
that a pulse reverberates within the low velocity layer producing multiple later arrivals, with
ever decreasing amplitude.
Example 5
Consider a gas reservoir with 10% porosity where the rock matrix has a vp = 3000 ms-1. The
overlying rock has vp = 2800 ms-1.
●What is velocity in the reservoir? (use time average equation)
●What value of R is expected for a reflection from the top of the reservoir?
●What value of R is expected where the 2800 ms-1 and 3000 ms-1 layers are in direct contact?
This is an example of a bright spot, which is a high amplitude reflection. In the Earth, typical
reflection coefficients are ±0.2 with the maximum values ±0.5
13
C1.5 Reflection and refraction at non-normal incidence
In a more general case, the seismic wave will be incident on an interface at some angle of
incidence, θi. Note that the angle of incidence is measured from the normal to the ray. Snell’s
Law was developed for optics, but can equally be applied to the seismic case. Consider a P-wave
that strikes the interface shown below. In this case v2 > v1
Reflected and refracted P-waves are generated from the incident P-wave. For the reflected P-
wave, θr = θi . Snell’s Law states that the ray parameter, p, for the incident and refracted waves
will be constant.
sin θ i sin θ t sin θ r
p= = =
v1 v2 v1
Thus with a velocity increase (v2 > v1) the ray is refracted away from the normal.
If the velocity decreases (v2 < v1) then the ray is refracted towards the normal.
Note that if v2 > v1 then there will be a value of θi which results in sin θt = 1. This gives a value
of θt = 90˚ and the refracted waves travels horizontally. In this configuration θi = θc and is called
the critical angle.
sin 90 o sin θ c ⎛ v1 ⎞
= and by rearranging we find that θ c = sin −1 ⎜
⎜ ⎟
v2 v1 v2 ⎟
⎝ ⎠
The wave travelling horizontally is called a head wave. For a head wave to develop, we must
have v2 > v1. Using Huyghen’s Principle, it can be shown that the head wave will generate
upward propagating wave at an angle θc to the normal. When these waves reach the geophones
they are called refracted arrivals. When θi > θc the wave is totally reflected.
14
In general we must also consider that the incident P-wave will generate a reflected S-wave and a
refracted S-wave through a process called mode conversion. Again Snell’s Law can be used to
calculate the angles of reflection and refraction.
sin θ i sin θ tP sin θ tS
p= = P
= S
v1 v2 v2
The amplitudes of the transmitted and reflected waves can be calculated from Zoeppritz’s
equations, which have a more complicated form than those listed in C1.4 (see Applied
Geophysics, Telford, 2nd Edition, page 157 for details and some examples). The
amplitude of the reflected wave varies with angle, and is sensitive to the change in
impedance across the boundary. This is the physical basis of the amplitude versus offset
(AVO) technique that we will discuss later on.
C1.6 Diffraction
Seismic energy can sometimes travel in regions where ray theory (Snell’s Law) does not predict
that it will go. This is typified by diffraction which occurs when a wave strikes an object that is
significantly smaller than a wavelength. A diffractor effectively radiates seismic energy in all
directions.
When a wave strikes a corner, Huyghens Principle shows that the corner will generate waves that
propagate in all directions. We will see in real seismic reflection data that these waves can have
significant amplitudes and are detected over a wide area at the surface. The mathematical process
of migration is used to trace the pattern of diffracted waves back the point source where they
originated. This allows subsurface geometry to be correctly imaged in 2-D and 3-D.
15
C1.7 Factors that cause the amplitude to change as wave propagates
Imagine a wave travelling outward from a point source. If the wave has travelled a distance r ,
then the wavefront covers an area A =4πr2.At this point the wave has an amplitude of X. The
energy is a wave is proportional to X2 . Thus
Total energy = X2 4πr2
Conservation of energy requires that this quantity remain constant as r increases. Thus
1 1
X2 α and Xα
r2 r
As a seismic wave travels in the Earth the amplitude will decrease as 1/r, even if no energy is
lost. This phenomena is known as spherical divergence or geometric spreading.
(b) Attenuation
As a wave passes through the Earth, the particle motion causes the material to be distorted and
the wave energy is converted in heat. This results in an additional loss of energy, that is described
by an exponential decay:
X = Xo e-kr
16
Where e = 2.718, Xo is the amplitude at r=0 and k is a constant. If k is small, the attenuation will
be small, as k increases, the attenuation becomes stronger. In a distance 1/k the amplitude falls
1
from Xo to Xo .
e
Another common definition is the absorption coefficient, α, expressed in decibels per
wavelength. This is based on the observation that the energy lost is dependent on the number of
oscillations per second produced by the wave. Thus high frequencies will attenuation more
quickly than low frequencies. This is illustrated in the MATLAB script waves_attenuation.m.
A consequence of frequency-dependent attenuation is that the shape of a pulse can change as it
propagates through the Earth. This occurs for the following reasons:
● The pulse can be decomposed into a set of sine or cosine curves with different wavelengths
(Fourier’s Theorem and illustrated in MATLAB script fourier_square_wave.m)
● As the pulse travels the short wavelength signals attenuate more quickly.
● The long wavelengths dominate, giving the pulse a smoother shape and longer duration
(example in Kearey, Figure 3.7)
(c) Scattering
Suppose a medium is inhomogeneous and contains some grains with a different seismic velocity
to the host rock. Seismic waves will be diffracted / scatteredfrom these grains and energy will be
lost from the coherent wavefronts and turned into random seismic energy. The net result is that
energy will be lost.
Footnote : Decibels
A seismic wave changes in amplitude from A1 to A2 as it travels from point 1 to point 2. The
corresponding intensity changes from I1 to I2. Note that I1 = A12. This change in decibels can be
expressed as :
⎛I ⎞ ⎛A ⎞
dB = 10 log10 ⎜ 2 ⎟ = 20 log10 ⎜ 2 ⎟
⎜I ⎟ ⎜A ⎟
⎝ 1⎠ ⎝ 1⎠
17
C1.8 Seismic energy sources
Basic requirements for an efficient seismic source include:
●Generate sufficient energy in an appropriate frequency band
●Economical
●Non-destructive
●Repeatability
1.8.1 Land seismic sources
Near surface exploration
● A high frequency seismic source is needed to generate short wavelength signals (the
wavelength should be less than thickness of layers to be studied). However, if the frequency is
too high, then attenuation will become a significant problem.
● A hammer striking a metal plate makes an
portable, repeatable source. A sledge
hammer gives good energy, but relatively
low frequency. A smaller hammer will
give high frequency signal for shallow
imaging. By stacking a number of hammer
blows, the signal-to-noise ratio can be
increased.
● Firearms give higher frequencies, but are not always permitted in civilized countries (e.g.
shotgun, semi-automatic etc.)
● Buffalo guns and Betsy guns allow a blank cartridge to be fired without using ammunition.
● Weight drops are also used and are commonly mounted on the back of a small truck. Larger
weight drops have included the use of helicopters.
Explosives
Explosives were once widely used, but this has declined with the development of the Vibroseis
method. Typically rotary drilling is used to place the shot in an 8-10 cm diameter hole at a depth
of 6-30 m. Drills can be mounted on a truck, or slung beneath a large helicopter. While
spectacular, blasting rocks and soil into the air represents a loss of energy and a safety hazard.
Sharp impulsive signal. Strong signals can be generated if needed for imaging the lower crust and
upper mantle.
18
Not repeatable, and quite slow. Most energy is in P-waves (why?).
Many environmental concerns and cannot be used in urban areas.
Drilling shot holes in Tibet, 1994 1000 kg shot
Vibroseis
Lithoprobe SNORCLE transect in B.C. Truck generating signal
For deeper exploration on land, the Vibroseis method is now the most widely used technique.
This does not generate a sharp impulsive signal, but generates a longer waveform that typically
lasts 10-20 seconds with frequencies 10-100 Hz.
Data processing requires that the field recording is analysed for correlation with the known
source signal. This can allow signal detection in areas with significant seismic noise (e.g. vehicle
traffic, wind etc)
Vibroseis trucks can generate forces up to 100,000 N and generally need a hard surface upon
which to operate. It is debatable whether they cause no damage to a road. For extra signal
strength, multiple trucks are used and they vibrate in phase. Both P-waves and S-waves can be
generated.
19
Example 0 : Two separate reflections
Example 1 : Two overlapping reflections
20
Example 2 : Two overlapping reflections and moderate noise levels
Example 3 : Two overlapping reflections and severe noise
21
Stacking is used to increase the signal-to-noise ratio. The inherent assumption of stacking is that
the signal is the same each time (coherent) and the noise is different (incoherent). Coherent
noise can not be removed by stacking.
22
1.8.2 Marine seismic sources
Air guns
An airgun works by releasing a bubble of high pressure air into the water. The rapid expansion of
the bubble generates seismic energy with a frequency content around 10-100 Hz.
Airgun array on Veritas ship
Operation of Bolt airgun. From Telford ch. 4.
The chamber is filled with air at 10-15 MPa
which is then released into the water.
Airgun array and streamers being towed
behind a survey vessel.
● Very repeatable, reliable source.
●The bubble pulse oscillates, generating a relatively long wave train. However, by using an
array of air guns at differing depths, the combined waveform can be made shorter in duration
(figure from Kearey).
● In water only P-waves are generated, but S-waves can be generated by mode conversion at
the seafloor.
● In recent years there has been a lot of concern about how marines seismic exploration affects
marine mammals.
23
Explosives
Once widely used in marine seismics, but now largely replaced by air guns. Bubble pulse problem
is difficult to overcome (Kearey figure 3.15), but can be minimized by detonating near the surface
(with some loss of energy). Occasionally large shots are used in deep studies but many restrictions
are in place to protect oil production facilities and fisheries.
Higher frequency marine sources
A range of higher frequency sources are used in marine exploration (100-10000 Hz). These
include pingers, sparkers and boomers. The high frequency gives a sharp image of shallow
sediment layers, but higher attenuation limits the depth of penetration. These methods are widely
used in offshore geotechnical surveying (e.g. choosing locations for seafloor pipes and cables, in
harbor construction).
1.8.3 Frequency content of seismic sources
Earthquake surface waves 0.1-0.01 Hz
Earthquake body waves 10-0.1 Hz
Vibroseis 10-100 Hz
Air guns 10-100 Hz
Explosives 10-300 Hz
24
C1.9 Seismic detectors
C 1.9.1 Onshore seismic exploration
Geophones
On land, the surface moves as a P-wave or S-wave arrives. Generally reflected signals arrive at
steep angles of incidence. Thus P-waves produce surface motion that is dominantly vertical.
Geophones measure ground motion by converting motion into electrical signals. Most geophones
measure a single component (vertical), but multiple component ones are sometimes used.
The most common design is the moving-coil
geophone. Ensuring good mechanical coupling
with the ground is essential, usually through a
metal spike. Sometimes a cluster of geophones
is planted and the output summed to improve
signal to noise ratio. For shallow studies,
geophone spacing can be as little as 1 metre. In
deeper studies a spacing of 10-50 metres is more
usual.
Geophones are manufactured to detect a particular frequency band. This should match the seismic
source being used in a particular survey.
Geophones are connected to telemetry cables that transmit the recorded signals back to a
recording unit.
Seismometers
For earthquake studies a more permanent installation is usually required. Three components are
usually recorded and the sensor is tuned to detect lower frequencies. Often the seismometer is
placed in a shallow vault to minimize wind and other forms of noise.
25
1.9.2 Offshore seismic exploration
Hydrophones and streamers
S-waves cannot travel in water, so only P-waves can be detected in the water. In marine
exploration seismic waves are detected by the change in pressure as a P-wave passes the detector.
This type of sensor is called a hydrophone and it converts a change in water pressure into an
electric signal through the piezoelectric effect.
In offshore exploration, many hydrophones are placed in neutrally buoyant, oil-filled streamers
that can be in excess of 6 km long. The depth is controlled by movable fins and a tail buoy is used
to determine the direction of the streamer relative to the ship.
Feathering occurs when ocean currents push the streamer at an angle to the direction in which the
survey ship is travelling. Modern ships can pull multiple streamers, and this overcomes feathering
and also provides 3-D coverage.
Ocean-bottom seismometers (OBS’s)
These are like land seismometers, but are dropped to the seafloor from the sea surface. Coupling
with the seafloor allows 3 components of motion to be recorded (i.e. P-waves and S-waves can be
detected). Generally slow to use, and most widely used in research surveys. Some applications in
hydrocarbon exploration.
Dalhousie University OBS on deck Scripps Institution of Oceanography OBS
Geophysics 224 MJU 2006
26
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# IQ Contest, Daily, Weekly & Monthly IQ & Mathematics Competitions
#### Question No 1
A car travels first half distance between two places with a speed of 40 kilometre per hour and rest of half distance with a speed of 60 km per hour. The average speed of car is:
Solution!
Let the total distance covered be"s" km.
Total time taken,
=(s/2)/30+(s/2)/60
=3s/120
Average speed,
=s/(3s/120)
=s(120/3s)
=40 km/hr
.
#### Question No 2
John is four times as old as Peter. If the sum of their ages is 60 years, how old is Peter?
Solution!
No explanation available for this question..
#### Question No 3
Three persons are walking from a place A to another place B their speeds are in ratio of 4 :3 :5. The time ratio to reach B by these person will be:
Solution!
Ratio of speed= 4:3:5
Therefore ratios of time taken= (1/4):(1/3):(1/5)=15:20:12
.
#### Question No 4
Find out the missing number in the following sequence:
1, 3, 3, 6, 7, 9, ?, 12, 21.
Solution!
Clearly, the given sequence is a combination of two series:
I. 1, 3, 7, ?, 21 and II. 3, 6, 9, 12
The pattern followed in I is + 2, + 4, ...; and
the pattern followed in IIis + 3.
Thus, missing number = 7 + 6 = 13.
#### Question No 5
Solve this?
3 2 ?
5 1 3
1 6 2
Solution!
All rows and columns are equal to 9.
So missing number is 4.
.
#### Question No 6
The sum of five consecutive numbers is 100. What are the smallest of these numbers?
Solution!
Let,s numbers are:
x, x+1, x+2, x+3, x+4
then
x+x+1+x+2+x+3+x+4=100
5x +10 = 100
5x=100-10
X=90/5
X=18
.
#### Question No 7
In farm there is 2 ducks,6 cats ,4 dogs . Each weight of duck 4 kg, cat 4 kg and dog 7 kg what is the average weight of all animals?
Solution!
={( number of ducks × weight)+( number of dogs x weight)}/ total animals
= 5 kilogram is average weight of all animals.
.
#### Question No 8
In following questions, one term in number series is incorrect.
Find out the incorrect number 16, 22, 30, 45, 52, 66
Solution!
The correct sequence is + 6, + 8, + 10, + 12, +14.
So, 45 is wrong and must be replaced by (30 + 10) i.e. 40..
#### Question No 9
7, 11, ___, 17, 19, 23
Find missing number?
Solution!
All are prime numbers..
#### Question No 10
In following alphabet series , one term missing as shown by question mark . Choose missing term from options.
A, Z, X, B, V, T, C, R, ?, ?
Solution!
The first, fourth and seventh letters are in alphabetical order. So, tenth letter would be the letter after C i.e. D. Also, the second and third letters are alternate and in reverse order and so are the fifth and sixth letters and the eighth and ninth letters.
.
#### Question No 11
What comes next in the series 15 , 23 , 30 , 46 , 45 , ?
Solution!
No explanation available for this question..
#### Question No 12
susan is 3 years older than Tom. 2 years ago Susan was twice as old as Tom. Find the present age of Susan.
Solution!
let's
t = x
S= x + 3
given condition
2(x - 2) = X + 3 - 2
x - 4 = x + 1
x = 5
Tom = 5
Susan= 8
.
#### Question No 13
Idiom and phrase.
To end in smoke
Solution!
Description is not available..
#### Question No 14
Find the odd one out
Solution!
No explanation available for this question..
#### Question No 15
If a boat travels with a speed of 8 km per hour in still water and the speed of stream is 3 kilometre per hour what would be the time taken by the boat to go 555 kilometre downstream?
Solution!
Downstream speed=(x+y)
8+3=11 km/hr
Time=distance/speed
55/11=5 hrs
.
#### Question No 16
Mrs Smith is 8 years more than twice the age of his son. The age of his son is 12. Find the difference between their age.
Solution!
age of son = 12
Age of mother:8 years more than twice age of son
age of mothers = 32
age difference:20
.
#### Question No 17
PBC, QDE, RFG, SHI....?
What's next?
Solution!
It is two Alphabetical series here.
First series is the first letter
only in sequence of P,Q,R,S,T,U,V...
2nd series is BC, DE, FG, HI, JK, LM...
.
#### Question No 18
Which word means the same as garbled?
Solution!
If something is garbled, it is unintentionally
distorted, jumbled or unintelligible.
.
#### Question No 19
A family consists of six members P, Q, R, S, T and U. There are two married couples. Q is a doctor and the father of T. U is grandfather of R and is a contractor. S is grandmother of T and is a housewife. There is one doctor, one contractor, one nurse, one housewife and two students in the family. Who is the sister of T?
Solution!
Q, the Doctor, is the father of T. S, the Housewife, is the grandmother of T and hence the mother of Q. Since there are only two married couples one being that of Q, the grandfather of R i.e., U must be married to S. Thus, R and T will be both children of Q and these must be the students. So, P, who remains, shall be the wife of Q and she alone can be the nurse. Thus, U must be the contractor.
Clearly, R and T are children of same parents. So, R will be the sister of T.
.
#### Question No 20
A horse is fleeing at speed of 20m/s. in 1 hour, he would cover:
Solution!
No explanation available for this question..
#### Question No 21
What comes next?
5, 17, 37, 65, ___?
Solution!
2^2+1, 4^2+1, 6^2+1, 8^2+1…. Is given series.
.
#### Question No 22
Time is to watch as pressure is to
Solution!
No explanation available for this question..
#### Question No 23
In following questions, one term in number series is incorrect.
Find out the incorrect number
1, 5, 5, 9, 7, 11, 11, 15, 12, 17
Solution!
The given sequence is a combination of two series:
I. 1, 5, 7, 11, 12 and
II. 5, 9, 11, 15, 17
The pattern in both I and II is + 4, + 2, + 4, + 2.
So, 12 is wrong and must be replaced by (11 + 2) i.e. 13..
#### Question No 24
____:nine:: pentagon:five
Solution!
A nanogo is a polygon with nine sides and a pentagon is a polygon with five sides..
#### Question No 25
Father in law of Aliya's mother's sister's daughter's husband is _____ of Aliya:
Solution!
No explanation available for this question..
#### Question No 26
In following number series, two terms have been put within brackets. Choose the appropriate option
Choose the appropriate option 3, 10, 29, (66), (127), 218
Solution!
The sequence is 13 + 2, 23 + 2, 33 + 2, 43 + 2, 53 + 2, 63 + 2.
Clearly, both the terms 66 and 127 are correct..
#### Question No 27
In alphabet series, some alphabets are missing which are given in that order as one of the alternatives below it. Choose the correct alternative.
a _ abbb _ ccccd _ ddccc _ bb _ ba
Solution!
The series is aaa / bbbb / cccc / dddd / cccc / bbbb / a..
#### Question No 28
30, 35, 55, 40, 45, 55, 50...?
Find next pair?
Solution!
It is alternative series with of difference 5 and every 3rd
difference number position number is 55 in series.
.
#### Question No 29
Miles is to distance as
Solution!
No explanation available for this question..
#### Question No 30
A + B means A is the father of B.
A -B means A is the wife of B.
A x B means A is the brother of B .
A ÷ B means A is the daughter of B.
If P + Q which of the following is true?
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## Curved Surface Area of Hemisphere given Diameter Solution
STEP 0: Pre-Calculation Summary
Formula Used
Curved Surface Area of Hemisphere = 2*pi*(Diameter of Hemisphere/2)^2
CSA = 2*pi*(D/2)^2
This formula uses 1 Constants, 2 Variables
Constants Used
pi - Archimedes' constant Value Taken As 3.14159265358979323846264338327950288
Variables Used
Curved Surface Area of Hemisphere - (Measured in Square Meter) - Curved Surface Area of Hemisphere is the quantity of plane enclosed on the curved surfaces (that is the bottom face is excluded) of the Hemisphere.
Diameter of Hemisphere - (Measured in Meter) - Diameter of Hemisphere is the distance from one point to other point on the Hemisphere which is collinear with the center of the Hemisphere.
STEP 1: Convert Input(s) to Base Unit
Diameter of Hemisphere: 10 Meter --> 10 Meter No Conversion Required
STEP 2: Evaluate Formula
Substituting Input Values in Formula
CSA = 2*pi*(D/2)^2 --> 2*pi*(10/2)^2
Evaluating ... ...
CSA = 157.07963267949
STEP 3: Convert Result to Output's Unit
157.07963267949 Square Meter --> No Conversion Required
157.07963267949 157.0796 Square Meter <-- Curved Surface Area of Hemisphere
(Calculation completed in 00.006 seconds)
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## < 6 Curved Surface Area of Hemisphere Calculators
Curved Surface Area of Hemisphere given Volume
Curved Surface Area of Hemisphere = 2*pi*((3*Volume of Hemisphere)/(2*pi))^(2/3)
Curved Surface Area of Hemisphere given Surface to Volume Ratio
Curved Surface Area of Hemisphere = 2*pi*(9/(2*Surface to Volume Ratio of Hemisphere))^2
Curved Surface Area of Hemisphere given Circumference
Curved Surface Area of Hemisphere = Circumference of Hemisphere^2/(2*pi)
Curved Surface Area of Hemisphere given Diameter
Curved Surface Area of Hemisphere = 2*pi*(Diameter of Hemisphere/2)^2
Curved Surface Area of Hemisphere
Curved Surface Area of Hemisphere = 2*pi*Radius of Hemisphere^2
Curved Surface Area of Hemisphere given Total Surface Area
Curved Surface Area of Hemisphere = 2/3*Total Surface Area of Hemisphere
## Curved Surface Area of Hemisphere given Diameter Formula
Curved Surface Area of Hemisphere = 2*pi*(Diameter of Hemisphere/2)^2
CSA = 2*pi*(D/2)^2
## What is a Hemisphere?
A sphere is defined as a set of points in three-dimension, and all the points lying on the surface is equidistant from the centre. When a plane cuts across the sphere at the centre or equal parts, it forms a Hemisphere. We can say, a Hemisphere is exactly half of a sphere.
## How to Calculate Curved Surface Area of Hemisphere given Diameter?
Curved Surface Area of Hemisphere given Diameter calculator uses Curved Surface Area of Hemisphere = 2*pi*(Diameter of Hemisphere/2)^2 to calculate the Curved Surface Area of Hemisphere, The Curved Surface Area of Hemisphere given Diameter formula is defined as the quantity of plane enclosed on the curved surfaces (that is the bottom face is excluded) of the Hemisphere, and calculated using the diameter of Hemisphere. Curved Surface Area of Hemisphere is denoted by CSA symbol.
How to calculate Curved Surface Area of Hemisphere given Diameter using this online calculator? To use this online calculator for Curved Surface Area of Hemisphere given Diameter, enter Diameter of Hemisphere (D) and hit the calculate button. Here is how the Curved Surface Area of Hemisphere given Diameter calculation can be explained with given input values -> 157.0796 = 2*pi*(10/2)^2.
### FAQ
What is Curved Surface Area of Hemisphere given Diameter?
The Curved Surface Area of Hemisphere given Diameter formula is defined as the quantity of plane enclosed on the curved surfaces (that is the bottom face is excluded) of the Hemisphere, and calculated using the diameter of Hemisphere and is represented as CSA = 2*pi*(D/2)^2 or Curved Surface Area of Hemisphere = 2*pi*(Diameter of Hemisphere/2)^2. Diameter of Hemisphere is the distance from one point to other point on the Hemisphere which is collinear with the center of the Hemisphere.
How to calculate Curved Surface Area of Hemisphere given Diameter?
The Curved Surface Area of Hemisphere given Diameter formula is defined as the quantity of plane enclosed on the curved surfaces (that is the bottom face is excluded) of the Hemisphere, and calculated using the diameter of Hemisphere is calculated using Curved Surface Area of Hemisphere = 2*pi*(Diameter of Hemisphere/2)^2. To calculate Curved Surface Area of Hemisphere given Diameter, you need Diameter of Hemisphere (D). With our tool, you need to enter the respective value for Diameter of Hemisphere and hit the calculate button. You can also select the units (if any) for Input(s) and the Output as well.
How many ways are there to calculate Curved Surface Area of Hemisphere?
In this formula, Curved Surface Area of Hemisphere uses Diameter of Hemisphere. We can use 5 other way(s) to calculate the same, which is/are as follows -
• Curved Surface Area of Hemisphere = 2*pi*Radius of Hemisphere^2
• Curved Surface Area of Hemisphere = 2/3*Total Surface Area of Hemisphere
• Curved Surface Area of Hemisphere = 2*pi*((3*Volume of Hemisphere)/(2*pi))^(2/3)
• Curved Surface Area of Hemisphere = 2*pi*(9/(2*Surface to Volume Ratio of Hemisphere))^2
• Curved Surface Area of Hemisphere = Circumference of Hemisphere^2/(2*pi)
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# EET 2261 Unit 8 Seven-Segment Displays; S19 Records; System Clocks
## Presentation on theme: "EET 2261 Unit 8 Seven-Segment Displays; S19 Records; System Clocks"— Presentation transcript:
EET 2261 Unit 8 Seven-Segment Displays; S19 Records; System Clocks
Read Almy, Appendix B and Chapter 16. Homework #8 and Lab #8 due next week. Quiz next week. Handouts: Quiz 7, Checksum practice sheet and Clock Frequencies practice sheet, and slides 8 and 11 on using two seven-segment displays.
Seven-Segment Displays
Recall that a seven-segment display has LEDs named a through g (and possibly one or two LEDs for decimal points) that can be lit up to display characters. This topic is discussed in the textbook’s Appendix B.
Common-Cathode or Common-Anode?
Every seven-segment display is either: Common-cathode: each LED has its own anode pin, and all LEDs share a single cathode pin. Or common-anode: each LED has its own cathode pin, and they all share a single anode pin. The Dragon12’s are common-cathode. So to show a 3, what logic level do we need to send to each of the pins shown above?
Current-Limiting Resistors
Recall that LEDs will burn out if you pass too much current through them. Each LED should always be connected in series with its own current-limiting resistor, typically between 220 Ω and 470 Ω. So if you wire a seven-segment display, you’ll need about eight of these current-limiting resistors: one for each segment and decimal point. (This is already done for us on the Dragon12 board.)
Banks of Seven-Segment Displays
Often we use banks of several seven-segment displays to show several digits. Example: For a clock that displays hours, minutes, and seconds, we’d need six seven-segment displays. The result: a lot of connections and a lot of current-limiting resistors (if you take the “brute-force” approach described below).
Two Approaches to Controlling a Bank of Seven-Segment Displays
Brute force approach Simple, but eats up a lot of resources (such as I/O ports on the HCS12). Multiplexing approach Harder to understand, but much more efficient in its use of limited resources (such as I/O ports). This approach is described on pages of Dragon12 manual and in the textbook’s Appendix B.
Approach #1: Brute Force
Connect a seven-segment display’s eight anode pins through current-limiting resistors to one of the HCS12 chip’s ports. Also connect the display’s common cathode pin to ground. Driving several displays in this way will use several of the I/O ports on our HCS12 chip.
Implementing Approach #1 (Brute Force) for Two Displays
Connect left a anode pin through a current-limiting resistor to the HCS12’s PA0. Connect left b anode pin through a resistor to the HCS12’s PA1. And so on for the other anode pins. Similarly for the right display, using PORTB.
Approach #2: Multiplexing
In the second approach, we use a single HCS12 port to drive the anode pins on all of the displays. We also connect the common-cathode pins to another HCS12 port. This lets us turn a particular display on (by making the common-cathode pin LOW) or off (by setting the common cathode pin HIGH).
Approach #2: Multiplexing (Cont.)
We sequence quickly through the displays, turning each one on for an instant: Send out the code for the first display’s digit. Turn the first display on by setting its cathode LOW and setting all of the other cathodes HIGH. Brief delay. Send out the code for the second display’s digit. Turn the second display on by setting its cathode LOW and setting all of the other cathodes HIGH. And so on for each of the other displays. Repeat the entire sequence from Step 1.
Implementing Approach #2 (Multiplexing) for Two Displays
Connect All a anode pins through a current-limiting resistor to the HCS12’s PB0. All b anode pins through a current-limiting resistor to the HCS12’s PB1. And so on for the other anode pins. The first display’s common-cathode pin to the HCS12’s PTP0. The second display’s common-cathode pin to the HCS12’s PTP1. And so on for any other displays.
Seven-Segment Displays on the Dragon12 Trainer
Our trainer board uses the multiplexing approach for its four seven-segment displays. See Schematic Diagram 4 and pages of Dragon12 manual. On the schematic diagram, note signal names “DIG0”, “DIG1”, etc. which are connected to outputs of the buffer/driver chip below it.
Reminder: You Must Configure Ports as Inputs or Outputs
We’ve seen that the seven-segment displays’ anode pins are wired to Port B and their common cathode pins are wired to Port P. So to use the seven-segment displays, we must first configure Ports B and P as outputs.
When CodeWarrior downloads a program to the HCS12, it does so using a widely used standard called the Motorola S19 file format. (http://en.wikipedia.org/wiki/S19_(file_format)) In Lab #3 you briefly examined an S19 file for a program that you were downloading. One line (or “record”) of that file looked like this: S B0520FE9F See next slide for analysis of this record.
Analyzing a Record from an S19 File
We can break the record into five pieces: S B0520FE 9F Address: The starting address in memory where the following data bytes are to be located. Record type: Identifies this record as a data sequence record containing a two-byte address. Data: The data bytes being downloaded. In our case, this is the machine code for a simple program. Checksum: For error-checking. It’s similar to the parity bits that are sometimes attached to data for error-checking. Byte Count: Tells how many bytes follow in this record.
Checksums A checksum byte is an extra byte appended to data that is being transmitted. Its purpose is to allow error checking by the receiver. (Similar to a parity bit.) There are different ways to compute checksum bytes. They’re all effective, as long as the sender and the receiver are using the same method. See next two slides for the checksum method used in S19 files.
Generating the Checksum in an S19 Record
The sender uses the following method to calculate the checksum byte: Add the byte count, the address bytes, and the data bytes, discarding any carries. Take the one’s-complement of the sum. The result of Step 2 is the checksum byte for this record. Do it for the record shown previously.
Checking the Checksum in an S19 Record
The receiver uses the following method to check for errors: Add the byte count, the address bytes, the data bytes, and the checksum byte, discarding any carries. The result should equal \$FF. If it does not equal \$FF, an error has occurred during transmission. Do checksum practice sheet.
Review: HCS12 Block Diagram
Up to now we’ve focused on the HCS12’s central processing unit, CPU12 in the block diagram (on page 6 of textbook or page 23 of Device User Guide). We’ve also looked at the memory blocks (Flash, RAM, EEPROM). And we’ve looked at the general-purpose I/O ports (PTA, PTB, etc. in the diagram).
Review: HCS12 Block Diagram
In coming weeks we’ll study other hardware “blocks” (or subsystems) in the HCS12, such as: Clock and Reset Generator Block Enhanced Capture Timer Block Interrupt Block And others Up to now our primary reference guide has been the CPU reference manual, but now we’ll need the reference guides for the other blocks, listed here.
Review: Special-Function Registers
Recall that the HCS12 has hundreds of special-function registers. Most of these registers are either: Data registers, which transfer data from place to place. (Example: PORTA) Control registers, which control various aspects of the chip’s operation. (Example: DDRA) Status registers, which hold status information about events that have occurred.
Review: How to Access the Special-Function Registers
In the HCS12’s memory map (page 26 of Device User Guide), addresses from \$0000 to \$03FF are assigned to the special- function registers. When you execute an LDAA or STAA instruction to one of these addresses, you’re not reading or writing to memory; instead, you’re reading from or writing to a special- function register.
Review: List of Special-Function Registers
Pages in the Device User Guide list all of the special-function registers and their addresses.
Finding Details on the Special-Function Registers
Up to now the only special-function registers we’ve used are the ones associated with I/O ports, such as PORTA, DDRA, etc. In coming weeks we’ll discuss many more special-function registers, which are associated with individual hardware blocks. For details on how these registers are used, refer to the block user guides.
Finding Details on the Special-Function Registers: Example
Example: The register at address \$0034, named SYNR, is one of many associated with the Clock and Reset Generator. For details on this register, refer to p. 15 of the Clock and Reset Generator Block User Guide.
Clock and Reset Generator (CRG) Block
Next we’ll look at the Clock and Reset Generator block, which includes an important circuit called a Phase-Locked Loop (PLL). Figure from p. 6 of textbook or p. 23 of Device User Guide).
Overview: Clock and Reset Generator (CRG) Block
As its name suggests, this hardware block has two major responsibilities: Generating the HCS12’s clock signals Resetting the HCS12 under certain conditions. See block diagram on p. 11 of the CRG Block User Guide.
Reset Generator Four conditions can cause the system to reset itself:
Applying power to the chip’s power pin. Pressing RESET button connected to pin 42. Clock Monitor failure. Computer Operating Properly (COP) timeout.
Clock Generator Three clock signals are used by different parts of the system: Core Clock. Bus Clock, whose frequency is ½ of the Core Clock. Oscillator Clock. The one we care most about is the Bus Clock, which tells us the instruction cycle time.
Oscillator and External Crystal
All three clock signals are derived ultimately from the external crystal connected to the chip’s oscillator. See diagram on p. 32 of the CRG Block User Guide.
Bypassing the Phase Locked Loop (PLL)
In the simplest case, we bypass the internal phase locked loop (PLL) circuit and just use the oscillator’s output. In this case, the Core Clock and Oscillator Clock will run at the crystal frequency, and the Bus Clock will run at ½ this frequency. Example: The Dragon12 has an 8-MHz crystal, so if we bypass the PLL, our clock frequencies will be: Oscillator Clock = 8 MHz Core Clock = 8 MHz Bus Clock = 4 MHz
Using the PLL The PLL circuit lets us increase or decrease the Core Clock and Bus Clock frequencies above or below what they would be if we just used the crystal and oscillator. Example: By default, CodeWarrior uses the Dragon12’s PLL to multiply the oscillator frequency by 6. Since we have an 8-MHz crystal, our clock frequencies will be: Oscillator Clock = 8 MHz Core Clock = 48 MHz Bus Clock = 24 MHz -Why might we want to increase or decrease the clock freqs? -Recall that our delay loop code was based on assumption of 41.7 ns cycle time (=1/24 MHz). If we change the frequency of the bus clock, our delay code will give us a longer or shorter delay than we designed it for.
Special-Function Registers Associated with the CRG Block
The twelve special-function registers located at addresses \$0034 to \$003F let us control the operation of the CRG block. Figure from p. 30 of the Device User Guide.
Special-Function Registers That Control the PLL
Of the registers associated with the CRG, we care most about these four: CRG PLL Control Register (PLLCTL) CRG Clock Select Register (CLKSEL) CRG Synthesizer Register (SYNR) CRG Reference Divider Register (REFDV) These let us tell the system whether we want to use the PLL or bypass it, and also let us specify how much we want to increase or decrease the clock frequency.
CRG PLL Control Register (PLLCTL)
The main bit we care about in this register is bit 6 (PLLON), which turns the PLL circuit on or off. Figure from p. 21 of CRG Block User Guide, which also provides detailed explanation.
CRG Clock Select Register (CLKSEL)
The main bit we care about in this register is bit 7 (PLLSEL), which says whether we want to bypass the PLL or use it. Figure from p. 19 of CRG Block User Guide, which also provides detailed explanation.
CRG Synthesizer Register (SYNR) & Reference Divider Register (REFDV)
These two registers determine how much we increase or decrease the clock frequency, according to this formula: See pp of CRG Block User Guide.
Example: Programming the PLL
LDAA # STAA CLKSEL ;Select oscillator, not PLL LDAA # STAA SYNR LDAA # STAA REFDV JSR Delay ;Need about 1 ms delay LDAA #\$ STAA CLKSEL ;Select the PLL LDAA #\$ STAA PLLCTL ;Turn on the PLL. Do clockFreqs practice sheet.
Measuring the Bus Clock Frequency
None of the HCS12’s clock signals are brought out to pins on the chip. So we can’t directly measure the clock frequencies. But we can indirectly measure the bus clock frequency. (We’ll do this in Lab 9.) Suppose we have a program that uses a delay loop to toggle a pin on a port. The time delay from the delay loop depends on the bus clock frequency. (A faster clock frequency results in a shorter delay.) So by measuring the toggle rate, we can figure out what the bus clock frequency is.
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Chapter 1
## Chapter 1
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##### Presentation Transcript
1. Chapter 1 The Study of Motion
2. Units • We can classify almost all quantities in terms of the fundamental physical quantities: • Length L • Mass M • Time T • For example: • Speed has units L/T (miles per hour)
3. Units, cont’d • SI (Système International) Units: • MKS: • L = meters (m) • M = kilograms (kg) • T = seconds (s) • CGS: • L = centimeters (cm) • M = grams (g or gm) • T = seconds (s)
4. Units, cont’d • British (or Imperial) Units: • L = feet (ft) • M = slugs or pound-mass (lbm) • T = seconds (s) • We will use mostly SI but we need to know how to convert back and forth.
5. Units, cont’d • The back of your book provides numerous conversions. Here are some: • 1 inch = 2.54 cm • 1 m = 3.281 ft • 1 mile = 5280 ft • 1 km = 0.621 mi
6. Units, cont’d • We can use these to convert a compound unit:
7. Converting units • Look at your original units. • Determine the units you want to have. • Find the conversion you need. • Write the conversion as a fraction that replaces the original unit with the new unit.
8. ExampleProblem 1.1 A yacht is 20 m long. Express this length in feet.
9. Example A yacht is 20 m long. Express this length in feet. ANSWER:
10. Example How many liters are in a five gallon bucket? There are four quarts in a gallon.
11. Example How many liters are in a five gallon bucket? There are four quarts in a gallon. ANSWER:
12. Metric prefixes • Sometimes a unit is too small or too big for a particular measurement. • To overcome this, we use a prefix.
13. Metric prefixes, cont’d
14. Metric prefixes, cont’d • Some examples: • 1 centimeter = 10-2 meters = 0.01 m • 1 millimeter = 10-3 meters = 0.001 m • 1 kilogram = 103 grams = 1,000 g
15. Frequency and period • We define frequency as the number of events per a given amount of time. • When an event occurs repeatedly, we say that the event is periodic. • The amount of time between events is the period.
16. Frequency and period, cont’d • The symbols we use to represent frequency are period are: • frequency: f • period: T • They are related by
17. Frequency and period, cont’d • The standard unit of frequency is the Hertz (Hz). • It is equivalent to 1 cycle per second.
18. ExampleExample 1.1 A mechanical stopwatch uses a balance wheel that rotates back and forth 10 times in 2 seconds. What is the frequency of the balance wheel?
19. ExampleExample 1.1 A mechanical stopwatch uses a balance wheel that rotates back and forth 10 times in 2 seconds. What is the frequency of the balance wheel? ANSWER:
20. Speed • Speed is the rate of change of distance from a reference point. • It is the rate of movement. • It equals the distance something travels divided by the elapsed time.
21. Speed, cont’d • In mathematical notation, • So we can write speed as
22. Speed, cont’d • The symbol D is the Greek letter delta and represents the change in. • As the time interval becomes shorter and shorter, we approach the instantaneous speed.
23. Speed, cont’d • If we know the average speed and how long something travels at that speed, we can find the distance it travels:
24. Speed, cont’d • We say that the distance is proportional to the elapsed time: • Using the speed gives us an equality, i.e., an equal sign, so we call v the proportionality constant.
25. Speed, cont’d • Note that speed is relative. • It depends upon what you are measuring your speed against. • Consider someone running on a ship.
26. Speed, cont’d • If you are on the boat, she is moving at
27. Speed, cont’d • If you are on the dock, she is moving at
28. Example When lightning strikes, you see the flash almost immediately but the thunder typically lags behind. The speed of light is 3 × 108 m/s and the speed of sound is about 345 m/s. If the lightning flash is one mile away, how long does it take the light and sound to reach you?
29. Example ANSWER: For the thunder: For the flash:
30. Velocity • Velocity is the speed in a particular direction. • It tells us not only “how fast” (like speed) but also how fast in “what direction.”
31. Velocity, cont’d • In common language, we don’t distinguish between the two. • This sets you up for confusion in a physics class. • During a weather report, you might be given the wind-speed is 15 mph from the west.
32. Velocity, cont’d • The speed of the wind is 15 mph. • The wind is blowing in a direction from the west to the east. • So you are actually given the wind velocity.
33. Vector addition • Quantities that convey a magnitude and a direction, like velocity, are called vectors. • We represent vectors by an arrow. • The length indicates the magnitude.
34. Vector addition, cont’d • Consider again someone running on a ship. • If in the same directions, the vectors add.
35. Vector addition, cont’d • Consider again someone running on a ship. • If in the opposite directions, the vectors subtract.
36. Vector addition, cont’d • What if the vectors are in different directions?
37. Vector addition, cont’d • The resulting velocity of the bird (from the bird’s velocity and the wind) is a combination of the magnitude and direction of each velocity.
38. c a b Vector addition, cont’d • We can find the resulting magnitude of the Pythagorean theorem.
39. Vector addition, cont’d • Let’s find the net speed of the bird? (Why didn’t I say net velocity?) 6 8 10
40. Vector addition, cont’d • Here are more examples, illustrating that even if the bird flies with the same velocity, the effect of the wind can be constructive or destructive.
41. Acceleration • Acceleration is the change in velocity divided by the elapsed time. • It measures the rate of change of velocity. • Mathematically,
42. Acceleration, cont’d • The units are • In SI units, we might use m/s2. • For cars, we might see mph/s.
43. Acceleration, cont’d • A common way to express acceleration is in terms of g’s. • One g is the acceleration an object experiences as it falls near the Earth’s surface: g = 9.8 m/s2. • So if you experience 2g during a collision, your acceleration was 19.6 m/s2.
44. Acceleration, cont’d • There is an important point to realize about acceleration: It is the change in velocity.
45. Acceleration, cont’d • Since velocity is speed and direction, there are three ways it can change: • change in speed, • change in direction, or • change in both speed & direction. • The change in direction is an important case often misunderstood.
46. Acceleration, cont’d • If you drive through a curve with the cruise control set to 65 mph, you are accelerating. • Not because your speed changes. • But because your direction is changing. • There must be an acceleration because items on your dash go sliding around. • More on this in chapter 2.
47. ExampleExample 1.3 A car accelerates from 20 to 25 m/s in 4 seconds as it passes a truck. What is its acceleration?
48. ExampleExample 1.3 ANSWER: The problem gives us The acceleration is:
49. ExampleExample 1.3 CHECK: Does this make sense? The car needs to increase its speed 5 m/s in 4 seconds. If it increased 1 m/s every second, it would only reach 24 m/s. So we should expect an answer slightly more than 1 m/s every second.
50. ExampleExample 1.4 After a race, a runner takes 5 seconds to come to a stop from a speed of 9 m/s. Find her acceleration.
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# Grade 3 EnVisions Math Topic 16 Common Core Version Inspired Power Point Lessons
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If you are teaching the Grade 3 Topic 16 EnVisions Math Common Core Version lessons, this power point is a great resource to complement your instruction. They allow you to integrate technology into your instruction and use the Common Core based EnVisions skills and terminology to help your students master the third grade math objectives. The power points are designed to be used as a teaching tool for each lesson.
Every lesson power point contains:
* Warm-up for spiral review of skills previously learned.
* Literature links for each lesson to help you integrate literacy into math to meet cross-curricular integration standards.
* A "Think and Write" to integrate math and writing skills and often make connections between math and daily life experiences of the students.
* Relevant vocabulary for the lesson or topic being taught
* Step by step breakdown of how to solve or perform the operation/skill being taught
* Practice problems that you can use to demonstrate on the board, on slates/communicators, or in notebooks to engage all students.
* Word problems to show whether students can solve problems using deeper analysis and evaluation levels of the skill. Some problems also incorporate multiple steps. By observing student responses, you can decide whether students need more review or are ready to move to independent work. It's a great method of informal assessment!
Topic 16 is focused on data, measurement and graphing. This includes interpretation of data, graphing data, drawing conclusions based on graphs, and creating word problems based on graphs. The graphs used are line plots, pictographs and bar graphs. This topic is aligned to CCSS 3.MD.3 and 3.MD.4.
Power point lessons are inspired by the following lessons:
16-1: Line Plots
16-2: Length and Line Plots
16-3: Reading Pictographs and Bar Graphs
16-4: Making Pictographs
16-5: Making Bar Graphs
16-6: Problem Solving: Use Tables and Graphs to Draw Conclusions
If you like this product, consider the other products in this series! They make teaching math as easy as, well, 1-2-3!
Grade 3 Topic 14 EnVisions Power Point Lessons
Grade 3 Topic 15 EnVisions Power Point Lessons
*** This product is inspired by the 2012 Common Core EnVisions series***
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47 pages
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## Pages
### Newton polynomial interpolation
Today, we will learn about polynomial interpolation.
Suppose we have the following polynomial $$P(x) = 2x^2 - 3x + 3$$
Let us calculate the values of the polynomial $P(x)$ for a few values of $x$ as follows $$P(1) = 2 - 3 + 3 = 2,$$ $$P(2) = 8 - 6 + 3 = 5,$$ $$P(3) = 18 - 9 + 3 = 12, \dots$$
Now, suppose that we are given the following information $$P(1) = 2, ~~P(2) = 5, ~~P(3) = 12,$$ can we reconstruct the polynomial $P(x)$?
The answer is, yes, we can. An interpolation formula enables us to reconstruct the polynomial $P(x)$ based on its values. Today, we will learn about Newton's interpolation formula, and in the next post, we will cover Lagrange's interpolation.
First, let us make the following important observations.
Observation 1. If we do not impose any constraint on the degree of the polynomial $P(x)$ then there will exist infinitely many polynomials $P(x)$ that satisfy the condition $$P(1) = 2, ~~P(2) = 5, ~~P(3) = 12.$$
Why is that? It is because if we can find one polynomial $P(x)$ that satisfies the condition, then we can form many other polynomials $G(x)$ that also satisfy the condition by setting
$$G(x) = P(x) + (x-1)(x-2)(x-3)H(x),$$ and we have $$G(1) = P(1), ~~G(2) = P(2), ~~G(3) = P(3).$$
Observation 2. If we impose a constraint on the degree of the polynomial and require that the degree of $P(x)$ must be less than or equal to $2$, then there exists a unique polynomial $P(x)$ that has degree less than or equal to $2$ that satisfies the condition $$P(1) = 2, ~~P(2) = 5, ~~P(3) = 12.$$
This is because, if $G(x)$ is another polynomial that has degree less than or equal to $2$ and satisfies the condition $$G(1) = 2, ~~G(2) = 5, ~~G(3) = 12,$$ then we can form $$D(x) = G(x) - P(x),$$ $D(x)$ will be a polynomial of degree less than or equal to $2$ and $$D(1) = D(2) = D(3) = 0,$$ this shows that the polynomial $D(x)$ has at least $3$ roots, however, its degree is less than or equal to $2$, so $D(x)$ must be the constant zero polynomial. So $G(x) = P(x)$, and this proves the uniqueness of the polynomial $P(x)$.
Let us state the above observation as a general theorem as follows.
Theorem. If $x_1, x_2, \dots, x_n, x_{n+1}$ are $n+1$ distinct real numbers. And $y_1$, $y_2$, $\dots$, $y_n$, $y_{n+1}$ are $n+1$ arbitrary real numbers. Then there exists a unique polynomial $P(x)$ that has degree less than or equal to $n$ and satisfies the following condition $$P(x_1) = y_1, ~~P(x_2) = y_2, \dots, ~~P(x_n) = y_n, ~~P(x_{n+1})=y_{n+1}.$$
The above theorem tells us that a polynomial of degree less than or equal to $n$ will be uniquely determined by its $n+1$ values.
Let us now formulate the Newton's polynomial interpolation. We will do that through our original example. Let us reconstruct the polynomial $P(x)$ from the following information $$P(1) = 2, ~~P(2) = 5, ~~P(3) = 12.$$
In a previous post, I say that one of the lessons I have learned is that, when facing a problem that we do not know where to start, the first thing we can do is to look at special cases of the problem and try to solve them first. In this case, we can start from a simple case and proceed up to more complex case.
First, if we have only one condition $P(1) = 2$, can we find the polynomial $P(x)$? Obviously, the simplest polynomial is the constant polynomial $A(x) = 2$.
Next, suppose we want to find polynomial $B(x)$ so that $B(1) = 2$ and $B(2) = 5$. We can consider polynomials of the following form $$B(x) = A(x) + \alpha (x-1) = 2 + \alpha (x-1)$$
This form is useful because we immediately have $B(1) = A(1) = 2$. Other requirement is $B(2) = 2 + \alpha = 5$, so we need to choose $\alpha = 3$, and we obtain $$B(x) = 2 + 3(x-1)$$
Similarly, if we want to find polynomial $P(x)$ such that $P(1) = 2$, $P(2) = 5$, and $P(3) = 12$, we consider polynomials of the form $$P(x) = B(x) + \alpha (x-1)(x-2) = 2 + 3(x-1) + \alpha (x-1)(x-2)$$
Since $P(x) = B(x) + \alpha (x-1)(x-2)$, we immediately have $$P(1) = B(1) = 2, ~~P(2) = B(2) = 5.$$ The other requirement $P(3) = 8 + 2 \alpha = 12$ makes $\alpha = 2$, and we obtain $$P(x) = 2 + 3(x-1) + 2 (x-1)(x-2)$$
So we have reconstructed the polynomial $P(x)$ that satisfies $$P(1) = 2, ~~P(2) = 5, ~~P(3) = 12.$$ It is $$P(x) = 2 + 3(x-1) + 2 (x-1)(x-2) = 2x^2 - 3x + 3$$
In general, if $x_1$, $x_2$, $\dots$, $x_n$, $x_{n+1}$ are $n+1$ distinct real numbers, and $y_1$, $y_2$, $\dots$, $y_n$, $y_{n+1}$ are $n+1$ arbitrary real numbers, then we can find a polynomial $P(x)$ of degree less than or equal to $n$ that satisfies the condition $$P(x_1) = y_1, ~~P(x_2) = y_2, \dots, ~~P(x_n) = y_n, ~~P(x_{n+1})=y_{n+1}.$$
By following the above construction, the polynomial $P(x)$ will have the form $$P(x) = \alpha_1 + \alpha_2 (x-x_1) + \alpha_3 (x-x_1)(x-x_2) + \dots + \alpha_{n+1} (x-x_1)(x-x_2)\dots(x-x_n)$$
The above equation is called the Newton's interpolation formula for polynomials. Substituting $x=x_1$ into the formula gives us the value of the coefficient $\alpha_1$. Next, if we substitute $x =x_2$ into the formula, we can determine the value of $\alpha_2$. Similarly, the last coefficient $\alpha_{n+1}$ is determined by substituting $x=x_{n+1}$.
Let us consider some examples.
Example 1. Find the polynomial $P(x)$ of degree less than or equal to $4$ such that $$P(1) = 1, ~~P(2) = 1, ~~P(3) = 2, ~~P(4) = 3, ~~P(5) = 5$$
We use Newton's interpolation formula $$P(x) = \alpha_1 + \alpha_2 (x-1) + \alpha_3 (x-1)(x-2) + \alpha_4 (x-1)(x-2)(x-3) + \alpha_5 (x-1)(x-2)(x-3)(x-4)$$
Substitute $x=1$ into the above formula, we have $P(1) = \alpha_1 = 1$, so $$P(x) = 1 + \alpha_2 (x-1) + \alpha_3 (x-1)(x-2) + \alpha_4 (x-1)(x-2)(x-3) + \alpha_5 (x-1)(x-2)(x-3)(x-4)$$
Take $x=2$, we have $P(2) = 1 + \alpha_2 = 1$, so $\alpha_2 = 0$, thus, $$P(x) = 1 + \alpha_3 (x-1)(x-2) + \alpha_4 (x-1)(x-2)(x-3) + \alpha_5 (x-1)(x-2)(x-3)(x-4)$$
Take $x=3$, we have $P(3) = 1 + 2 \alpha_3 = 2$, so $\alpha_3 = \frac{1}{2}$, thus, $$P(x) = 1 + \frac{1}{2} (x-1)(x-2) + \alpha_4 (x-1)(x-2)(x-3) + \alpha_5 (x-1)(x-2)(x-3)(x-4)$$
Take $x=4$, we have $P(4) = 4 + 6 \alpha_4 = 3$, so $\alpha_4 = -\frac{1}{6}$, thus, $$P(x) = 1 + \frac{1}{2} (x-1)(x-2) -\frac{1}{6} (x-1)(x-2)(x-3) + \alpha_5 (x-1)(x-2)(x-3)(x-4)$$
Take $x=5$, we have $P(5) = 3 + 24 \alpha_5 = 5$, so $\alpha_5 = \frac{1}{12}$.
The required polynomial is $$P(x) = 1 + \frac{1}{2} (x-1)(x-2) - \frac{1}{6} (x-1)(x-2)(x-3) + \frac{1}{12} (x-1)(x-2)(x-3)(x-4).$$
Example 2. Find the polynomial $P(x)$ of degree less than or equal to $4$ such that $$P(1) = 1, ~~P(2) = 4, ~~P(3) = 9, ~~P(4) = 16, ~~P(5) = 25$$
Again, we use Newton's interpolation formula $$P(x) = \alpha_1 + \alpha_2 (x-1) + \alpha_3 (x-1)(x-2) + \alpha_4 (x-1)(x-2)(x-3) + \alpha_5 (x-1)(x-2)(x-3)(x-4)$$
Substituting $x=1$, we have $P(1) = \alpha_1 = 1$, so $$P(x) = 1 + \alpha_2 (x-1) + \alpha_3 (x-1)(x-2) + \alpha_4 (x-1)(x-2)(x-3) + \alpha_5 (x-1)(x-2)(x-3)(x-4)$$
With $x=2$, $P(2) = 1 + \alpha_2 = 4$, so $\alpha_2 = 3$, thus, $$P(x) = 1 + 3 (x-1) + \alpha_3 (x-1)(x-2) + \alpha_4 (x-1)(x-2)(x-3) + \alpha_5 (x-1)(x-2)(x-3)(x-4)$$
With $x=3$, $P(3) = 7 + 2 \alpha_3 = 9$, so $\alpha_3 = 1$, thus, $$P(x) = 1 + 3 (x-1) + (x-1)(x-2) + \alpha_4 (x-1)(x-2)(x-3) + \alpha_5 (x-1)(x-2)(x-3)(x-4)$$
With $x=4$, $P(4) = 16 + 6 \alpha_4 = 16$, so $\alpha_4 = 0$, thus, $$P(x) = 1 + 3 (x-1) + (x-1)(x-2) + \alpha_5 (x-1)(x-2)(x-3)(x-4)$$
With $x=5$, $P(5) = 25 + 24 \alpha_5 = 25$, so $\alpha_5 = 0$.
The required polynomial is $$P(x) = 1 + 3(x-1) + (x-1)(x-2) = x^2$$
From the above two examples, we can see that the polynomial $P(x)$ determined by the condition $$P(x_1) = y_1, ~~P(x_2) = y_2, \dots, ~~P(x_n) = y_n, ~~P(x_{n+1})=y_{n+1},$$ may have degree equal to $n$ (as in the example 1), and also may have degree less than $n$ (as in the example 2).
Let us stop here for now. There is another interpolation formula called the Lagrange's interpolation formula that we will study in the next post. See you again then.
Homework.
1. Find the polynomial $P(x)$ that has degree less than or equal to $4$ such that $$P(1) = 2, ~~P(2) = 4, ~~P(3) = 6, ~~P(4) = 8, ~~P(5) = 10$$
2. The Fibonacci sequence is determined as: $F_0=0$, $F_1=1$, $F_{n+1}=F_n+F_{n−1}$. So $$F_0=0, ~F_1=1, ~F_2=1, ~F_3=2, ~F_4=3, ~F_5=5, ~F_6=8, \dots$$
Let $P(x)$ be a polynomial that satisfies the following condition $$P(0) = 2011^{F_{2012}}, ~~P(1) = 2011^{F_{2011}}, ~~P(2) = 2011^{F_{2010}}, \dots$$ $$P(2010) = 2011^{F_{2}}, ~~P(2011) = 2011^{F_{1}}.$$
Prove that the degree of the polynomial $P(x)$ must be greater than or equal to $2011$.
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## jedidiah 2 years ago An official volleyball measures 27 inches in circumference. find surface area.
• This Question is Open
1. jedidiah
that's for the circumference.
2. campbell_st
so you need to find the radius so then the circumference formula is $C = 2\pi r$ in your question you have $27 = 2 \pi r$ make r the subject... when you have the value of r substitute it into the formula for surface area of a sphere $SA = 4\pi r^2$
3. heathernelly
4pir^2 4*3.14(27/2)^2 12.56(13.5)^2 12.56*182.25 2,289.06
4. jedidiah
thanks :)
5. campbell_st
6. jedidiah
a bathroom floor tile measures 6 in on a side. if the whole floor contains 100 tiles, what is the area of the bathroom floor in square ft. Note: 1 ft = 12in
7. campbell_st
well go back to the circumference formula and make r the subject... correctly...
8. jedidiah
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# Practical to find out the concentration of cell sap.
Extracts from this document...
Introduction
01/10/01 Practical to find out the concentration of cell sap by Emma Walters Aim: My aim is to find the concentration of cell sap. Apparatus: * 1m sugar solution * distilled water * 6 boiling tubes * 2 test tube racks * ruler * measuring cylinder * balance * cork borer * 18 pieces of potato * stop clock * paper towels * scalpel Diagram: Method: 1. In the six boiling tubes, sort out the 6 different concentrations with the components: * 1m - 20ml sugar solution * 0.8m - 16ml sugar solution, 4ml distilled water * 0.6m - 12ml sugar solution, 8ml distilled water * 0.4m - 8ml sugar solution, 12ml distilled water * 0.2m - 4ml sugar solution, 16ml distilled water * 0m - 20ml distilled water 2. Use the cork borer to cut the potatoes into 18 pieces. 3. Put them into groups of 3, so there are 6 groups. 4. Dry all the pieces of potato off, so that when you weigh and measure them it is a fair test. ...read more.
Middle
I have tried to draw a line of best fit from the results, and as you can see from the weight results, the concentration of cell sap should be 0.57m. From the length results, the concentration of cell sap should be 0.41m. The average of these is 0.485m, so this should be the concentration of the cell sap in a potato. As I said in my prediction, if the solution is isotonic, the potatoes weight and length should stay the same, and this nearly happened, as for the 0.6m solution, the length of the potato did not change, but the weight did by only 0.2g. This could have been partly to me weighing them wrong, or not leaving them in long enough, but this solution was nearly isotonic. If the solution was Hypotonic, the potatoes length and weight should have gone up. For the solutions 0.2m and 0m, this happened, as they were hypotonic solutions. The Hypertonic solutions, 1m and 0.8m should have lost weight and gone down in length, as they did. ...read more.
Conclusion
The amounts of the solutions I thought were quite accurate, but when you put the boiling tubes next to each other, you could tell that some solutions had slightly more in than the next. This could also have affected my results. Not only this, but I do not think that I left the potatoes in the solutions for long enough for them to conduct osmosis to a reasonable degree of accuracy. I think that this was my biggest error that affected my results the most, and this is why the information that I got was not as good as it potentially could have been. If I were to do this experiment again, this time I would make sure that I am more accurate with all my measuring and weighing, and that I leave the potatoes in the solutions for over 15 minutes, so that they have time to conduct osmosis. This should make my results reliable and accurate, and let me get a better graph. This should also mean that I get a more accurate result for the concentration of cell sap. ...read more.
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# Related GCSE Life Processes & Cells essays
1. ## Find the concentration of Potato Cell Sap.
-4.0 50.0 46.0 -4.0 Part 4; Analysis; The Rseults suggest the following conclusion; Where the line crosses the x-axis, there is no change in the length of the potato and therefor there is no movement of the water molecules. If there is no movement of the water molecules, then the
2. ## To find out the concentration of cell sap
and into the solution outside and they potato piece will become flaccid and lose mass. Fair Test Below are the variables and non-variables of the experiment: Variables Non-Variables Solution concentration Weighing scales Conditions Surface area of potato Solution volume Duration of experiment Temperature Ruler Type of potato In order for
1. ## Enzyme Practical
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Therefore, we will try our best to weigh the potatoes as quickly and as safely as we can. Getting and experimenting with the exact measurements of molars and water is vital to this task. If the volume of one solution in a petri dish is higher or lower than another, will affect the pattern of results later on.
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# sicp-ex-1.31
a.
``` (define (product term a next b)
(if (> a b) 1
(* (term a) (product term (next a) next b))))
```
factorial function, in terms of product function above, can be written as below.
```
(define (identity x) x)
(define (next x) (+ x 1))
(define (factorial n)
(product identity 1 next n))
```
approximations to pi using john wallis' formula, can be found by defining a new term to be used by product function as below
``` (define (pi-term n)
(if (even? n)
(/ (+ n 2) (+ n 1))
(/ (+ n 1) (+ n 2))))
```
And it can be used as follows.
``` (* (product pi-term 1 next 6) 4) ;;= 3.3436734693877552
(* (product pi-term 1 next 100) 4) ;;= 3.1570301764551676
```
b.
``` (define (product term a next b)
(define (iter a res)
(if (> a b) res
(iter (next a) (* (term a) res))))
(iter a 1))
```
Another approach would be to recognize the series as 4*4*6*6*.../3*3*5*5*... The leading factor of 2 in the numerator must be dealt with.
``` (define (pi n)
(define (term x) (* x x))
(define (next x) (+ x 2))
; since we are increasing the numbers by two on every iteration
(define limit (* n 2))
; upper term: - 2 always goes first, start building product from 4
; - as 2 numbers are skipped, the limit must respect that, too
; - since we are squaring one time too often at the end,
; we have to divide that back out of the result
; lower term: start with 3, which is 1 more than the upper term
; -> so increase limit by 1
;
(* 4 (/ (/ (* 2 (product term 4 next (+ limit 2)))
(+ limit 2))
(product term 3 next (+ limit 1)))))
```
Yielding:
``` (pi 6) ;;= 3.255721745
(pi 100) ;;= 3.149378473
```
n defines steps in terms of pairs of factors. So where the above solution does 6 steps when n=6, this solution does 12 steps when n=6:
``` (pi 3) ;;= 3,343673469
(pi 50) ;;= 3,157030176
```
A third way of defining the pi-term function, using integer arithmetic instead of the even? predicate:
``` (define (pi n)
(define (pi-term n)
(/ (* 2 (ceiling (/ (+ 1 n) 2)))
(+ 1 (* 2 (ceiling (/ n 2))))))
(product pi-term 1 inc n))
```
A fourth way of doing it is to recognize pi = 4 * (2 * (1/3) *4 * (1/5) ...) * ((1/3) * 4 * (1/5) ...)
Arithmetic progression formula was used to generate the nth number of the sequence
namely a_k = a_1 + (k-1)d
``` (define (pi-prod n)
(define (inv k)
(if (even? k)
k
(/ 1 k)))
(* 4.0 (* (product 2 (+ 2 (- n 1)) inc inv )
(product 3 (+ 3 (- n 1)) inc inv))))
```
Another way is to recognise that the formula for pi is composed of two separate products:
pi = 4 * ((2/3 * 4/5 * 6/7 ... * n/(n+1)) * (4/3 * 6/5 * ... * n/(n-1))
This gives:
``` (define (approx-pi n)
(* 4
(product
(lambda (n) (/ n (+ 1 n)))
2
(lambda (n) (+ n 2))
n)
(product
(lambda (n) (/ n (- n 1)))
4
(lambda (n) (+ n 2))
n)))
```
master
Here's how I visualize the problem: If you recognize that the denominator follows the pattern 3,3,5,5,7,7 and that for any given pair of n's, the numerators will be n-1 and n+1, then there is no need to take into account the "stray" 2 at the start of the numerator sequence.
``` (define (product-iter f a next b)
(define (iter a result)
(if (> a b)
result
(iter (next a) (* (f a) result))))
(iter a 1))
(define (pi-term n)
(* (/ (- n 1.0) n) (/ (+ n 1.0) n)))
(define (next-pi n)
(+ n 2.0))
(define (pi accuracy)
(* 4 (product-iter pi-term 3.0 next-pi accuracy)))
```
My method is the same as the above. A clarification though, this method works by splitting the formula into:
(2⋅4/3⋅3) ⋅ (4⋅6/5⋅5) ⋅ (6⋅8/7⋅7)
We can then see the general expression for each pair of terms:
((n-1)⋅(n+1))/n^2 --> (n^2 - 1)/n^2 --> n^2/n^2 - 1/n^2 --> 1 - 1/n^2
This translates into code as:
``` (define (pi-term a)
(- 1 (/ 1 (square a))))
(define (pi-next a)
(+ a 2))
(define (pi accuracy)
(* 4.0
(product-iter pi-term 3 pi-next (+ (* accuracy 2) 1))))
```
The method I used for the Wallis Product recognizes the pairs as following:
(2/3) -> 2.5
(4/3) -> 3.5
(4/5) -> 4.5
(6/5) -> 5.5
...
We first of all define our sum (named Pi as in the mathematical notation):
``` (define (pi term next n upper_bound)
(if (> n upper_bound)
1
(* (term n)
(pi term next (next n) upper_bound)
)
)
)
```
Then we define the Wallis Product:
``` (define (wallis a b pi)
(define (wallis-term x)
(if (= (remainder (inexact->exact (floor x)) 2) 0)
(/ (floor x) (ceiling x))
(/ (ceiling x) (floor x))
)
)
(define (wallis-next x)
(+ x 1)
)
(* 4 (pi wallis-term wallis-next a b))
)
```
And then we can test it:
``` (display (wallis 2.5 1000 pi))
```
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167 views
If $152$ is divided into four parts proportional at $3$, $4$, $5$ and $7$, then the smallest part is
1. $29$
2. $26$
3. $25$
4. $24$
retagged | 167 views
Let say $3x,4x,5x,7x$ are the four parts of the $152$
we can write like this $3x+4x+5x+7x = 152$
$\implies 19x=152$
$\implies x = 8$
So, four parts is $: 3\times 8,4\times 8,5\times 8,7\times 8 = 24,32,40,56$
So, the smallest part is $24.$
Option $(D)$ is the correct answer.
by Veteran (60.6k points)
Smallest number in the series is 3.
So the number which is multiplicative by 3 in the options is 24 and this would be considered as smallest part.
by (47 points)
Option A 24
by (11 points)
Take sum of ratio 3+4+5+7=29
Smallest part is : 152*3%19=24
by (11 points)
Option D 24
Because it is the only option which is multiple of 3 as smallest ratio value is 3.
by (11 points)
D: 24
because
if we take the smallest number =x
then the number will be 3x,4x,5x and 7x
and adding 3x,4x,5x,7x to each other then it will be 19x
so,19x=152
or,x=8
3x=3 x 8=24
4x= 4 x 8=32
5x=5 x 8=40
and 7x=7 x 8=56
so the smallest number is 24.
by (167 points)
edited by
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# NCERT Solutions for Class 8 Maths
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NCERT - Mathematics
NCERT Solutions for Class 8 Maths cover all the back-exercise questions that are present in Class 8 Maths NCERT book. These solutions have been prepared by our expert teachers having a thorough knowledge of Maths subject. Our maths experts do extensive research to come up with clear and well-explained solutions for NCERT Class 8 Maths Book.
Our NCERT Maths Class 8 Solutions include clear explanations of complex Maths topics which can help you better your performance in the exam. In any case, you can check out the chapter-wise solutions for CBSE Class 8th Maths Textbook below.
## NCERT Class 8 Maths Solutions - All Chapters
Our NCERT Class 8 Maths Book Solutions include answers and explanations for all the questions given in the NCERT Maths Textbook. These solutions will help students understand complex Maths topics and questions properly and solve them easily. The CBSE 8th Class Maths Solutions provided by us are based on the latest curriculum.
## NCERT Solutions for Class 8 Maths (Chapter-wise description)
### Chapter 1: Rational Numbers
Introduction: Let us first recall real numbers from earlier classes. Real numbers consist of natural numbers, whole numbers, and integers. In Class 7 NCERT Maths textbook, you first came across the concept of rational numbers. You learnt that these are also part of real numbers.
To define, rational numbers can be expressed in the form of p/q, where q cannot be equal to zero. In NCERT Class 8 Maths, you will learn to work out with rational numbers. You will also study how these numbers can be closed under the operations of addition, subtraction and multiplication.
In the latter half of this chapter, you will get to understand how rational numbers are represented on a number line. This chapter consists of 2 exercises and 18 questions based on the following concepts.
Topics
1. Properties of Rational Numbers
(a). Closure
(b). Commutativity
(c). Associativity
(d). The role of zero
(e). The role of 1
(f). Negative of a number
(g). Reciprocal
(h). Distributivity of multiplication over addition for rational numbers
2. Representation of Rational Numbers on the Number Line
3. Rational Numbers between Two Rational Numbers
### Chapter 2: Linear Equations in One Variable
Introduction: In the last two grades, you were taught that an algebraic equation is equality that involves variables. So far, you have only covered linear equations in one variable. An equation is said to be linear when its highest power of the variable is 1.
This year, you will learn to transpose variables from one side of the equation to the other. In an equation, the expression on the left-hand side is always equal to the expression on the right-hand side.
The utility of linear equations is in various applications such as the combination of currency notes, different problems on numbers, ages, perimeters etc. Initially, you may find some word problems tricky. However, you can verify or rectify your answers by looking at our NCERT Solutions for Class 8 Maths Chapter 2.
Topics
1. Equations With Linear Expressions On One Side And The Numbers On The Other Side
2. Some Applications
3. Solving Equations Having The Variable On Both Sides
4. Applications Of Linear Equations
5. Reducing Equations To Simpler Form
6. Equations Reducible To The Linear Form
### Chapter 3: Understanding Quadrilaterals
Introduction: Before you understand quadrilaterals, try to recall different varieties of curves such as plane curve, simple curve, simple closed curve etc. In NCERT Class 8 Maths Chapter-3, our topic of concern is a simple closed curve made up of line segments, which is called a polygon.
Based on the number of sides or vertices, polygons can be classified as a triangle (3 sides), quadrilateral (4 sides), pentagon (5 sides) and so on. After completing this chapter, you will have a clear understanding of properties of different types of quadrilaterals such as parallelogram, rhombus, rectangle, square and kite.
There are a total of 33 questions in 4 exercises based on the following concepts.
Topics
1. Polygons
(a). Classification Of Polygons
(b). Convex And Concave Polygons
(c). Regular And Irregular Polygons
(d). Angle Sum Property
2. Sum of the Measures of the Exterior Angles of a Polygon
3. Types of Quadrilaterals
(a). Trapezium
(b). Kite
(c). Parallelogram
(d). Elements of a Parallelogram
(e). Angles of a Parallelogram
(f). Diagonals of a Parallelogram
4. Some Special Parallelograms
(a). Rhombus
(b). Rectangle
(c). Square
Important Concepts of Understanding Quadrilaterals
1. Properties of Parallelogram
• Opposite sides are equal
• Opposite angles are equal
• Diagonals bisect each other
2. Properties of Rhombus
• Opposite sides are equal
• Opposite angles are equal
• Diagonals bisect each other
• Diagonals are perpendicular to each other
3. Properties of Rectangle
• Opposite sides are equal
• Opposite angles are equal
• Diagonals bisect each other
• Each of the angle is a right angle
• Diagonals are equal
4. Properties of Square
It includes all the properties of a parallelogram, rhombus and a rectangle.
5. Properties of Kite
• The diagonals are perpendicular to one another
• Diagonals bisect each other
### Chapter 4: Practical Geometry
Introduction: In the previous edition of NCERT Maths textbook, you have learnt how to draw triangles. It took three measurements to draw a unique triangle. Having understood the properties of different types of quadrilaterals, we will now proceed with the construction of quadrilaterals.
In this chapter, you will come across five different known conditions for the construction of a unique quadrilateral. These conditions are mentioned below.
Topics
A quadrilateral can be constructed when-
• Four sides and one diagonal are given
• Two diagonals and three sides are given
• Two adjacent sides and three angles are given
• Three sides and two included angles are given
• When other special properties are known
### Chapter 5: Data Handling
Introduction: When we extract data from sources, it is available in an unorganized form, also called raw data. In this chapter, you will learn to draw meaningful inferences and organize the data systematically. As studied earlier in pictograph, a frequency gives the number of times a particular entry occurs.
With the help of grouped frequency distribution, raw data can be grouped and presented systematically. Grouped data can be represented in various ways, including histograms, pie charts, bar graphs, etc.
As you proceed further, you will study about random experiments and the probability of an event. Check out the complete list of topics covered in NCERT Class 8th Maths Chapter- 5.
Topics
1. Organizing Data
2. Grouping Data
(a). Bars with a Difference
3. Circle Graph on Pie Chart
4. Chance and Probability
(a). Getting a Result
(b). Equally Likely Outcomes
(c). Linking Chances to Probability
(d). Outcomes as Events
(e). Chance and Probability
Important Formulas of “Probability”
Probability of an event (P) = Number of outcomes that make an event/ Total number of outcomes of the experiment
### Chapter 6: Squares and Square Roots
Introduction: Suppose, a natural number m can be expressed as n2, where n is a natural number, then m is a square number. One can identify a square number by observing its unit place. Numbers that end with 0,1, 4, 5, 6 or 9 are called square numbers.
Another way to identify a square number is by counting the number of zeros at the end. Numbers that have an even number of zeros at the end are called square numbers. The square root is the inverse operation of the square.
This chapter is divided into four exercises, with a total of 31 questions. By the end of the chapter, you will have a clear understanding of the following topics.
Topics
1. Properties of Square Numbers
2. Finding the Square of a Number
3. Square Roots
4. Finding Square Root by Division Method
5. Square Roots of Decimals
6. Estimating Square Root
### Chapter 7: Cubes & Cube Roots
Introduction: S.Ramanujan, who is regarded as one of the finest mathematical geniuses, came out with a unique concept when he met Prof. G.H. Hardy. Professor came to visit him in a taxi whose number was 1729. While talking to each other, Ramanujan suddenly thought of it as the smallest number that can be expressed as a sum of two cubes.
Similarly, there are infinitely many such numbers such as 4104, 13832 etc. These numbers are known as Hardy-Ramanujan Numbers. Cube numbers such as 8, 27, 64, 125 etc. can be obtained when 2, 3, 4, 5 etc. are multiplied by themselves three times respectively.
Cube root is the inverse operation of the cube. For example ∛27 = 3. When doing prime factorisation of any number, if you find each factor appearing three times, then the number is a perfect cube.
Topics
1. Cubes
2. Cube Roots
(a). Cube root through prime factorisation method
(b). Cube root of a cube number
### Chapter 8: Comparing Quantities
Introduction: Before you start-off this chapter, it is recommended that you revisit ratios and percentages. In the first exercise of this chapter, you can strengthen your previous concepts related to ratios and percentages. While the second exercise of the chapter will include questions based on the increase or decrease percent, discounts, profit and loss, and goods and services tax.
In the last exercise, you will deal with word problems involving the use of compound interest annually or half-yearly. A few questions will be based on the applications of compound interest formula.
This chapter requires you to have utmost accuracy that can only come with regular practice. When you are practising questions, it is important that you cross-check your answers from our NCERT Class 8 Maths Chapter-8 Solutions. In total, there are 28 questions across 3 exercises based on the following concepts.
Topics
1. Ratios and Percentages
2. Finding the Increase or Decrease Percent
3. Finding Discounts
4. Prices Related to Buying and Selling
5. Sales Tax/ Value Added Tax/ Goods and Services Tax
6. Compound Interest
7. Deducing a Formula for Compound Interest
8. Rate Compounded Annually or Half Yearly
9. Applications of Compound Interest Formula
Important Formulas of “Comparing Quantities”
1. Discount = Marked price - Sale price
2. Discount = Discount % of the marked price
3. Cost Price = Buying price + Overhead expenses
4. Compound Interest (A) = Principal amount + Interest
5. Compound Interest when interest is compounded annually
= P (1 + R/100)n, where P is principal, R is the rate of interest, n is time period
6. Compound Interest when interest is compounded half-yearly
= P (1 + R/200)2n, where R/2 is half yearly rate, 2n = number of half-years
### Chapter 9: Algebraic Expressions and Identities
Introduction: In earlier editions of NCERT Maths textbook, you learnt that terms are formed from variables and constants. When you add or subtract these terms, you will get an expression. Expressions that contain exactly one, two and three terms are called monomials, binomials and trinomials, respectively.
In general, a polynomial can be termed as an expression with one or more terms having non-zero coefficients.
When we speak of Like and Unlike terms, like terms are formed from the same variables having a similar degree. However, coefficients of Like terms need not be the same.
Later in the chapter, you will be introduced to three standard identities and an additional identity. These identities will be useful in solving squares and products of algebraic expressions. Check out the topics and identities of this chapter below.
Topics
1. Expressions
2. Terms, Factors and Coefficients
3. Monomials, Binomials and Polynomials
4. Like and Unlike Terms
5. Addition and Subtraction of Algebraic Expressions
6. Multiplication of Algebraic Expressions
7. Multiplying a Monomial by a Monomial
8. Multiplying a Monomial by a Polynomial
9. Multiplying a Polynomial by a Polynomial
10. What is Identity?
11. Standard Identities
12. Applying Identities
Important Identities of “Algebraic Expressions and Identities”
1. (a + b)2 = a2 + 2ab + b2
2. (a - b)2 = a2 - 2ab + b2
3. (a + b) (a - b) = a2 - b2
4. (x + a) (x + b) = x2 + (a + b)x + ab
### Chapter 10: Visualising Solid Shapes
Introduction: In class 7, you have learnt about two-dimensional (plane) shapes and three-dimensional (solid) shapes. You may recall that circle, rectangle, triangle etc. are classified under 2-D figures while cubes, cylinders, cones, spheres etc. are 3-D figures.
This year, you will study 3D objects that have different views from different positions. The main highlight of this chapter is Euler’s formula, in which you will get to understand the relationship between faces, vertices, and edges.
Topics
1. Views of 3D-Shapes
2. Mapping Space Around Us
3. Faces, Edges and Vertices
Euler’s Formula
F + V - E = 2, where
‘F’ stands for the number of faces
‘V’ stands for the number of vertices
‘E’ stands for the number of edges
### Chapter 11: Mensuration
Introduction: In the previous edition of NCERT Maths, you have learnt to find the area and perimeter of various plane figures such as triangles, rectangles, circles, etc. Now, your next task will be to solve problems related to perimeter and area of various plane closed figures.
While the last exercise of this chapter will include word problems based on surface area and volume of solids such as a cube, cuboid and cylinder. With the help of NCERT Solutions for Class 8 Maths Chapter 11, you will get to practice 34 questions across 4 exercises. These solutions will enable you to grasp the following concepts.
Topics
1. Area and Perimeter of Rectangle, Square, Triangle, Parallelogram, Circle
2. Area of Trapezium
3. Area of a General Quadrilateral
4. Area of a Polygon
5. Solid Shapes
6. Surface Area of Cube, Cuboid and Cylinder
7. Volume of Cube, Cuboid and Cylinder
8. Volume & Capacity
Important Formulas of “Mensuration”
1. Area of Trapezium = ½ x (Sum of the length of parallel sides) x Distance between them
2. Area of Rhombus = ½ x Product of its diagonals
3. Total Surface Area of a Cuboid = 2(lb + bh + hl)
4. Total Surface Area of a Cube = 6l2
5. Total Surface Area of a Cylinder = 2πr (r + h)
6. Volume of a Cuboid = l x b x h
7. Volume of a Cube = l3
8. Volume of a Cylinder = πr2h
9. Some Important Conversions
(a). 1 cm3 = 1 mL
(b). 1 L = 1000 cm3
(c). 1 m3 = 1000000 cm3 = 1000 L
### Chapter 12: Exponents and Powers
Introduction: You should be well-versed with the laws of exponents before you start solving questions of chapter-12. Earlier, you dealt with only positive exponents, but now you will extend your knowledge to negative exponents. The laws of exponents are applicable for numbers with negative exponents.
Basic laws of exponents and important topics of this chapter are given below.
Topics
1. Powers with Negative Exponents
2. Laws of Exponents
3. Use of Exponents to Express Small Numbers in Standard Form
Important Formulas/ laws of “Exponents and Powers”
(a). am x an = am+n
(b). am ÷ an = am-n
(c). (am)n = amn
(d). am x bm = (ab)m
(e). a0 = 1
(f). am/ bm = (a/b)m
### Chapter 13: Direct and Inverse Proportions
Introduction: Direct and inverse proportion is one of the most important chapters of NCERT Class 8 Maths as it will come to play in higher classes, especially in Physics and Chemistry. Two variables x and y are said to be in direct proportion when the value of y increases with respect to increase in the value of x such that the ratio x/y does not change.
On the other hand, two variables x and y are said to be in inverse proportion, when x increases y decreases and vice-versa. This chapter features 21 questions in its 2 exercises.
Topics
1. Direct Proportion
2. Inverse Proportion
Formulas of Direct and Indirect Proportion
1. Direct Proportion
x1/y1 = x2/y2
2. Indirect Proportion
x1/ x2 = y2/ y1
### Chapter 14: Factorisation
Introduction: Previously, we learnt to obtain factors using the prime factorisation method. When we factorise an algebraic expression, we represent it in the form of the product of factors. In NCERT Solutions for Class 8 Maths Chapter- 14, you will study four different methods of factorisation given below in the list of topics.
In the latter half of this chapter, you will study division of a polynomial by a monomial. You can carry out the division either by dividing each term of the polynomial by a monomial or by the common factor method.
In the last exercise of this chapter, your task will be to spot errors in algebraic equations. In total, there are 34 questions across its 4 exercises.
Topics
1. Factorization
(a). Method of Common Factors
(b). Regrouping Terms
(c). Using Identities
(d). Factors of the Form (x + a) (x + b)
2. Division of Algebraic Expressions
(a). Division of a Monomial by Another Monomial
(b). Division of a Polynomial by a Monomial
3. Division of Algebraic Expressions
4. Error Spotting
### Chapter 15: Introduction to Graphs
Introduction: Graphical representation of data is carried out after collecting raw data from the authentic sources. Three basic methods of graphical representation include a bar graph, a pie chart and a histogram.
Further, you will learn to display data using a line graph. A line graph is alternatively called a linear graph. To locate a point on the graph sheet we need, x-coordinate and y-coordinate. This chapter consists of 13 questions.
Topics
1. Bar Graph
2. Pie Chart
3. Histogram
4. Line Graph
5. Linear Graphs
(a). Location of a Point
(b). Coordinates
6. Applications of Graphs
## Benefits of studying with NCERT Class 8 Maths Book Solutions
There are several benefits that you can experience by studying maths with the help of NCERT Solutions. To begin with, you can practise each topic and question by following a detailed and meticulous approach. Moreover, you get the opportunity to learn the topics and concepts sitting at your home with the help of the Internet. Continue to read below, to learn about additional benefits that you can get by relying on CBSE Class 8 Maths Solutions from Gradeup School.
• Well researched and easy to understand solutions
• Accurate and Reliable explanations of questions
• Exercise and topic-based solutions covering the majority of questions
• Based on the latest curriculum released by the CBSE
• Step by step solutions for a better understanding of the topics
• Help to score better marks in the exam.
• These solutions can be accessed at anytime of the day anywhere
## FAQs | NCERT Class 8 Maths Textbook
Students looking to study with the help of NCERT Solutions for Class 8 often come up with some doubts regarding the applicability and reliability of these solutions. Therefore, to address the concern of students, we have compiled a few questions that are frequently asked by them.
### Can I score more marks by studying NCERT Solutions for class 8?
Obviously, you can score more marks by relying on NCERT Solutions for Class 8 Maths Textbook. However, for that to happen, you need to practise important questions regularly. Our NCERT Solutions cover explanations for all topics and questions, and thus by practising them, you can definitely score better marks.
### Are CBSE NCERT Class 8 Maths Solutions reliable?
Since these solutions are prepared by qualified and expert teachers of Gradeup School, they are 100% reliable. Moreover, the solutions are based on the curriculum of CBSE for Class 8, which further confirms their validity and reliability.
### Can I get NCERT Solutions for Class 8 Maths for Free?
Yes, NCERT Solutions for 8th Class Maths are made available by us for free of cost. You can access and download these solutions at anytime and anywhere for practice purpose.
### How long will it take to practise all the questions in CBSE 8th Class Maths Solutions?
It entirely depends upon you and the amount of time you are willing to devote to practise these solutions. You can practise these solutions simultaneously with your NCERT Maths Textbook questions, or you can use them at the time of examination for supplementing your preparation.
Given the fact that Maths is always a tough subject to prepare and you cannot rely on your class textbook only to score good marks in the exam. NCERT Solutions provided by Gradeup School can go a long way in helping you clear your doubts and understand the complex concepts thoroughly. This would certainly enable you to perform better in the exam and secure more marks.
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# FREE 6th Grade MCAS Math Practice Test
We know that the time of your student to prepare for the 6th Grade MCAS Math exam is limited! So, in this article, we have decided to select 20 real 6th Grade MCAS Math questions from the past exams for your student so that the student is fully prepared for the 6th Grade MCAS Math exam. Using these questions will increase your student’s chances of getting a high score on the 6th Grade MCAS Math test.
Make sure to follow some of the related links at the bottom of this post to get a better idea of what kind of mathematics questions students need to practice.
The Absolute Best Book to Ace the MCAS Math Test
## 10 Sample 6th Grade MCAS Math Practice Questions
1- There are 55 blue marbles and 143 red marbles. We want to place these marbles in some boxes so that there is the same number of red marbles in each box and the same number of blue marbles in each of the boxes. How many boxes do we need?
A. 8
B. 9
C. 10
D. 11
2- What is the value of the following expression?
$$2,205÷315$$
A. 5
B. 6
C. 7
D. 8
3- Solve the following equation.
$$112=22+x$$
A. $$x=-90$$
B. $$x=90$$
C. $$x=-134$$
D. $$x=134$$
4- Car A travels 221.5 km at a given time, while car B travels 1.2 times the distance car A travels at the same time. What is the distance car B travels during that time?
A. 222.7 km
B. 233.5 km
C. 241.5 km
D. 265.8 km
5- The perimeter of the trapezoid below is 38. What is its area?
A. 198 cm$$^2$$
B. 162 cm$$^2$$
C. 99 cm$$^2$$
D. 81cm$$^2$$
6- Which of the following expressions has the greatest value?
A. $$3^1+12$$
B. $$3^3-3^2$$
C. $$3^4-60$$
D. $$3^5-218$$
7- Alfred has $$x$$ apples. Alvin has 40 apples, which is 15 apples less than a number of apples Alfred owns. If Baron has $$\frac{1}{5}$$ times as many apples as Alfred has. How many apples does Baron have?
A. 5
B. 11
C. 55
D. 275
8- In the following triangle find $$α$$.
A. $$100^\circ$$
B. $$90^\circ$$
C. $$60^\circ$$
D. $$30^\circ$$
9- The price of a laptop is decreased by $$15\%$$ to $425. What is its original price? A.$283
B. $430 C.$500
D. $550 10- Find the perimeter of shape in the following figure? (all angles are right angles) A. 21 B. 22 C. 24 D. 20 11- What are the values of mode and median in the following set of numbers? $$1,3,3,6,6,5,4,3,1,1,2$$ A. Mode: 1, 2, Median: 2 B. Mode: 1, 3, Median: 3 C. Mode: 2, 3, Median: 2 D. Mode: 1, 3, Median: 2.5 12- Which expression equivalent to $$x × 92$$? A. $$(x×90)+2$$ B. $$x×9×2$$ C. $$(x×90)+(x×2)$$ D. $$(x×90)+2$$ 13- The ratio of pens to pencils in a box is 3 to 5. If there are 96 pens and pencils in the box altogether, how many more pens should be put in the box to make the ratio of pens to pencils 1 : 1? A. 22 B. 23 C. 24 D. 25 14- If point A placed at $$-\frac{24}{3}$$ on a number line, which of the following points has a distance equal to 5 from point A? A. $$-13$$ B. $$-3$$ C. $$-2$$ D. A and B 15- Which of the following shows the numbers in increasing order? A. $$\frac{3}{13}, \frac{4}{11}, \frac{5}{14}, \frac{2}{5}$$ B. $$\frac{3}{13}, \frac{5}{14}, \frac{4}{11}, \frac{2}{5}$$ C. $$\frac{3}{13}, \frac{5}{14}, \frac{2}{5}, \frac{4}{11}$$ D. $$\frac{5}{14}, \frac{3}{13}, \frac{2}{5}, \frac{4}{11}$$ 16- If $$x=- 4$$, which of the following equations is true? A. $$x(3x-1)=50$$ B. $$5(11-x^2 )=-25$$ C. $$3(-2x+5)=49$$ D. $$x(-5x-19)=-3$$ 17- What is the missing prime factor of number 450? $$450=2^1×3^2×…$$ _________ 18- What is the perimeter of the following shape? (it’s a right triangle) A. 14 cm B. 18 cm C. 24 cm D. 32 cm 19- 65 is what percent of 50? A. $$50 \%$$ B. $$77 \%$$ C. $$130 \%$$ D. $$140 \%$$ 20- Which of the following expressions has a value of $$-23$$? A. $$-10+(-8)+ \frac{5}{2}×(-2)$$ B. $$5×3+(-2)×18$$ C. $$-10+6×8÷(-4)$$ D. $$(-3) × (-7) + 2$$ ## Best 6th Grade MCAS Math Workbook Resource for 2024 ## Answers: 1- D First, we need to find the GCF (Greatest Common Factor) of 143 and 55. $$143=11×13$$ $$55=5×11→$$ GFC$$= 11$$ Therefore, we need 11 boxes. 2- C $$2205÷315=\frac{2205}{315}=\frac{441}{63}=\frac{147}{21}= 7$$ 3- B $$112=22+x$$ Subtract 22 from both sides of the equation. Then: $$x=112-22=90$$ 4- D Distance that car B travels $$=1.2 ×$$ distance that car A travels =$$1.2×221.5=265.8$$ km 5- D The perimeter of the trapezoid is 38. Therefore, the missing side (height) is $$= 38 – 8 – 10 – 11 = 9$$ Area of the trapezoid: $$A = \frac{1}{2} h (b_1 + b_2) = \frac{1}{2}1 (9) (8 + 10) = 81$$ 6- D A. $$3^1+12=3+12=15$$ B. $$3^3-3^2=27-9=18$$ C. $$3^4-60=81-60=21$$ D. $$3^5-218=243-218=25$$ 7- B Alfred has $$x$$ apple which is 15 apples more than the number of apples Alvin owns. Therefore: $$x-15=40→x=40+15=55$$ Alfred has 55 apples. Let $$y$$ be the number of apples that Baron has. Then: $$y=\frac{1}{5}×55=11$$ 8- A Complementary angles add up to 180 degrees. $$β+150^\circ=180^\circ→β=180^\circ-150^\circ=30^\circ$$ The sum of all angles in a triangle is 180 degrees. Then: $$α+β+50^\circ=180^\circ→α+30^\circ+50^\circ=180^\circ$$ $$→α+80^\circ=180^\circ→α=180^\circ-80^\circ=100^\circ$$ 9- C Let $$x$$ be the original price. If the price of a laptop is decreased by $$15\%$$ to$425, then:
$$85 \% \space of \space x=425⇒ 0.85x=425 ⇒ x=425÷0.85=500$$
10- C
Let $$x$$ and $$y$$ be two sides of the shape. Then:
$$x+1=1+1+1→x=2$$
$$y+6+2=5+4→y+8=9→y=1$$
Then, the perimeter is:
$$1+5+1+4+1+2+1+6+2+1=24$$
11- B
First, put the numbers in order from least to greatest: $$1, 1, 1, 2, 3, 3, 3, 4, 5, 6, 6$$
The Mode of the set of numbers is: 1 and 3 (the most frequent numbers)
Median is: 3 (the number in the middle)
12- C
$$x×92=x×(90+2)=(x×90)+(x×2)$$
13- C
The ratio of pens to pencils is $$3 : 5$$. Therefore there are 3 pens out of all 8 pens and pencils. To find the answer, first dived 96 by 8 then multiply the result by 3.
$$96÷8=12→12×3=36$$
There are 36 pens and 60 pencils $$(96-36)$$. Therefore, 24 more pens should be put in the box to make the ratio $$1 : 1$$
14- D
If the value of point A is greater than the value of point B, then the distance of two points on the number line is: the value of A- the value of B
A. $$-\frac{24}{3}-(-13)=-8+13=5=5$$
B. $$-3-(-\frac{24}{3})=-3+8=5=5$$
C. $$-2-(-\frac{24}{3})=-2+8=6≠5$$
15- B
$$\frac{3}{13}≅0.23 , \frac{5}{14}≅0.357 , \frac{4}{11}≅0.36 , \frac{2}{5}=0.4$$
16- B
Plugin the value of $$x$$ in the equations. $$x = -4$$, then:
A.$$x(3x-1)=50→-4(3(-4)-1)=-4(-12-1)=-4(-13)=52≠50$$
B. $$5(11-x^2 )=-25→5(11-(-4)^2 )= 5(11-16)=5(-5)=-25$$
C. $$3(-2x+5)=49→3(-2(-4)+5)=3(8+5)=39≠49$$
D. $$x(-5x-19)=-3→-4(-5(-4)-19=-4(20-19)=-4≠-3$$
17- 5
Let $$x$$ be the missing prime factor of 450.
$$450= 2 × 3 × 3 × x ⇒ x =\frac{450}{18} ⇒ x = 25=5×5$$
18- C
Use the Pythagorean theorem to find the hypotenuse of the triangle.
$$a^2+b^2=c^2→6^2+8^2=c^2→36+64=c^2→100=c^2→c=10$$
The perimeter of the triangle is: $$6+8+10=24$$
19- C
Use the percent formula:
$$Part = \frac{percent}{100} × whole$$
$$65= \frac{percent}{100} × 50⇒ 65 = \frac{percent ×50}{100}⇒ 65=\frac{percent ×5}{10}$$
multiply both sides by 10.
$$650 =percent ×5, \space divide \space both \space sides \space by \space 5.$$
130 = percent
The answer is $$130\%$$
20- A
Let’s check the options provided.
A. $$-10+(-8)+ (\frac{5}{2})×(-2)=-10+(-8)+(-5)=-10-13=-23$$
B. $$5×3+(-2)×18=15+(-38)=-21$$
C. $$-10+6×8÷(-4)=-10+48÷(-4)=-10-12=-22$$
D. $$(-3)× (-7)+ 2=21+2=23$$
The Most Comprehensive Review for 6th-Grade Students
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# counting rules
• May 21st 2006, 06:27 PM
skhan
counting rules
A research study was conducted to determine how much the populat "Hukt on Fonics" reading program actually helps young children learn how to read. Three classes, with each class having 10 students, participated in the study. One class did not use the program all the time. After the program, all 30 students rook a reading test, which showed a wide range of scores and no two scores the same.
If you randomly line up the 30 students in a row, what is the probability that they would end up arranged in order of their reading scores from lowest to highest? [3.77E-33]
--> there was a part a and a part B to this question as well which i was able to answer but i cant seem to figure out how to do this last part
• May 21st 2006, 06:55 PM
ThePerfectHacker
Quote:
Originally Posted by skhan
A research study was conducted to determine how much the populat "Hukt on Fonics" reading program actually helps young children learn how to read. Three classes, with each class having 10 students, participated in the study. One class did not use the program all the time. After the program, all 30 students rook a reading test, which showed a wide range of scores and no two scores the same.
If you randomly line up the 30 students in a row, what is the probability that they would end up arranged in order of their reading scores from lowest to highest? [3.77E-33]
--> there was a part a and a part B to this question as well which i was able to answer but i cant seem to figure out how to do this last part
$\displaystyle \mbox{probability }=\frac{\mbox{favorable outcomes}}{\mbox{possible outcomes}}$
$\displaystyle _{30}P_{30}=30!$
$\displaystyle p=\frac{1}{30!}\approx 3.77\times 10^{-33}$
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# Lines and Angles Worksheet
The Lines and Angles Worksheet are a guide to help you organize and do math in the classroom. This worksheet is designed to help students learn and apply basic mathematical concepts, to help them think logically about things, to work on independent learning, and to develop a sense of responsibility. Using this worksheet will also help students develop and improve their problem solving skills. Below are some tips on how to use this worksheet.
First, introduce the use of the Lines and Angles Worksheet to the class. Have each student bring in a small-sized crayon or pencil so that they can color their own areas on the page. Have each student color a different area on the page to make it easy for them to write in the Numbers, and Percentages box as well as the box when you are writing the main information in the Math Worksheet.
Second, let students do some exercises as part of the introduction. Students will need to select an area of the page and then fill in the word numbers and make sure to fill in the number of times the word occurs on the page as well. The students will need to perform these exercises so that they can figure out the answers to the next sections.
Third, use the Line and Angle Worksheet to explain how different angles can be expressed. This worksheet will help students practice new words and phrases that they learn as they read the book. In this way, they can develop their reading skills as well as their math skills by reading the book and using the book as a reference when they do the assignments.
Fourth, use the Lines and Angles Worksheet to explain how the line is measured. A straight line can be measured using a compass as well as with a ruler. Having students measure a straight line from one place to another will help them develop the ability to measure angles and convert them to other measurements. They will also learn the properties of a circle and how to use a pen and paper to solve math problems.
Fifth, use the Lines and Angles Worksheet to show students how to calculate the slope of a line. Slopes can be calculated using formulas, or students can do the calculation themselves. Either way, having students do the calculations is a fun activity that develops the understanding of a concept in a fun way.
Sixth, use the Lines and Angles Worksheet to show students how to use the x and y coordinates to figure out slopes and directions. Having students solve a problem and then ask questions while doing so is a great way to reinforce this understanding. Students can use the blue and red to figure out slopes while using the red and blue to figure out directions.
These are just a few tips on how to use the Lines and Angles Worksheet. A student can use it in a number of ways as long as they remember that it is an effective worksheet to help students learn the basics of math. Remember that this worksheet is used for learning, not for being really good at it!
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### Chapter 3: Stoichiometry
```Chapter 3: Stoichiometry
• Stoichiometry - The study of
quantities of materials
consumed and produced in
chemical reactions.
• In order to understand
stoichiometry, we must first
understand atomic masses
Atomic Masses
• Elements occur in nature as
mixtures of isotopes
• Carbon = 98.89% 12C
1.11% 13C
<0.01% 14C Carbon
atomic mass = 12.01 amu
Try this…
Naturally occurring lithium is:
7.42% 6Li (6.015 amu)
92.58% 7Li (7.016 amu)
• Find the average atomic mass
of lithium.
7.42(6.015 ) 92.58 7.016
100
6 .9 4 1a m u
Fill in the blanks.
Isotope
28
Si
29
Si
32
Si
Mass (amu)
Abundance
27.98
29.97
4.70%
3.09%
The mole
• A unit to count the number of
molecules
Dozen = 12
Pair = 2
• 1 mol = 6.022 x 1023
Samples of 1 mole of
Cu, Al, Fe, S, I, and Hg
eggs
Molar mass is the mass of 1 mole of shoes in grams
marbles
atoms
1 mole 12C atoms = 6.022 x 1023 atoms = 12.00 g
1 12C atom = 12.00 amu
1 mole 12C atoms = 12.00 g 12C
1 mole lithium atoms = 6.941 g of Li
For any element
atomic mass (amu) = molar mass (grams)
Try this!
Which of the following 100.0 g
samples contains the greatest
number of atoms?
a) Magnesium
b) Zinc
c) Silver
Rank the following according to number of
atoms (greatest to least):
a) 100.0 g of silver
b) 62.0 g of zinc
c) 21.0 g of magnesium
Molar mass
• My expectation is that you do
NOT need a refresher in molar
mass! So try this!!!
• What is the molar mass of
isopentyl acetate (C7H14O2)?
If 1 g of C7H14O2 is released in a bee sting,
how many molecules are released?
• HINT:
– g = 1x10-6g
– g mol molecules
• How many atoms of carbon are
present?
How many C atoms in bee sting?
• HINT: each molecule of
isopentyl acetate has 7 atoms
of carbon
Percent Composition or Mass %
• There are two ways to describe
the composition of a
compound.
– The number of constituents
atoms
– The percent of each of its
elements (by mass)
isopentyl acetate (C7H14O2)
• Let’s look at mass percent of
an element
m ass%
n m olar m ass of elem ent
100%
m olar m ass of com poun d
• Find the mass % of C
m a ss%
7 1 2 .0 1
100%
1 3 0 .1 8
m a ss% o f C 6 4 .5 8 %
Try this one.
Consider separate 100.0 gram
samples of each of the
following:
H2O, N2O, C3H6O2, CO2
• Rank them from highest to
lowest percent oxygen by
mass.
Formulas
• molecular formula = (empirical
formula)n
[n = integer]
• molecular formula = C6H6 = (CH)6
• empirical formula = CH
– The lowest ratio of atoms in a
molecule (mole ratio)
Working backwards
• From the percent
composition, you can
determine the empirical
formula.
Determine the empirical formula of a compound
that has the following percent composition by
mass: K 24.75%, Mn 34.77% , O 40.51%
Assume 100 g sample of the compound
n K 24.75g
n M n 34.77g
1m ol K
0.6330m olK
39.10g K
1m ol M n
0.6329m olM n
54.94g M n
n O 40.51g
1m ol O
16.00g O
2.532m ol O
nK = 0.6330, nMn = 0.6329, nO = 2.532
K
0 .6 3 3 0
1 .0
0 .6 3 2 9
Mn
0 .6 3 2 9
1 .0
0 .6 3 2 9
O
2 .5 3 2
4 .0
0 .6 3 2 9
KMnO4
This one has a twist!
• A compound is made of only
sulfur and nitrogen. It is 69.6%
S by mass. Its molar mass is
184 g/mol. What is its
empirical and molecular
formula?
Equations
2 Mg + O2 2 MgO
2 atoms Mg + 1 molecule O2 makes 2
formula units MgO
2 moles Mg + 1 mole O2 makes 2 moles
MgO
48.6 grams Mg + 32.0 grams O2 makes
80.6 g MgO
NOT
2 grams Mg + 1 gram O2 makes 2 g MgO
• They are sentences
• They describe what happens in
a chemical reaction
• Reactants Products
• They should be balanced
• They have the same number of
each kind of atoms on both
sides
Try this
Chromium compounds exhibit a variety of bright
colors. When solid ammonium dichromate
(NH4)2Cr2O7, a vivid orange compound is ignited
a spectacular reaction occurs. Assume that the
products are chromium (III) oxide, nitrogen gas,
and water vapor. Write and balance the
equation.
And this one
• At 1000oC, ammonia gas, (NH3)
reacts with oxygen gas to form
gaseous nitric oxide, and water
vapor. This reaction is the first
step in the commercial production
of nitric acid by the Ostwald
process. Write and balance the
equation for this reaction.
Stoichiometry
How much do you remember?
• Methanol burns in air according to the
equation
2CH3OH + 3O2 2CO2 + 4H2O
If 209 g of methanol are used up in the
combustion, what mass of water is produced?
A good FRQ for class.
• Methane (CH4) reacts with the oxygen in
the air to produce carbon dioxide and
water.
• Ammonia (NH3) reacts with the oxygen
in the air to produce nitrogen monoxide
and water.
• What mass of ammonia would produce
the same amount of water as 1.00 g of
methane reacting with excess oxygen?
1.4g ammonia
Limiting Reactant
• Reactant that determines the
amount of product formed.
• The one you run out of first.
• Makes the least product.
• Three methods to solve:
– Ratio method shown in text
– Have versus need
– Do 2 stoichiometry problems to
find the amount of product. The
reactant that produces the LEAST
amount is the limiting reagent.
Let’s try together!
• Nitrogen gas can be prepared by passing
gaseous ammonia over solid copper(II) oxide
at high temperatures. The other products of
the reaction are solid copper and water vapor.
If a sample containing 18.1g of NH3 is reacted
with 90.4g of CuO, which is the limiting
reactant? How many grams of nitrogen gas will
be formed?
• Have versus need method:
g NH3mol NH3 mol CuO neededg CuO needed
2NH3 + 3CuO N2 +3Cu + 3H2O
1 8 .1g N H 3
1m o lN H 3
1 7 .0 4 g N H 3
3m o l C u O
2m o lN H 3
7 9 .5 5 g C u O
127g CuO needed
1m o l C u O
• The problem says the we HAVE 90.4g of CuO.
• Using ALL of the NH3 given we NEED 127g of
CuO; therefore, the CuO is the limiting
reactant since we don’t have enough of CuO
• You can do this starting with the CuO as well.
Let’s double check our work.
2NH3 + 3CuO N2 +3CuO + 3H2O
9 0 .4 g C u O
1m o l C u O
7 9 .5 5 g C u O
2m o lN H 3
3m o l C u O
1 7 .0 4 g N H 3
1m o lN H 3
1 2 .9 g N H 3 n e e d e d
• The problem says the we HAVE 18.1g of NH3.
• Using ALL of the CuO given we NEED 12.9g of
NH3; therefore, the CuO is the limiting reactant
since we have an excess of NH3.
• There is an excess of 5.2g of NH3
2NH3 + 3CuO N2 +3CuO + 3H2O
• Do the stoichiometry problem twice. The
reactant that gives you the LEAST amount of
product is the limiting reactant.
9 0 .4 g C u O
1 8 .1g N H 3
1m o l C u O
1m o lN 2
2 8 .0 2g N 2
7 9 .5 5C u O
3m o l C u O
1m o lN 2
1m o lN H 3
1m o lN 2
2 8 .0 2g N 2
1 7 .0 4 g N H 3
2m o lN H 3
1m o lN 2
1 0 .6 g N 2 m a d e
1 4 .9 g N 2 m a d e
• The CuO is the limiting reactant. This confirms
our results from the have versus need method.
• When 124 grams of aluminum are
reacted with 601 grams of iron(III) oxide,
aluminum oxide and solid iron are
produced. Calculate the mass of
aluminum oxide formed.
Excess Reactant (reagent)
• The reactant you don’t run out of.
• The amount of stuff you make is the
yield.
• The theoretical yield is the amount you
would make if everything went perfect.
• The actual yield is what you make in the
lab.
Percent Yield
% yie ld
a ctu a l
100%
th e o re tica l
OR
% yie ld
w h a t yo u g o t
w h a t yo u co u ld h a v e
100%
Try this
Methanol is the simplest alcohol. It is used
as a fuel in race cars and is a potential
replacement for gasoline. Suppose
68.5kg CO(g) is reacted with 8.6kg H2(g).
Calculate the theoretical yield of
methanol.
If 3.57x104g CH3OH is actually produced,
what is the percent yield?
```
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### Seyaua's blog
By Seyaua, 11 years ago, translation,
## Problem A. Morning run
We were asked to find the expected value of meetings between the runners. How to do that? As the first step, expected value is lineal, so we can split the initial problems into the different ones: find the expected value of meetings between the fixed pair of runners. We will solve these problems. To do that we need to make some observations:
1. Let x be the distance between the two runners and they run face to face for infinite amount of time (probability of that obviously equals to 0.5·0.5 = 0.25). Then the first meeting will happen at time , the next one — , the next — and so on.
2. Let us assume that every run ran for l time units. Then if two runners meet — they meet exactly two times. The probability of the meeting equals to 0.5, because in two cases they run in the same direction and in two cases in the different ones.
We will build our solution based on these two observations. As the first step let us represent t as t = k·l + p, where 0 ≤ p < l. Then each runner will run k full laps. What does that mean? Because we have pairs of runners, then in those k laps each pair will have 2k meetings with probability equals to 0.5. So, we need to add to the answer.
Now we need to take into account p seconds of running. Let us assume that the distance between two runners is x and they run towards each other. Then they will meet if , or x ≤ 2t. They will meet once more if , ir x ≤ 2t - l. They cannot meet more than twice, because p < l.
Let us fix one of the runners, then using binary search we can find all other runners at distance no more than x from the fixed one. Let us choose x as x = 2t, and then the number of runners at the distance no more than x stands for the number of runners which meet with the fixed one at least once. If x = 2t - l, we will find the number of runners which meet with the fixed one exactly two times. Multiplying these numbers by 0.25 — probability of the meeting, and add it to the answer.
The complexity of this solution is . We can reduce it using two pointers method.
We were asked to find the maximal number of words we can fit into the block of size r × c. Let's first solve such problem: what is the maximal number of consecutive words which can fit in a row of lenth c, if the first word has index i. We can solve it using binary search or moving the pointer. Now let us build a graph, where vertices are the words and there is a directional edge between i and j, if words from i to j - 1 fit in one row of length c, but words from i to j don't. The weight of the edge is j - i. The we have the following problem: we need to find the path of length k, which has the maximal weight. Easy to solve it with complexity saving weights of all the paths with lengthes equal to the powers of two, or in O(n) time using dfs.
The other problems competitors faced — that we were asked to print the whole text, not only the length.
## Problem C. Memory for Arrays
We were asked to find the maximal number of arrays we can fit into the memory. A small observation first, let the answer be k, then one of the optimal solutions fits the k smallest arrays into the memory. We can assume that we have arrays of size 1 and we want to arrange the memory for the maximal arrays as possible. Then if we have parts of memory of odd size, if we fit array of size 1 at this part we will obtain part of even size. From other hand, if we put arrays of bigger size we will not change the parity and if we don't fill it with arrays of size one and initially it's of odd size then in the end we will have at least one empty cell. So it's reasonable to put the arrays of size one into the memory of odd size. Let's do it until we can do it. We have three possible situations:
1. We don't have memory parts of odd size anymore.
2. We don't have arrays of size 1 anymore.
3. We don't have neither arrays of size 1 neither memory parts of size 1.
Let us start from the first case. Suppose that there are some arrays of size 1 left, but there are no memory parts of odd size. Easy to see then in such case we need to group arrays of size 1 in pairs and then consider them as the same array. So we can divide every number by two and reduce the problem to the initial one.
In the second case if we divide every number by two we will obtain the same problem (and that cannot increase the answer).
The third case is similar to the second one.
When implementing this we need to remember that first we have to fill the memory with arrays which are build from the maximal numbers of initial arrays.
The complexity of the given algorithm is .
## Problem D. Tennis Rackets
We were asked to find the number of obtuse triangles which satisfy the problem statement. Author's solution has complexity O(n2), but it has some optimizations, so it easily works under the TL.
Every triangle has only one obtuse angle. Due to symmetry reasons we can fix one of the sides and assume that obtuse angle is tangent to this side. Then we only need to find the number of such triangles and multiple the answer by 3.
Every side is also symmetric, so we can consider only one half of it and then multiple the answer by 2.
Let us assume that vertex A of the triangle has coordinates (0,0). Vertex B (0,) and C(2,0). Then we can find the coordinates of every single point at each of the sides and apply cosine theorem. We obtain the inequality which guarantee us that the triangle is obtuse. It can be written in many ways, on of them is following: If 1 ≤ i, j, k ≤ n — indices of points which are fixed at each of the sides, then triangle is obtuse iff: f(i, j, k) = 2i2 - i(n + 1) + 2j(n + 1) - ij - k(i + j) < 0. We can see that monotonically increases, so we can use moving pointer method applied to variable k. Then just go over all i from m + 1 to , then j from m + 1 till upper bound for k is less than or equal to n - m + 1 and just sum the results.
We should mention that all of the operations have to be done in int type, it significantly increases the speed.
## Problem E. Sheep
Author's supposed greedy algorithm as a solution for this problem. Let us follow this algorithm. Let us create to label for every interval Positionv and MaximalPositionv.
Positionv — stands for position of v in the required permutation.
MaximalPositionv — stands for maximal possible position of v in the particular moment of the algorithm.
Also let's consider count as a counter with initial value of 0 and set of unprocessed vertices S. The algorithm is following.
1. Using binary search we find the maximal possible distance between the farthest sheep. And then check whether exists the permutation with maximal distance no more than K.
2. Sort all the intervals in the increasing order of ri.
3. Positioni = 0, 1 ≤ i ≤ n, MaximalPositioni = n, 1 ≤ i ≤ n, current = 1, count = 0.
4. Do count = count + 1, Positioncurrent = count, erase current from S, if S — is empty, required permutation has been found.
5. Look at every interval connected to current, and update MaximalPositionv = min(MaximalPositionv, Positioncurrent + K)
6. Build sets S(count, j) = {v|MaximalPositionv ≤ count + j}. If for every j ≥ K - count + 1 holds |S(count, j)| ≤ j go to the step 7, otherwise there is no such permutation.
7. Choose the minimal j such that |S(count, j)| = j. Choose from it the interval with the smallest ri and consider it as a new value for current, go to the step 4.
First let us discuss the complexity. Let us fix K (in total there are iterations with fixed K).
Every step from 4 to 7 will be done at most n times (every time size of S decreases by one). Every step can be implemented in O(n) time. The most difficult one — step 6. But we can see that it's not necessary to actually build the sets, all we need to know — their sizes. This can be done in linear time just counting the number of intervals that MaximalPositionv = i. Let if be Ci — then size of S(count, j) equals to C1 + C2 + ... + Ccount + j, which can be easily calculated with partial sums.
Now let us discuss why this algorithm works. If we have Position labels for every interval — we obviously have the solution. Now let us assume that we ended up earlier. Then we will show that there is no such permutation. If algorithm ended, it means that for some count (consider the smallest such count), exists j0, that |S(count, j0)| > j0 at this step. Then |S(count, k)| > k. Let us prove that from contradiction. From the definition of count we have |S(count - 1, j)| ≤ j for every j ≥ k - count + 2. Then |S(count, j)| = |S(count - 1, j + 1)| - 1 ≤ j for every j ≤ k - 1. And S(count, j) = S(count, k) for k ≤ j < n - count = |S(count, j)| = |S(count, k)| ≤ j. Finally |S(count, n - count)| = n - count. Then |S(count, j)| ≤ j for every j, so we obtain contradiction. That means if algorithm stops at step 6 we have |S(count, k)| > k. So exist at least k + 1 interval, which still don't have assigned label Position and they should be assigned after count. So one of the intervals in S(count, k) has to have the value of Position at least count + k + 1. But every intervals in S(count, k) connected to at least one interval with Position ≤ count. So, we obtain that there is now such permutation.
Editorial was prepared by sdya and Seyaua.
• +43
| Write comment?
» 11 years ago, # | 0 in E: why you just find the smallest ri as current? Can you explain it?
• » » 2 years ago, # ^ | -10 same question: why?
» 10 years ago, # | 0 I have read your solution of problem E. There is a very big problem with the last part. You proved, that |S(Count, K)| > K. Okay. After that, you have a logic mistake: you're telling, that there is no way to end your permutation. Oh, yes, you're right! But you didn't prove, that there was no way to build the permutation in a different way with all conditions. Once again in another words: that's right, you have K+1 segments to put into K places, it's hard to make it. But you should prove, that in any other different beginning of the permutation (first Count positions of the array) there still will be some problems with ending it. And you didn't do it, cause these K+1 vertexes could be connected to the different vertexes before, could not be all connected to each other, etc... I couldn't really even prove that if there is a solution then there is a solution with the first segment with the smallest R_i.P.S. Maybe I didn't get anything or someone has the right solution? I am trying to solve for a very long time... P.P.S The only condition for j in 6 statement is j >= 0, not what you have.
• » » 4 years ago, # ^ | ← Rev. 2 → 0 It's too old so nobody will prolly care, but the proof in editorial is just a copy of some lemmas in this paper. The paper contains a full proof. I tried to understand it, but it looks quite complicated.Also the condition for $j$ is okay. You don't have to care cases about $MaximalPosition_i \le k$ because you can't make it. Of course, $j \geq 0$ is also correct.
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• Get Stata homework help for tasks related to heteroscedasticity
• Affordable Stata assignment help for students who need assistance with correlation
• Hypothesis testing explained by our Stata tutors
To estimate the CEO salaries, a linear regression analysis was carried out. The results are presented in Table 1 below.
Table 1: Regression Model 1
From the outcome:
Salary = 273.81 + 12.76(age) – 0.10(agesq) -40.70(grad) + 11.45(ceoten) + 0.02(mktval) +0.22(profits)
The average age from the results in Table 2 was 56 years.
Table 2: Age Summary
The researcher tested whether the age of 65 years had a statistically significant effect on salaries, starting at the average value of age. The results are presented in Table 3.
Table 3: T-Test – Comparing Salaries for 56 Years and 65 Years
From the findings, t(176) = 18.13; p<0.05. Since the p-value was less than 0.05, this meant that there was a statistically significant difference in the salaries between the average age of 56 and the age of 65.
StatisticsHomeworkHelper.com provides academic support to students who find the concept of heteroscedasticity difficult to grasp. We have an entire team of experts who provide both Stata homework help and online tutoring services to make it possible for students to excel in this area. Whether you are looking to pay someone to do Stata homework for you or teach you the basics of heteroscedasticity, you can rest assured you will get the best academic support from us.
## Affordable Stata assignment help for students who need assistance with correlation
Table 4: Correlation and VIFs
Thus, to deal with multicollinearity, it was imperative to remove one of the variables, particularly profits.
Table 5: Regression Results
Forage (t = 0.31; p>0.05) and for agesq (t = -0.29; p>0.05). Since both beta coefficients were not statistically significant, this meant that the relationship between age and CEO salaries was neither linear nor quadratic in nature.
The corresponding model is presented in Table 5.
Table 6: Regression Results – Excl. Age
From the outcome, only the CEO's tenure at the firm had a statistically significant influence on salary.
Students struggling with the concept of correlation can always take our Stata assignment help and have this area explained in depth by our experts. Those who have trouble completing an econometrics assignment using Stata can also contact us for help and we will provide the best possible assistance. In other words, StatisticsHomeworkHelper.com is a one-stop solution provider for all your assignment needs. You can always count on us whenever you find yourself stuck on a statistics assignment.
## Hypothesis testing explained by our Stata tutors
H0: The coefficients for firm profits and market value are jointly equal to zero (βprofits = βmktvalue = 0)
H1: The estimated coefficients for firm profits and market value are not jointly equal to zero
Table 6 presents the results.
Table 7: Joint significance
From the outcome, F(2, 172) = 35.98; p<0.05. This meant that the estimated coefficients for firm profits and market value were not jointly equal to zero
The major reason why profits and market value have joint significance but lack individual statistical significance is that both are confounding variables. The profits influence both the dependent variable, CEO salary, and the independent variable, market value. This causes a spurious association.
. generate lg_salary = log(salary)
. generate lg_mktval = log(mktval)
. regress lg_salary grad ceoten lg_mktval profits, robust
Our Stata tutors summarize the regression results as below.
Table 8: Regression Results
From the outcome, again, the market value was now a statistically significant variable.
The logarithm function log(x) is defined only for values greater than zero. So the logarithm of a negative number would be not defined. The profits variable included negative numbers (loss) and this meant the need to avoid using the natural log for the profits variable.
Market value had a statistically significant impact on the CEO salary (t = 3.83; p<0.05). The profits did not have a statistically significant impact on the CEO salary (t = 0.60; p>0.05).
Yes, we should drop profits from the model because it is highly correlated with market value and this. From Table 8 below, their correlation coefficient was 0.918 and this was a very high correlation. Further, the VIF statistic for both market value and profits was observed to be higher (VIF> 5) than the minimum acceptable threshold range of between 3 and 5.
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# What does it mean when a function is finite?
When someone says a real valued function $f(x)$ on $\mathbb{R}$ is finite, does it mean that $|f(x)| \leq M$ for all $x \in \mathbb{R}$ with some $M$ independent of $x$?
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What you are describing is what is usually called a bounded function. I have not seen the English word finite used in this context. Perhaps you could mention the context in which the term was used. – André Nicolas May 17 '12 at 4:03
As we can see from the contradictory answers (I would choose George's answer), the OP must provide some context to get something useful. – GEdgar Feb 10 '13 at 17:42
Just pointing out that "finite" sometimes also means nonzero (with a finite measure, usually $>0$). In the sense that $dx$ can be understood as infinitesimal, but still finite interval. Most likely not in this case, but if we are discussing semantics, we should include all the cases. – orion Aug 7 '15 at 7:12
For example $x\mapsto \frac{1}{x}$ is finite valued on $]0,\infty [$ since for all $x\in ]0,\infty [$, $-\infty <f(x)<\infty$, but it's not valued on $[0,\infty [$ since $f(0)=+\infty$. – Surb Oct 5 '15 at 15:28
## 6 Answers
In Elias Stein's Real Analysis, at the beginning of Chapter 4.1, it reads "We shall say that $f$ is finite-valued if $-\infty<f(x)<\infty$ for all $x$."
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Since a valued function may have $\mathbb R \cup \{\infty\}$ as target, it's possible that finite function $f$ corresponds to cases where $\forall x \in \mathbb R \quad f(x) \neq \infty$ like $f(x)=x$ or $f(x)= \frac x {x^2+6}$ while $f(x)=\frac 1x$ , for example, is not finite according to this meaning, because $f(0)=\infty$ (thing that can be taken by defintion or convention)
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A function is finite if it never asigns infinity to any element in its domain. Note that this is different than bounded as $f(x):\mathbb R \to \mathbb R \cup\{\infty\}: f(x)=x^2$ is not bounded since $lim_{x \to \infty}=\infty$. However, $f$ is finite since it does not assign $\infty$ to any real number.
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This doesn't sound right. No $\mathbb R\to\mathbb R$ function can assign infinity to any element in its domain, because infinity is not a real number. – Rahul Feb 10 '13 at 17:43
It's not a problem since the point of this example is to show that not all finite functions are bounded. – Student Feb 10 '13 at 20:49
A convex function $f$ is said to be proper if its epigraph is non-empty and contains no vertical lines, i.e., if $f(x)<+\infty$ for at least one $x$ and $f(x)>-\infty$ for every x. (Section 4, Chapter 1, Convex analysis, Rockafellar, 1997)
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I am not familiar with the term finite in this context. One possible definition would be this.
A function $f: A \to B$ is finite if and only if $f(A) \subseteq B$ is finite.
However, I would not use this definition because the relation $f \subseteq A \times B$ is still an infinite set (if $A$ is infinite).
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No, It means there are only finitely many $y$ such that $f(x)=y$, for example $f(x)=0,1,2,\dots,n$ where $n\in\mathbb{N}$ is finitely valued, although it is also bounded, but not all bounded functions are finitely valued for example $f(x)=x$ on $\mathbb{R}$ or $\mathbb{Q}$ takes uncountably many values within any $|x_i-x_j|<\epsilon$
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Amusingly, one might also use the term for the opposite condition -- for each $y$, $f(x)=y$ only has finitely many solutions for $x$. (c.f. a finite morphism of schemes) – Hurkyl Jun 23 '12 at 1:48
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# A factorisation question
• Oct 1st 2009, 03:03 AM
anshulbshah
A factorisation question
Factorise : (b-c)(b^3 + c^3) + (c-a)(c^3 + a^3) + (a-b)(a^3 + b^3)
• Oct 1st 2009, 05:52 AM
Soroban
Hello, anshulbshah!
I expanded it, then repeatedly factored "by grouping".
If there is a more direct way, I hope someone finds it.
Quote:
Factor: . $(b-c)(b^3 + c^3) + (c-a)(c^3 + a^3) + (a-b)(a^3 + b^3)$
$b^4 - b^3c + bc^3 - c^4 + c^4 - ac^3 + a^3c - a^4 + a^4 - a^3b + ab^3 - b^4$
. . $= \;ab^3 - ac^3 - a^3b + a^3c - b^3c + nbc^3$
. . $= \;a(b^3-c^3) - a^3(b-c) - bc(b^2-c^2)$
. . $=\; a{\color{red}(b-c)}(b^2+bc+c^2) - a^3{\color{red}(b-c)} - bc{\color{red}(b-c)}(b+c)$
. . $=\; {\color{red}(b-c)}\bigg[a(b^2+bc+c^2) - a^3 - bc(b+c)\bigg]$
. . $=\; (b-c)\bigg[ab^2 + abc + ac^2 - a^2 - b^2c - bc^2\bigg]$
. . $=\; (b-c)\bigg[ac^2 - a^3 - b^2c + ab^2 - bc^2 + abc\bigg]$
. . $=\; (b-c)\bigg[a(c^2-a^2) - b^2(c-a) - bc(c-a)\bigg]$
. . $= \;(b-c)\bigg[a{\color{red}(c-a)}(c+a) - b^2{\color{red}(c-a)} - bc{\color{red}(c-a)}\bigg]$
. . $=\; (b-c){\color{red}(c-a)}\bigg[a(c+a) - b^2 - bc\bigg]$
. . $=\; (b-c)(c-a)\bigg[ac + a^2 - b^2 - bc\bigg]$
. . $=\; (b-c)(c-a)\bigg[a^2-b^2 + ac - bc\bigg]$
. . $=\; (b-c)(c-a)\bigg[{\color{red}(a-b)}(a+b) + c{\color{red}(a-b)}\bigg]$
. . $=\; (b-c)(c-a){\color{red}(a-b)}\bigg[a+b+c\bigg]$
Answer: . ${\color{blue}(a-b)(b-c)(c-a)(a+b+c)}$
• Oct 1st 2009, 06:13 AM
masters
Quote:
Originally Posted by Soroban
Hello, anshulbshah!
I expanded it, then repeatedly factored "by grouping".
If there is a more direct way, I hope someone finds it.
$b^4 - b^3c + bc^3 - c^4 + c^4 - ac^3 + a^3c - a^4 + a^4 - a^3b + ab^3 - b^4$
. . $= \;ab^3 - ac^3 - a^3b + a^3c - b^3c + nbc^3$
. . $= \;a(b^3-c^3) - a^3(b-c) - bc(b^2-c^2)$
. . $=\; a{\color{red}(b-c)}(b^2+bc+c^2) - a^3{\color{red}(b-c)} - bc{\color{red}(b-c)}(b+c)$
. . $=\; {\color{red}(b-c)}\bigg[a(b^2+bc+c^2) - a^3 - bc(b+c)\bigg]$
. . $=\; (b-c)\bigg[ab^2 + abc + ac^2 - a^2 - b^2c - bc^2\bigg]$
. . $=\; (b-c)\bigg[ac^2 - a^3 - b^2c + ab^2 - bc^2 + abc\bigg]$
. . $=\; (b-c)\bigg[a(c^2-a^2) - b^2(c-a) - bc(c-a)\bigg]$
. . $= \;(b-c)\bigg[a{\color{red}(c-a)}(c+a) - b^2{\color{red}(c-a)} - bc{\color{red}(c-a)}\bigg]$
. . $=\; (b-c){\color{red}(c-a)}\bigg[a(c+a) - b^2 - bc\bigg]$
. . $=\; (b-c)(c-a)\bigg[ac + a^2 - b^2 - bc\bigg]$
. . $=\; (b-c)(c-a)\bigg[a^2-b^2 + ac - bc\bigg]$
. . $=\; (b-c)(c-a)\bigg[{\color{red}(a-b)}(a+b) + c{\color{red}(a-b)}\bigg]$
. . $=\; (b-c)(c-a){\color{red}(a-b)}\bigg[a+b+c\bigg]$
Answer: . ${\color{blue}(a-b)(b-c)(c-a)(a+b+c)}$
Wow!
• Oct 1st 2009, 06:55 AM
Hello anshulbshah
Quote:
Originally Posted by anshulbshah
Factorise : (b-c)(b^3 + c^3) + (c-a)(c^3 + a^3) + (a-b)(a^3 + b^3)
Having seen Soroban's answer, I suppose this is being wise after the event, but here's a quicker method, if you spot it!
If we put $a = b$ in the expression, we get:
$(b-c)(b^3+c^3) + (c-b)(c^3+b^3) + 0 = 0$
so $(a-b)$ is a factor.
Similarly $(b-c)$ and $(c-a)$ are also factors.
Since the expression is of degree $4$ in $a,b,c$, the remaining factor is linear, and by symmetry is therefore $k(a+b+c)$ for some constant $k$.
So $(b-c)(b^3+c^3)+(c-a)(c^3+a^3)+(a-b)(a^3+b^3) = k(a-b)(b-c)(c-a)(a+b+c)$
Compare coefficients of, say, $bc^3: 1 = k$ (taking the $-b$ from the first factor, the $-c$ from the second, the $c$ from the third and the $c$ from the fourth)
$\Rightarrow (b-c)(b^3+c^3)+(c-a)(c^3+a^3)+(a-b)(a^3+b^3) = (a-b)(b-c)(c-a)(a+b+c)$
Or did I cheat a bit?
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# Lecture 6: Borsuk-Ulam and some combinatorial applications
Now we’ll move away from spectral methods, and into a few lectures on topological methods. Today we’ll look at the Borsuk-Ulam theorem, and see a stunning application to combinatorics, given by Lovász in the late 70’s. A great reference for this material is Matousek’s book, from which I borrow heavily. I’ll also discuss why the Lovász-Kneser theorem arises in theoretical CS.
### The Borsuk-Ulam Theorem
We begin with a statement of the theorem. We will think of the n-dimensional sphere as the subset of ${\mathbb R^{n+1}}$ given by
${S^n = \left\{ (x_1, \ldots, x_{n+1}) : x_1^2 + \cdots + x_{n+1}^2 = 1\right\}.}$
Borsuk-Ulam Theorem: For every continuous mapping ${f : S^n \to \mathbb R^n}$, there exists a point ${x \in S^n}$ with ${f(x)=f(-x)}$.
Pairs of points ${x,-x \in S^n}$ are called antipodal.
There are a couple of common illustrative examples for the case ${n=2}$. The theorem says that if you take the air out of a basketball, crumple it (no tearing), and flatten it out, then there are two points directly on top of each other which were antipodal before. Another common example states that at every point in time, there must be two points on the earth which both have exactly the same temperature and barometric pressure (assuming, of course, that these two parameters vary continuously over the surface of the eath).
The n=1 case is completely elementary. For the rest of the lecture, let’s use ${N = (0,0,\ldots,1)}$ and ${S=(0,0,\ldots,-1)}$ to denote the north and south poles (the dimension will be obvious from context). To prove the n=1 case, simply trace out the path in ${\mathbb R}$ starting at ${f(N)}$ and going clockwise around ${S^1}$. Simultaneously, trace out the path starting at ${f(S)}$ and going counter-clockwise at the same speed. It is easy to see that eventually these two paths have to collide: At the point of collision, ${f(x)=f(-x)}$.
We will give the sketch of a proof for ${n \geq 2.}$ Let ${g(x)=f(x)-f(-x)}$, and note that our goal is to prove that ${g(x)=0}$ for some ${x \in S^n}$. Note that ${g}$ is antipodal in the sense that ${g(x)=-g(-x)}$ for all ${x \in S^n}$. Now, if ${g(x) \neq 0}$ for every ${x}$, then by compactness there exists an ${\epsilon > 0}$ such that ${\|g(x)\| > \epsilon}$ for all ${x \in S^n}$. Because of this, we may approximate ${g}$ arbitrarily well by a smooth map, and prove that the approximation has a 0. So we will assume that ${g}$ itself is smooth.
Now, define ${h : S^n \to \mathbb R^n}$ by ${h(x_1, \ldots, x_{n+1}) = (x_1, \ldots, x_n)}$, i.e. the north/south projection map. Let ${X = S^n \times [0,1]}$ be a hollow cylinder, and let ${F : X \to \mathbb R^n}$ be given by ${F(x,t)=t g(x) + (1-t)h(x)}$ so that ${F}$ linearly interpolates between ${g}$ and ${h}$.
Also, let’s define an antipodality on ${X}$ by ${\nu(x,t) = (-x,t)}$. Note that ${F}$ is antipodal with respect to ${\nu}$, i.e. ${F(\nu(x,t))=-F(x,t)}$, because both ${g}$ and ${h}$ are antipodal. For the sake of contradiction, assume that ${g(x) \neq 0}$ for all ${x \in S^n}$.
Now let’s consider the structure of the zero set ${Z = F^{-1}(0)}$. Certainly ${(N,0), (S,0) \in Z}$ since ${h(N)=h(S)=0}$, and these are h’s only zeros. Here comes the sketchy part: Since ${g}$ is smooth, ${F}$ is also smooth, and thus locally ${F}$ can be approximated by an affine mapping ${F_{\mathrm{loc}} : \mathbb R^{n+1} \to \mathbb R^n}$. It follows that if ${F_{\mathrm{loc}}^{-1}(0)}$ is not empty, then it should be a subspace of dimension at least one. By an arbitrarily small perturbation of the initial ${g}$, ensuring that ${F}$ is sufficiently generic, we can ensure that ${F_{\mathrm{loc}}^{-1}(0)}$ is either empty or a subspace of dimension one. Thus locally, ${F^{-1}(0)}$ should look like a two-sided curve, except at the boundaries ${t=0}$ and ${t=1}$, where ${F^{-1}(0)}$ (if non-empty) would look like a one-sided curve. But, for instance, ${F^{-1}(0)}$ cannot contain any Y-shaped branches.
It follows that ${Z}$ is a union of closed cycles and paths whose endpoints must lie at the boundaries ${t=0}$ and ${t=1}$. (${Z}$ is represented by red lines in the cylinder above.) But since there are only two zeros on the ${t=0}$ sphere, and none on the ${t=1}$ sphere, ${Z}$ must contain a path ${\gamma}$ from ${(N,0)}$ to ${(S,0).}$ Since ${F}$ is antipodal with respect to ${\nu}$, ${\gamma}$ must also satisfy this symmetry, making it impossible for the segment initiating at N to ever meet up with the segment initiating at S. Thus we arrive at a contradiction, implying that ${g}$ must take the value 0.
Notice that the only important property we used about ${h}$ (other than its smoothness) is that is has a number of zeros which is twice an odd number. If ${h}$ had, e.g. four zeros, then we could have two ${\nu}$-symmetric paths emanating from and returning to the bottom. But if ${h}$ has six zeros, then we would again reach a contradiction.
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mersenneforum.org Gaps between non-consecutive primes
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2022-04-29, 23:22 #45 Bobby Jacobs May 2018 32·29 Posts Forbidden prime gap combinations Let an n-prime gap be the gap between a prime p and the prime n primes after p. Then, 2 and 3 are always the start of a maximal n-gap for all n. 5 is the start of a maximal n-gap if and only if the (n+2)nd prime and the (n+3)rd prime are not twin primes. 7 is the start of a maximal n-gap if and only if the (n+3)rd prime and the (n+4)th prime are not twin primes. 11 is the start of a maximal n-gap if and only if the (n+4)th and (n+5)th primes have a gap greater than 4. 13 is the start of a maximal n-gap if and only if the (n+5)th and (n+6)th primes are not twin primes, and the last 2 gaps before the (n+6)th prime are not (2, 4). Basically, every prime has a set of "forbidden prime gap combinations" such that the mth prime is the start of a maximal n-gap if and only if the last gaps before the (m+n)th prime are not one of the forbidden gap combinations. Here are the forbidden gap combinations of the first few primes. Code: 2 [] 3 [] 5 [[2]] 7 [[2]] 11 [[2], [4]] 13 [[2], [2, 4]] 17 [[2], [4]] 19 [[2], [2, 4]] 23 [[2], [4]] 29 [[2], [4], [6]] 31 [[2], [2, 4], [2, 6], [2, 6, 4], [2, 4, 6]] 37 [[2], [4], [6], [2, 4, 8], [2, 4, 2, 10]] 41 [[2], [4], [2, 6], [4, 6], [2, 4, 6, 6], [4, 2, 4, 8], [2, 4, 2, 10]] 43 [[2], [2, 4], [2, 6, 4], [4, 2, 6], [2, 4, 6], [2, 4, 6, 2, 6], [2, 4, 2, 4, 8]] 47 [[2], [4], [4, 2, 4, 6], [4, 2, 4, 6, 2, 6], [4, 2, 4, 2, 4, 8]] Notice that 29 just has the forbidden gaps 2, 4, 6, but 37 has the extra combinations (2, 4, 8) and (2, 4, 2, 10). That is why 29 is more common than 37.
2022-05-01, 14:56 #46
mart_r
Dec 2008
you know...around...
3·11·23 Posts
Quote:
Originally Posted by Bobby Jacobs Here are the forbidden gap combinations of the first few primes.
Very good! That's the sort of analysis I was looking for.
Do you have a program for these gap combinations?
2022-05-11, 16:28 #47 Bobby Jacobs May 2018 4058 Posts Yes. I have a program, but it is slow for primes above 47. We basically want admissible k-tuples where the total of the gaps is less than or equal to the total of the k gaps before p. Let pm be the mth prime. Suppose the k gaps before the (m+n)th prime are one of these forbidden k-tuples. If pm+n-pm+n-k<=pm-pm-k, then pm+n-k-pm-k>=pm+n-pm. Then, the (m-k)th prime will have at least as big of an n-gap as the mth prime. Therefore, the forbidden gaps are minimal admissible k-tuples >= the k gaps before pm.
2022-05-17, 17:02 #48 mart_r Dec 2008 you know...around... 10111101112 Posts CSGmax for p<=1014: Code: k gap CSG_max p 1 766 0.81776202 19581334192423 2 900 0.89182288 21185697626083 3 986 0.92092951 21185697625997 4 1134 0.93874248 66592576389587 5 1170 0.91718026 66592576389551 6 1154 0.89752827 30103357357379 7 1148 0.88499578 14580922576079 8 790 0.92651781 11878096933 9 1316 0.95316163 14580922575911 10 726 0.95096666 866956873 11 754 0.94094924 866956873 12 784 0.93630856 866956873 13 1448 0.95644952 5995661470529 14 1496 0.95744289 5995661470481 15 1322 0.95352216 396016668869 16 1358 0.94653445 396016668833 17 1688 0.98369275 8281634108801 18 1722 0.97105216 8281634108767 19 1812 1.01651543 8281634108677 20 1830 0.98808150 8281634108677 21 2134 1.02168813 78736011999913 22 2148 0.99072269 78736011999913 23 2166 0.96394446 78736011999913 24 2310 1.04764008 78736011999913 25 2322 1.01591301 78736011999901 26 2338 0.98829568 78736011999913 27 2376 0.98009540 78736011999847 28 2432 0.98752862 78736011999791 29 2454 0.96623635 78736011999769 30 2494 0.96053115 78736011999913 31 2478 0.97621396 38986211476403 32 2524 0.97682408 38986211476357 33 2560 0.96892955 38986211476321 34 2286 0.97036452 2481562496471 35 2320 0.96392716 2481562496437 36 2616 0.98341715 17931997861517 37 2396 0.98957750 1933468592177 38 2444 0.99810203 1933468592129 39 2472 0.98638661 1933468592101 40 2538 0.98219566 2481562496219 41 2760 0.98030051 10631985435829 42 2380 0.99919669 327076778191 43 2392 0.97198960 327076778179 44 2442 0.98739166 327076778129 45 2470 0.97842905 327076778101 46 2762 0.97061179 2481562496219 47 2520 0.95456660 327076778051 48 2776 0.94157086 1933468592101 49 3038 0.94152718 10026387088493 50 3092 0.95310074 10026387088439 51 2946 0.94609699 2796148447381 52 2976 0.93822027 2796148447381 53 3450 0.93208471 60681682061173 54 3224 0.92791606 10026387088493 55 3278 0.93965214 10026387088439 56 3096 0.92379571 2481562495661 57 3390 0.94611179 11783179421371 58 3560 0.93957475 29077945916363 59 3808 0.96141677 90210824580841 60 3764 0.95339422 55956455554739 61 3798 0.94719704 55956455554739 62 3852 0.95602954 55956455554651 63 3942 0.99181087 55956455554561 64 3976 0.98566033 55956455554561 65 4004 1.00012038 45921691543349 66 4020 0.98072956 45921691543333 67 4086 0.99893031 45921691543267 68 4140 1.00814094 45921691543213 69 3854 1.02429119 6215409275249 70 4292 1.05955757 45921691543061 71 4310 1.04178765 45921691543043 72 4332 1.02721666 45921691543061 73 4386 1.03648387 45921691543061 74 4062 1.03664125 6215409275041 75 4078 1.01808582 6215409275041 76 4128 1.02764140 6215409275041 77 4150 1.01426227 6215409275407 78 4200 1.02384705 6215409275357 79 4308 1.08099942 6215409275249 80 4328 1.06590295 6215409275249 81 4340 1.04448708 6215409275237 82 4380 1.04597955 6215409275177 83 4414 1.04265662 6215409275143 84 4516 1.09443534 6215409275041 85 4536 1.07968013 6215409275041 86 4548 1.05867025 6215409275029 87 4556 1.03473951 6215409275021 88 4578 1.02218676 6215409275041 89 4596 1.00663763 6215409275041 90 4620 0.99596010 6215409275041 91 4642 0.98385445 6215409275041 92 5020 0.96845804 36683716323913 93 5058 0.97814135 33994032583531 94 5146 1.00067267 36683716323913 95 5194 1.00631376 36683716323913 96 5278 1.03712167 36683716324039 97 5404 1.09772451 36683716323913 98 5418 1.07925696 36683716323899 99 5470 1.08766762 36683716323847 100 5482 1.06802709 36683716323847 101 5526 1.07087308 36683716323791 102 5590 1.08768345 36683716323913 103 5638 1.09332314 36683716323913 104 5656 1.07811268 36683716323847 105 5704 1.08378894 36683716323847 106 5758 1.09362393 36683716323913 107 5772 1.07585272 36683716323899 108 5824 1.08431548 36683716323847 109 5830 1.06128699 36683716323841 And just above 1014, these 22 new records showed up: Code: 10 1528 0.96314466 102591551174059 11 1560 0.94298881 102591551174027 50 3450 0.97333053 102267713449991 51 3480 0.96260938 102267713449991 52 3562 0.99122668 102267713449879 53 3592 0.98063297 102267713449879 54 3634 0.97918812 102267713449807 55 3684 0.98379105 102267713449757 56 3714 0.97357591 102267713449757 57 3768 0.98125523 102267713449673 58 3798 0.97126377 102267713449673 59 3834 0.96582204 102267713449607 60 3874 0.96340363 102267713449567 61 3904 0.95381038 102267713449567 62 3958 0.96169379 102267713449483 66 4186 1.00199403 102267713449117 68 4324 1.03945196 102267713449117 69 4354 1.03013486 102267713449117 76 4658 1.03478754 101562452774609 77 4694 1.03029216 101562452774609 92 5304 1.01634058 102267713449117 93 5328 1.00471893 102267713449093 @ Bobby: I'm working on a program to look for the forbidden gap combinations. If it works, it should be fast enough for primes up to at least 97 (well at least I hope so).
2022-05-19, 17:28 #49 mart_r Dec 2008 you know...around... 3×11×23 Posts It appears my VBA code for "forbidden gap combinations" (for getting a heuristic grip on the generalized maximal gap candidates) works as it should: Code: 5: [ 2] 7: [ 2] 11: [ 2], [ 4] 13: [ 2], [ 2, 4] 17: [ 2], [ 4] 19: [ 2], [ 2, 4] 23: [ 2], [ 4] 29: [ 2], [ 4], [ 6] 31: [ 2], [ 2, 4], [ 2, 6], [ 2, 6, 4], [ 2, 4, 6] 37: [ 2], [ 4], [ 6], [ 2, 4, 8], [ 2, 4, 2, 10] 41: [ 2], [ 4], [ 2, 6], [ 4, 6], [ 2, 4, 6, 6], [ 4, 2, 4, 8], [ 2, 4, 2, 10] 43: [ 2], [ 2, 4], [ 2, 6, 4], [ 4, 2, 6], [ 2, 4, 6], [ 2, 4, 6, 2, 6], [ 2, 4, 2, 4, 8] 47: [ 2], [ 4], [ 4, 2, 4, 6], [ 4, 2, 4, 6, 2, 6], [ 4, 2, 4, 2, 4, 8] 53: [ 2], [ 4], [ 6], [ 2, 4, 2, 4, 8], [ 4, 2, 4, 2, 10], [ 2, 4, 2, 4, 6, 2, 10] 59: [ 2], [ 4], [ 6], [ 4, 8], [ 2, 10], [ 2, 4, 2, 4, 6, 8], [ 2, 4, 2, 4, 6, 2, 6, 10] 61: [ 2], [ 2, 4], [ 2, 6], [ 2, 6, 4], [ 2, 4, 6], [ 2, 6, 6], [ 2, 4, 8], [ 2, 6, 6, 4], [ 2, 6, 4, 6], [ 2, 4, 6, 6], [ 2, 4, 2, 10], [ 2, 4, 2, 4, 8, 6, 4], [ 2, 4, 2, 4, 6, 8, 4], [ 4, 2, 4, 2, 4, 8, 6], [ 2, 4, 2, 4, 6, 2, 10], [ 2, 4, 2, 4, 6, 2, 6, 6, 4, 6], [ 2, 4, 2, 4, 6, 2, 6, 4, 6, 6], [ 2, 4, 2, 4, 6, 2, 6, 4, 2, 10] 67: [ 2], [ 4], [ 6], [ 2, 4, 8], [ 2, 6, 4, 8], [ 2, 4, 6, 8], [ 2, 4, 2, 10], [ 2, 4, 2, 12], [ 2, 6, 4, 2, 10], [ 2, 4, 6, 2, 10], [ 2, 4, 2, 4, 6, 2, 6, 10], [ 2, 4, 2, 4, 6, 2, 6, 4, 2, 4, 12] 71: [ 2], [ 4], [ 2, 6], [ 4, 6], [ 2, 4, 6, 6], [ 4, 2, 4, 8], [ 2, 4, 2, 10], [ 4, 6, 2, 6, 6], [ 2, 6, 4, 6, 6], [ 4, 2, 4, 8, 6], [ 4, 6, 2, 4, 8], [ 4, 2, 6, 4, 8], [ 4, 2, 4, 6, 8], [ 2, 6, 4, 2, 10], [ 2, 4, 6, 2, 10], [ 4, 2, 4, 2, 12], [ 2, 4, 2, 4, 6, 2, 6, 6, 4, 6, 6], [ 2, 4, 2, 4, 6, 2, 6, 4, 6, 6, 6], [ 2, 4, 2, 4, 6, 2, 6, 4, 2, 10, 6], [ 2, 4, 2, 4, 6, 2, 6, 4, 2, 4, 12] 73: [ 2], [ 2, 4], [ 2, 6, 4], [ 4, 2, 6], [ 2, 4, 6], [ 2, 4, 6, 2, 6], [ 2, 4, 2, 4, 8], [ 2, 6, 4, 6, 2, 6], [ 2, 4, 6, 6, 2, 6], [ 2, 4, 2, 10, 2, 6], [ 2, 4, 6, 2, 6, 6], [ 2, 4, 2, 4, 8, 6], [ 2, 4, 2, 4, 6, 8], [ 2, 4, 6, 2, 6, 6, 4], [ 2, 4, 2, 4, 8, 6, 4], [ 2, 4, 2, 4, 6, 8, 4], [ 2, 6, 4, 2, 6, 4, 6], [ 2, 4, 6, 2, 6, 4, 6], [ 2, 6, 4, 2, 4, 6, 6], [ 2, 6, 4, 2, 4, 2, 10], [ 2, 4, 2, 4, 6, 2, 10], [ 2, 4, 2, 4, 6, 2, 6, 6, 4, 6], [ 2, 4, 2, 4, 6, 2, 6, 4, 6, 6], [ 2, 4, 2, 4, 6, 2, 6, 4, 2, 10], [ 2, 4, 2, 4, 6, 2, 6, 4, 2, 4, 6, 8] 79: [ 2], [ 4], [ 6], [ 4, 2, 4, 8], [ 2, 4, 2, 10], [ 2, 4, 2, 4, 6, 8], [ 2, 4, 6, 2, 6, 4, 8], [ 2, 6, 4, 2, 4, 6, 8], [ 2, 4, 2, 4, 6, 2, 10], [ 2, 6, 4, 2, 4, 2, 12], [ 2, 4, 2, 4, 6, 2, 12], [ 2, 4, 6, 2, 6, 4, 2, 10], [ 2, 4, 2, 4, 6, 2, 6, 10], [ 2, 4, 2, 4, 6, 2, 6, 4, 2, 4, 12], [ 2, 4, 2, 4, 6, 2, 6, 4, 2, 4, 6, 6, 8] 83: [ 2], [ 4], [ 2, 6], [ 4, 6], [ 4, 2, 4, 6, 6], [ 2, 4, 2, 4, 8], [ 4, 2, 4, 2, 10], [ 4, 2, 4, 6, 2, 6, 6], [ 4, 2, 4, 2, 4, 8, 6], [ 2, 4, 2, 4, 6, 2, 10], [ 4, 6, 2, 4, 6, 2, 6, 6], [ 4, 2, 4, 6, 6, 2, 6, 6], [ 2, 4, 6, 2, 6, 4, 6, 6], [ 4, 2, 4, 6, 2, 6, 6, 6], [ 2, 4, 2, 4, 6, 2, 10, 6], [ 4, 2, 4, 6, 2, 6, 4, 8], [ 2, 4, 6, 2, 6, 4, 2, 10], [ 2, 4, 2, 4, 6, 2, 6, 10], [ 2, 4, 2, 4, 6, 2, 6, 4, 2, 4, 6, 8] 89: [ 2], [ 4], [ 6], [ 2, 4, 8], [ 4, 2, 10], [ 2, 4, 2, 4, 6, 8], [ 4, 2, 4, 6, 2, 10], [ 4, 2, 4, 6, 2, 6, 4, 8], [ 2, 4, 2, 4, 6, 2, 6, 10], [ 4, 2, 4, 6, 6, 2, 6, 4, 8], [ 4, 2, 4, 2, 4, 8, 6, 4, 8], [ 4, 6, 2, 6, 4, 2, 4, 6, 8], [ 2, 4, 6, 2, 6, 4, 6, 2, 10], [ 2, 6, 4, 2, 4, 6, 6, 2, 10], [ 2, 6, 4, 2, 4, 2, 10, 2, 10], [ 2, 4, 2, 4, 6, 2, 10, 2, 10], [ 2, 4, 6, 2, 6, 4, 2, 6, 10], [ 2, 4, 2, 4, 6, 2, 6, 6, 10], [ 4, 6, 2, 6, 4, 2, 4, 2, 12], [ 2, 4, 6, 2, 6, 4, 2, 4, 12], [ 2, 4, 2, 4, 6, 2, 6, 4, 12], [ 2, 4, 2, 4, 6, 2, 6, 4, 2, 4, 6, 6, 8] 97: [ 2], [ 4], [ 6], [ 8], [ 2, 10], [ 2, 12], [ 2, 6, 10], [ 2, 4, 12], [ 2, 6, 6, 10], [ 2, 4, 8, 10], [ 2, 6, 4, 12], [ 4, 2, 6, 12], [ 2, 4, 6, 12], [ 4, 2, 4, 14], [ 2, 4, 2, 4, 8, 6, 10], [ 2, 4, 2, 4, 6, 8, 10], [ 4, 2, 4, 6, 2, 6, 12], [ 2, 4, 2, 4, 6, 2, 16], [ 2, 4, 2, 4, 6, 2, 6, 6, 4, 12], [ 2, 6, 4, 2, 4, 6, 6, 2, 6, 12], [ 2, 4, 2, 4, 6, 2, 6, 4, 6, 12], [ 2, 4, 2, 4, 6, 2, 10, 2, 4, 14], [ 2, 4, 2, 4, 6, 2, 6, 6, 4, 14], [ 2, 4, 6, 2, 6, 4, 2, 4, 6, 14], [ 2, 4, 2, 4, 6, 2, 6, 4, 6, 14], [ 2, 4, 2, 4, 6, 2, 6, 4, 2, 16], [ 2, 4, 2, 4, 6, 2, 6, 4, 2, 18], [ 2, 4, 6, 2, 6, 4, 2, 4, 6, 8, 10], [ 2, 4, 6, 2, 6, 4, 2, 4, 2, 12, 10], [ 2, 4, 6, 2, 6, 4, 2, 4, 6, 6, 12], [ 2, 4, 6, 2, 6, 4, 2, 4, 2, 10, 12] Computation time: less than a minute, but I believe it is possible to do it in less than a second with some really optimised code.
2022-05-22, 15:15 #50 Bobby Jacobs May 2018 32×29 Posts Very good! It seems like the popularity of a prime as the start of a maximal gap is based upon the gaps before p. However, there are weird exceptions like 29 and 37. The 2 gaps before 29 are (4, 6) and the gaps before 37 are (2, 6), but 29 is more popular than 37. If (2, 8) was an admissible gap combination, then that would be a forbidden gap combination for 29, but not 37. However, (2, 8) is not admissible. By the way, I meant to use <= instead of >= in my previous post. The forbidden gap combinations are minimal admissible k-tuples <= the k gaps before pm. How do I make the correct symbols for <= and >=?
2022-06-02, 20:15 #51
mart_r
Dec 2008
you know...around...
10111101112 Posts
I have finally fully figured out how to tackle the behaviour of $$o_n(x)$$ - i.e. the number of occurrences of primes $$p_n$$ as initial members of maximal gaps between non-consecutive primes $$p_n$$ and $$p_{n+k}$$ for all $$k<=x$$.
Again, sincere thanks to Bobby for pushing me in the right direction. Although, "forbidden gap constellations" sounds kind of illegal, anyone mind if I call them "blocking patterns" or similar? Suggestions are welcome.
So, let $$B(p_n)$$ be the set of blocking patterns for the prime $$p_n$$, for example $$B(31)=\lbrace\lbrace0,2\rbrace,\lbrace0,2,6\rbrace,\lbrace0,2,8\rbrace,\lbrace0,2,8,12\rbrace,\lbrace0,2,6,12\rbrace\rbrace$$. (Correspondingly, the blocking gap patterns are $$\lbrace\lbrace2\rbrace,\lbrace2,4\rbrace,\lbrace2,6\rbrace,\lbrace2,6,4\rbrace,\lbrace2,4,6\rbrace\rbrace$$.)
These patterns form a minimal set of sorts. I got temporarily addicted to try and find as many of them as possible. Much to my surprise, I recently even managed to get up to p=97 in less than a second even though my code is far from being optimised, but computation time is ballooning exponentially for larger p. The list in the attachment is not guaranteed to be exhaustive.
To evaluate $$o_n(x)$$ directly, we subtract from x the number of occurrences of all patterns in $$B(p_n)$$ in the range [$$p_{n-k+2}$$, $$p_{n+x}$$], where k is the cardinality of the pattern.
By looking at the table of blocking patterns, we can now see right away, for example, that 29 occurs more often than 37 for large x by a margin equivalent to the number of occurrences of the patterns {0,2,6,14} and {0,2,6,8,18} below x. This answers post # 44.
$$o_n(x)$$ remains large if there are very few blocking patterns. $$n=2$$ has none because $$B(p_n)=B(3)=\lbrace0,1\rbrace$$, a non-admissible prime pattern for $$p>=3$$, hence $$o_1(x)=o_2(x)=x$$.
$$n=3$$ and $$n=4$$ only have $$\lbrace0,2\rbrace$$ as blocking patterns, all larger n have at least one pattern more, thus $$o_4(x)>o_n(x)$$ for all $$n>4$$ and $$x>18$$ (particularly, $$o_4(x)=x+2-\#(twin\:primes\:below\:p_{4+x})$$).
There's $$n=8$$ with $$B(19)=\lbrace\lbrace0,2\rbrace,\lbrace0,2,6\rbrace\rbrace$$, a minimum for its kind, only twins and the first kind of triplets are blocked, and all $$n>8$$ have at least one more blocking pattern, or one that is more common, like $$\lbrace0,6\rbrace$$. We have $$o_8(x)>o_n(x)$$ for all $$n>8$$ and $$x>496$$ (i.e. p=19 "cannot be beaten" above that point).
The asymptotic growth rate of $$o_n(x)$$ can be obtained via the blocking patterns with additional regard to the open residue classes in each pattern. For x large, $$o_{29}(x)$$ ($$p_{29}=109$$) differs from $$o_8(x)$$ only by a margin of the number of occurrences of sextuples $$\lbrace0,2,8,12,14,20\rbrace$$, $$\lbrace0,2,6,12,14,20\rbrace$$, $$\lbrace0,2,6,8,12,18\rbrace$$, and $$\lbrace0,2,6,8,12,20\rbrace$$ with a total of 8 open residue classes mod 210; in terms of error this is $$O(x\cdot(\log x)^{-6})$$. We can leave septuples or longer patterns out of the game as the have $$O(x\cdot(\log x)^{-7})$$ or smaller. If p=109 should be beaten in the long run, it requires, apart from the minimum of $$\lbrace0,2\rbrace$$ and $$\lbrace0,2,6\rbrace$$ as blocking patterns, either sextuples with less open residue classes in total, or no sextuples at all. And of course, no quadruple or quintuple blocking pattern as well. The next candidate for this is p=5659: only $$\lbrace0,2,6,8,12,18\rbrace$$ gets blocked, and this pattern has only one open residue class mod 210.
Regarding p=5659 vs. p=9439 (my fallacy in post # 40), the latter seems to be in the lead judging by the small numbers because of the millions of possible blocking patterns in favor of p=9439, but these have a cardinality of as small as 5. At $$x=10^6$$, p=9439 is in the lead by more than 30,000 - it takes at least as many quintuples of the forms $$\lbrace0,4,6,12,16\rbrace$$, $$\lbrace0,4,6,10,16\rbrace$$, $$\lbrace0,6,8,14,18\rbrace$$, $$\lbrace0,2,8,14,18\rbrace$$, $$\lbrace0,6,10,12,18\rbrace$$, $$\lbrace0,4,10,12,18\rbrace$$, $$\lbrace0,6,8,12,18\rbrace$$, $$\lbrace0,2,8,12,18\rbrace$$, $$\lbrace0,2,6,12,18\rbrace$$, $$\lbrace0,4,6,10,18\rbrace$$, or $$\lbrace0,2,6,8,18\rbrace$$ until p=5659 can overtake p=9439, we expect this not to happen before $$x=10^8$$.
To conclude, the primes for which a local maximum as described in post # 35 is reached for $$\lim x\to\infty$$, or rather, for sufficiently large x, should be equal to $$2\:(3), 7,$$ and $$19$$, with infinitely many $$o_n(x)$$ for $$n>8$$ coming arbitrarily close to $$o_8(x)-O(x\cdot(\log x)^{-6})$$ (e.g. the primes $$5659$$ ($$n=746$$), $$88819$$ ($$n=8605$$), $$855739$$ ($$n=68032$$), $$74266279$$ ($$n=4353833$$), $$964669639$$ ($$n=49141276$$), $$9853497769$$ ($$n=448687813$$), etc. each move toward this upper bound from below). So my previous implicit assumption that the list of primes with local maxima, bounded from above, is infinite was wrong.
Phew, that took long enough. And it's only framework, sort of. Also poorly worded at times, but I really need to finish this off now, one way or another.
I flip out if now someone gives me a link to some obscure 19th century work that covers all this...
Attached Files
blocking_gap_patterns.txt (80.5 KB, 10 views)
2022-06-09, 17:26 #52 mart_r Dec 2008 you know...around... 3·11·23 Posts "Sitting target / sitting, waiting / anticipating / nothing / nothing." CSG looks well-behaved even for k <= 1024 (p in range < 10^14): Code: k gap CSG_max p 112 5940 1.05550107 36683716323847 116 6052 1.02516052 36683716323619 120 6220 1.03269957 36683716323283 124 6388 1.04043735 36683716323283 128 6510 1.01858817 36683716323161 132 6642 1.00390061 36683716323167 136 6742 0.96976743 36683716323109 140 6658 0.94384648 17674627574311 144 6840 0.96488178 17674627574369 148 6992 0.96688912 17674627574141 152 7126 0.95779790 17674627574083 156 7460 0.97452838 30512335335437 160 7614 0.97643675 30512335335437 164 7732 0.95708911 30512335335319 168 7946 0.99499058 30512335334951 172 8100 0.99726110 30512335334797 176 8254 0.99967015 30512335334797 180 8364 0.97661645 30512335335299 184 8510 0.97483498 30512335335059 188 8736 1.01921576 30512335334927 192 8892 1.02319197 30512335334771 196 9004 1.00215904 30512335334797 200 9148 1.01448166 28330683392731 204 9324 1.03039259 28330683392659 208 9492 1.04177291 28330683392597 212 9630 1.03626675 28330683392353 216 9778 1.03654152 28330683392371 220 9856 0.99828866 28330683392371 224 9974 0.98269325 28330683392147 228 10058 0.94929275 28330683392129 232 8294 0.94835143 185067241757 236 9700 0.95641246 5185992136441 240 9850 0.96394780 5185992136453 244 10626 0.94205155 28330683392597 248 10818 0.96644566 28330683392371 252 10596 0.94771341 12666866223047 256 11310 0.93908073 52248744686339 260 11476 0.94818065 52248744686197 264 11604 0.93866201 52248744686069 268 11724 0.92547977 52248744686197 272 11264 0.93001522 12666866223047 276 12106 0.91167574 68182243872601 280 11752 0.91251084 21947823205027 284 11920 0.92535164 21947823205027 288 12096 0.94216306 21947823204943 292 12310 0.94178965 25698372297889 296 12460 0.94533825 25698372297691 300 12704 0.99505355 25698372297029 304 12920 1.03124573 25698372297029 308 13170 1.08475215 25698372297029 312 13308 1.08194976 25698372297029 316 13482 1.09729159 25698372297029 320 13616 1.09257083 25698372296963 324 13728 1.07704353 25698372296873 328 13878 1.08048911 25698372297029 332 13986 1.06336781 25698372297007 336 14136 1.06693878 25698372296963 340 14234 1.04538481 25698372296873 344 14336 1.02612587 25698372296243 348 14466 1.02043217 25698372295733 352 14642 1.03657564 25698372297029 356 14778 1.03379163 25698372295733 360 14890 1.01983645 25698372295711 364 15044 1.02563857 25698372296963 368 15222 1.04265118 25698372295019 372 15360 1.04099593 25698372294839 376 15546 1.06172759 25698372295033 380 15694 1.06474083 25698372294457 384 15832 1.06313720 25698372294409 388 15968 1.06066576 25698372294611 392 16158 1.08316800 25698372294421 396 16242 1.05682221 25698372294337 400 16344 1.03908815 25698372294563 404 16536 1.06228245 25698372294457 408 16678 1.06277691 25698372294421 412 16762 1.03722225 25698372294337 416 16852 1.01477000 25698372294457 420 16974 1.00672310 25698372295033 424 17160 1.02688378 25698372294421 428 17302 1.02766540 25698372294421 432 17396 1.00751345 25698372294611 436 17586 1.02931311 25698372294421 440 17724 1.02843581 25698372294409 444 17810 1.00513477 25698372294323 448 17886 0.97797632 25698372294253 452 17972 0.95549900 25698372294281 456 18114 0.95669262 25698372293557 460 18234 0.94875594 25698372293809 464 18390 0.95581939 25698372293597 468 18536 0.95873888 25698372293597 472 19506 0.94029942 93152147737543 476 19770 0.98628553 93152147737279 480 19878 0.97192308 93152147737199 484 19954 0.94554667 93152147737237 488 19192 0.94334862 25698372294421 492 20322 0.97552201 93152147736727 496 20490 0.98440496 93152147736559 500 20598 0.97040570 93152147736451 504 20748 0.97245252 93152147736301 508 20850 0.95643446 93152147736199 512 21004 0.96004428 93152147736073 516 21260 1.00210684 93152147735789 520 21390 0.99658464 93152147735659 524 21478 0.97538922 93152147735599 528 21592 0.96413821 93152147735371 532 21726 0.96039757 93152147735351 536 21874 0.96185682 93152147735203 540 21964 0.94210589 93152147735113 544 22076 0.93058797 93152147734973 548 22224 0.93216222 93152147733739 552 22486 0.97513465 93152147732647 556 22628 0.97445048 93152147734421 560 22792 0.98180763 93152147734171 564 22958 0.98989507 93152147734091 568 23130 1.00017583 93152147733919 572 23346 1.02661845 93152147733703 576 23524 1.03914014 93152147733553 580 23610 1.01786364 93152147733467 584 23706 1.00050490 93152147733553 588 23912 1.02309904 93152147733137 592 24068 1.02754489 93152147732981 596 24240 1.03781136 93152147732723 598 24402 1.07077256 93152147732647 600 24436 1.05684370 93152147732641 604 24540 1.04234724 93152147732509 608 24676 1.03956218 93152147732401 612 24798 1.03177448 93152147732251 616 24880 1.00980556 93152147732197 620 25008 1.00437652 93152147732069 624 25164 1.00888491 93152147731913 628 25264 0.99368357 93152147731813 632 25348 0.97310855 93152147731729 636 25500 0.97626664 93152147731549 640 25578 0.95395507 93152147731499 644 25696 0.94554893 93152147731381 648 25860 0.95284382 93152147731217 652 26004 0.95333346 93152147731073 656 26252 0.98935642 93152147730797 660 26412 0.99528904 93152147730637 664 26606 1.01294463 93152147730443 668 26706 0.99822651 93152147730371 672 26826 0.99048613 93152147730223 676 26938 0.98010919 93152147730139 680 27094 0.98468729 93152147729983 684 27186 0.96770698 93152147729891 688 27276 0.95027202 93152147729983 692 27368 0.93371537 93152147729891 696 27516 0.93569220 93152147729561 700 27582 0.91092486 93152147729467 704 27698 0.90261024 93152147729561 708 27820 0.89629956 93152147729143 712 27948 0.89196235 93152147729143 716 28048 0.87877867 93152147729143 720 27710 0.89275685 54116590394771 724 27860 0.89636501 54116590394621 728 27998 0.89606095 54116590394483 732 28172 0.90750139 54116590393157 736 28332 0.91437757 54116590394149 740 28536 0.93570243 54116590393991 744 28666 0.93272540 54116590393861 748 28800 0.93108486 54116590393777 752 28982 0.94518150 54116590393499 756 29130 0.94812537 54116590393447 760 29370 0.98143984 54116590393157 764 29456 0.96387023 54116590393121 768 29630 0.97537452 54116590392947 772 29706 0.95469068 54116590393157 776 29826 0.94853391 54116590392947 780 29964 0.94825668 54116590392929 784 30076 0.93959664 54116590392451 788 30192 0.93229670 54116590392947 792 30288 0.91869099 54116590392289 796 30456 0.92805971 54116590392121 800 30654 0.94703768 54116590391873 804 30740 0.93024784 54116590391837 808 30816 0.91051339 54116590391861 812 30990 0.92171618 54116590391873 816 31176 0.93673527 54116590391351 820 31368 0.95371131 54116590391113 824 31516 0.95671400 54116590391011 828 31596 0.93820121 54116590391011 832 31734 0.93807554 54116590391113 836 31852 0.93169439 54116590391011 840 31936 0.91480473 54116590391077 844 32062 0.91104489 54116590391077 848 32158 0.89810577 54116590391011 852 32880 0.89162466 93152147732647 856 33006 0.88736277 93152147732641 860 32594 0.90483328 54116590389887 864 32714 0.89936117 54116590389863 868 32790 0.88065023 54116590389887 872 32960 0.89035364 54116590389887 876 33068 0.88139835 54116590389473 880 33158 0.86716391 54116590389419 884 33276 0.86135519 54116590389863 888 33420 0.86326354 54116590389473 892 33550 0.86104325 54116590388977 896 33738 0.87595633 54116590388789 900 34052 0.86261430 65480290959731 904 34264 0.88413463 65480290959547 908 34380 0.87757423 65480290959403 912 34474 0.86468039 65480290959403 916 35030 0.86491644 93152147730443 920 35156 0.86099754 93152147730497 924 34932 0.87719975 65480290958651 928 35160 0.90333082 65480290958651 932 35254 0.89038566 65480290958557 936 35504 0.89448345 70981263873617 940 35646 0.89544859 70981263873617 944 35654 0.88610601 65480290958129 948 35814 0.89239499 65480290957997 952 36080 0.90067205 70981263873617 956 36204 0.89647169 70981263873617 960 36294 0.88259124 70981263872969 964 32722 0.87857798 3529553758999 968 36550 0.87667128 70981263873109 972 36728 0.88790229 70981263872969 976 36864 0.88721696 70981263872257 980 37006 0.88823910 70981263872257 984 37086 0.87179868 70981263872257 988 37224 0.87172780 70981263872257 992 37440 0.89358331 70981263872257 996 37564 0.88954266 70981263872257 1000 36346 0.87913969 25264345114117 1004 36534 0.89520659 25264345113919 1008 36604 0.87646499 25264345113919 1012 36740 0.87719292 25264345113713 1016 36876 0.87792446 25264345113613 1020 37294 0.87603752 31618998499597 1024 37074 0.85792826 25264345113613 Update on blocking patterns (see previous post): p = 157: 1195 patterns on my watch p = 163: at least 2125 patterns p = 167: at least 4000 patterns p = 173: at least 5733 patterns p = 179: at least 7357 patterns p = 181: at least 16345 patterns p = 191: at least 11710 patterns But the number of patterns is not terribly important (and probably impossible to compute in full for p > 179 or 181) - for some decent comparisons between values of on(x), it should be sufficient to know the patterns with cardinality <= 7 or 8 or thereabouts, these are not too hard to figure out if p is not too large. This is getting boring, I'm going to watch some episodes of PJ Masks now.
2022-06-12, 18:13 #53
Bobby Jacobs
May 2018
32·29 Posts
Quote:
Originally Posted by Bobby Jacobs What is the pattern with the sequence of primes with record low numbers of occurrences? It seems like the sequence is 2, 5, 11, 29, 37, 59, 97, 127, 223, 307, 541, 907, 1151, 1361, ... This is similar to the primes at the end of maximal prime gaps, but not exactly. I wonder what the pattern is.
I believe that 223 will eventually beat 127. It starts out behind because there are two consecutive gaps of 12 before 223. However, 223 will eventually catch up with 127 because 127 has a gap of 14. Therefore, the sequence of record lows will start 2, 5, 11, 29, 37, 59, 97, 127, 307, 541, 907, 1151, 1361, ...
2022-06-13, 18:00 #54
mart_r
Dec 2008
you know...around...
3·11·23 Posts
Quote:
Originally Posted by Bobby Jacobs I believe that 223 will eventually beat 127. It starts out behind because there are two consecutive gaps of 12 before 223. However, 223 will eventually catch up with 127 because 127 has a gap of 14. Therefore, the sequence of record lows will start 2, 5, 11, 29, 37, 59, 97, 127, 307, 541, 907, 1151, 1361, ...
That's what I would assume as well. These would be the primes at the end of a maximal gap, including ones where there is a tie to the previous maximal gap, if the blocking patterns cover more common patterns.
My search is still running, slowly approaching 2e14 for k <= 109.
Does anyone think CSG > 1.1 is possible to find?
Last fiddled with by mart_r on 2022-06-13 at 18:04
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𝛢𝛼 𝛣𝛽 𝛤𝛾 𝛥𝛿 𝛦𝜀𝜖 𝛧𝜁 𝛨𝜂 𝛩𝜃𝜗 𝛪𝜄 𝛫𝜅 𝛬𝜆 𝛭𝜇 𝛮𝜈 𝛯𝜉 𝛰𝜊 𝛱𝜋 𝛲𝜌 𝛴𝜎𝜍 𝛵𝜏 𝛶𝜐 𝛷𝜙𝜑 𝛸𝜒 𝛹𝜓 𝛺𝜔
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# The sum of the four numbers which are in arithmetic progression is 20 and product is 625. Find the numbers?
1
by lkjh07
let 4 no be a-3d,a-d,a+d,a+3d
Comment has been deleted
## Answers
2014-04-04T22:54:14+05:30
Let us take the the four values as a-2d,a-d,a+d,a+2d
Sum of these numbers is
a-2d+a-d+a+d+a+2d=20
we can find a values by cancelling d's so a=5
now product of these values is
(a-2d)*(a-d)*(a+d)*(a+2d)=625
taking a common we get
a[(1-2d)*(1-d)*(1+d)*(1+2d)]=625
since a=5;
after computing we get
1-5d^2+4d^4=125
4d^4-5d^2=124
we can find the d value and on substituting that we can get the four values due to time i cant complete the problem but this is the concept.
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# all integers are rational numbers
0, 1, 2, 3, 4â¦, Integers include all whole numbers and their negative counterpart e.g. Accordingly, it is a rational number. Does whmis to controlled products that are being transported under the transportation of dangerous goodstdg regulations? Integers include all whole numbers and their negative counterpart e.g. So they are 1, 2, 3, 4, 5, ... (and so on). "All whole numbers are rational numbers" Answer : True. For instance, we can write 1.5 as 3/2, 6/4, 9/6, and more. 2 See answers morganarmstrong380 morganarmstrong380 Answer: D. Step-by-step explanation: i hope this helps. Natural numbers are all numbers 1, 2, 3, 4⦠They are the numbers you usually count and they will continue on into infinity. ⢠All integers are rational numbers. 0, 1, 2, 3, 4â¦. Opposites have the same absolute value since they are both at the same distance from 0. The number 4 is an integer as well as a rational number. For integers the denominator is 1. .9 is also a rational number because .9 comes from the ratio 9/10. quotient of two integers, the divisor not being zero. A Rational Number can be made by dividing two integers. This is true Each integer can be written as a fraction of two integers. Like the integers, the rational numbers are closed under addition, subtraction, and multiplication. So integers are actually a subset of rational numbers (cases when the numerator is a multiple of the denominator). All Rights Reserved. A) Always! All integers are rational numbers, not all rational numbers are integers. Hereâs a hint: if youâre working with a number with a long line of different decimals, then your number is irrational! As it can be written without a decimal component it belongs to the integers. Integers include all whole numbers and their negative counterpart e.g. All fractions are rational numbers. Clearly, 3/2,-5/3, etc. 5 is an integer, 2/3 is a fraction, both are rational. (true or false) All rational numbers are whole numbers. However, numbers like 1/2, 45454737/2424242, and -3/7 are also rational, since they are fractions whose numerator and denominator are integers. â¦-4, -3, -2, -1, 0,1, 2, 3, 4,â¦, All integers belong to the rational numbers. If two numbers are at the same distance from 0 as in the case of 10 and -10 they are called opposites. Integers and rational numbers. All the Whole numbers, negative numbers and also zero are integers. Therefore, the rational numbers are closed under division. What is the balance equation for the complete combustion of the main component of natural gas? First, the number '1' is an integer. Thus, every integer is a rational number. Moreover, several floating-point numbers can be expressed as fractions. When did organ music become associated with baseball? conclude that all natural numbers, all whole numbers and all integers are also rational numbers. In other words, it is a fraction whose denominator is not zero, and both the denominator and numerator are integers. Write this statement in the form x, if then , but x such that . Write these numbers under the appropriate category. The material on this site can not be reproduced, distributed, transmitted, cached or otherwise used, except with prior written permission of Multiply. Real numbers include the integers (Z). It is shown with the symbol. Rational numbers can be expressed as fractions, p/q, where q is not equal to zero. A rational number is a number that can. In other words, most numbers are rational numbers. Write the given ⦠$$\left | 10 \right |=10=\left | -10 \right |$$. Let Ratl(x) be x is a rational number and Int(x) be x is an integer. is False, so the rules of logic allow any conclusionâmeaning the implication is True. 1. a. Since the For integers the denominator is 1. jojorereference jojorereference Answer: d. Step-by-step explanation: New questions in Mathematics. It is a rational number because it can be written as: A rational number written in a decimal form can either be terminating as in: All rational numbers belong to the real numbers. ad/bc is represented as a ratio of two integers, which is the exact definition of a rational number. All integers are rational numbers, not all rational numbers are integers. The answer is . Yes. Hence, they are also rational numbers. No, not all rational numbers are integers. What did women and children do at San Jose? All integers are rational numbers. Whole numbers are a subset of the set of rational numbers and can be written as a ratio of the whole number to 1. Real numbers (R) include all the rational numbers (Q). Why don't libraries smell like bookstores? Integers involves the natural numbers(N). 2. How much money do you start with in monopoly revolution? all integers are rational numbers but all rational numbers need not be integers give an example to satisfy the following statement 2 See answers cathrin12 cathrin12 AN INTEGER IS LIKE 1,2,3 SO THEY CAN BE WRITTEN IN P/Q FORM BUT RATIONAL NO. These numbers include all the above (natural, whole, integers) PLUS some types of fraction/decimal. ðNerdstudy.com - check out our website for the most clear and detailed math lessons! They are the numbers you usually count and they will continue on into infinity. However, the premise of ... ⢠0.278254â¦.. is a terminating decimal, therefore it is a rational number. "Some integers are irrational numbers" Answer : False. Rational numbers can be expressed as fractions, p/q, where q is not equal to zero. Below diagram helps us to understand more about the number sets. Mathplanet is licensed by Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 Internationell-licens. Find an answer to your question âThere are whole numbers that are not integers. All integers are rational numbers. A number can be considered as a rational number if it express as the ratio of two integers. why is Net cash provided from investing activities is preferred to net cash used? Any rational number, expressed as the quotient of an integer a and a (non-zero) natural number b, satisfies the above definition because x = a / b is the root of a non-zero polynomial, namely bx â a.; The quadratic surds (irrational roots of a quadratic polynomial ax 2 + bx + c with integer coefficients a, b, and c) are algebraic numbers. They are the numbers you usually count and they will continue on into infinity. Rational numbers are numbers that can be represented as fractions. Both 4 and 5 are rational numbers. The set of rational numbers includes all integers and all fractions. In mathematics, a rational number is a number such as -3/7 that can be expressed as the quotient or fraction p/q of two integers, a numerator p and a non-zero denominator q. How long will the footprints on the moon last? When did sir Edmund barton get the title sir and how? Because you can't \"count\" zero. 0, 1, 2, 3, 4â¦. All integers are rational numbers. Because an integer should not have any decimals, they should be a whole number. The thing that sets integers apart from these first two groups is that they also include negative numbers. 5 is an integer, 2/3 is a fraction, both are rational. Who is the longest reigning WWE Champion of all time? What was the weather in Pretoria on 14 February 2013? A rational number can be expressed in the form where and are integers and. All integers are rational numbers. divisor of an integer is 1, then all integers are rational. This distance between a number x and 0 is called a number's absolute value. Also check out all of our Shakespeare lessons! b. Natural numbers are all numbers 1, 2, 3, 4â¦. True. where a and b are both integers. ⢠Terminating decimals are rational numbers. So 5 for example can be written as 5/1. So, what makes a fraction/decimal rational? Answers: 1 Show answers Integers include all natural and whole numbers. All rational numbers are algebraic. Correct answers: 2 question: Which of the following is not true regarding sets of numbers? Which of these rational numbers are integers? The ancient greek mathematician Pythagoras believed that all numbers were rational, but one ⦠Since the denominator can be equal to 1, all integers are rational numbers. a. all whole numbers are rational numbers b. all integers are rational numbers c. all rational numbers are integers d. all irrational numbers are real numbers ⦠Another way to say this is that the rational numbers are closed under division. Every whole numberis a rational number because every whole number can be expressed as a fraction. All integers are whole numbers, but a non-whole number can be rational if the numbers after the decimal point either 1. end or 2. repeat. (An integer is a number with no fractional part.) â¦-4, -3, -2, ⦠So the set of all rational numbers will contain the numbers 4/5, -8, 1.75 (which is 7/4), -97/3, and so on. All whole numbers are integers; Negative numbers are integers only; Rational Numbers. (true or false) All integers are ...â in ð Mathematics if you're in doubt about the correctness of the answers or there's no answer, then try to use the smart search and find answers to the similar questions. Explain your choice. If 1.) This is because of two main facts. Counting Numbers are Whole Numbers, but without the zero. Whole numbers are all natural numbers including 0 e.g. Discovering expressions, equations and functions, Systems of linear equations and inequalities, Representing functions as rules and graphs, Fundamentals in solving equations in one or more steps, Ratios and proportions and how to solve them, The slope-intercept form of a linear equation, Writing linear equations using the slope-intercept form, Writing linear equations using the point-slope form and the standard form, Solving absolute value equations and inequalities, The substitution method for solving linear systems, The elimination method for solving linear systems, Factor polynomials on the form of x^2 + bx + c, Factor polynomials on the form of ax^2 + bx +c, Use graphing to solve quadratic equations, Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 Internationell-licens. In other words, any integer a can be written as a = a/1, which is a rational number. Copyright © 2021 Multiply Media, LLC. A number is rational if we can write it as a fraction, where both denominator and numerator are integers. Consider the statement All integers are rational numbers but some rational numbers are not integers. Question 6 : Tell whether the given statement is true or false. A rational number is a number that can be expressed as the ARE ALWAYS IN THE P/Q FORM AND THE NO. â¦-4, -3, -2, ⦠4. Examples. are rational numbers but they are not integers. Rational numbers are which can be written in a p/q form. So all integers are rational (but not all rational numbers are integers). Integers and rational numbers. Some examples of such fractions are You can always express an integer as a fraction, for example itself over 1. Natural numbers are all numbers 1, 2, 3, 4â¦. Some integers are not rational numbers. Whole numbers are all natural numbers including 0 e.g. a/b and c/d are rational numbers, meaning that by definition a, b, c, and d are all integers. Furthermore, when you divide one rational number by another, the answer is always a rational number. âRationalâ comes from the word âratio.â In this case, the ratio is made up of integers. Do This Consider the following collection of numbers 1, 1 2, â 2, 0.5, 4 1 2, 33 7 â, 0, 4 7, 0.3 , 22, â 5, 2 19, 0.125. Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website! You notice that all integers, as well as all rational numbers, are at a specific distance from 0. -2/5 is a rational no.but it is not an INTEGER All integers are rational numbers true or false explain. Some fractions, however, may contain a numerator or denominator that is not an integer. This equation shows that all integers, finite decimals, and repeating decimals are rational numbers. 3. Since integers are rational numbers, you can be assured that the negative counterparts of natural and whole numbers are considered rational, too. A rational number is a number. Hence, every integer is a rational number but a rational number need not be an integer. Rational Numbers are everything that includes decimals, integers (positive and negative whole numbers), fractions -- the numbers after the decimal should terminate or be repetitive. Whole numbers are all natural numbers including 0 e.g. So all the integers are rational numbers but not all the rational numbers are integers. Since the integers are closed under multiplication, ad and bc are also integers. , but x such that groups is that the negative counterparts of natural and whole and... Main component of natural and whole numbers are at the same distance from 0 clear detailed! In this case, the premise of... ⢠0.278254â¦.. is a no.but... Monopoly revolution they should be a whole number can be written without a component! Such that not integers are called opposites Each integer can be written in a p/q form that definition... Such that âThere are whole numbers and their negative counterpart e.g,... ( and on! Of 10 and -10 they are 1, 2, 3, 4⦠should not have any,! Of a rational number 1 ' is an integer rational numbers are all natural numbers including 0 e.g it a! A specific distance from 0 your website a ratio of the whole.. We can write it as a rational number is rational if we can it! So on ) | on the moon last ratio 9/10 the. Natural, whole, integers include all whole numbers are considered rational, too that also!, 45454737/2424242, and d are all natural numbers are integers 4 is an integer as a number! To Net cash used the moon last of all time this distance between a number that can be without!: D. Step-by-step explanation: New questions in Mathematics footprints on the moon last whole... Balance equation for the most clear and detailed math lessons count\ ''.! Integers ; negative numbers and also zero are integers only ; rational numbers, x. If youâre working with a number that can be assured that the numbers... 1, 2, 3, 4⦠cash provided from investing activities is preferred to Net cash used of allow... Integer rational numbers can be represented as a fraction whose denominator is not equal to zero between! Should be a whole number to 1 be x is a rational number true Each integer can be written a... And can be written in a p/q form and the No some integers are numbers... And both the denominator ) fractions, p/q, where q is not integer. Numerator and denominator are integers notice that all integers are rational number x and 0 is called a with! 4¦, integers include all whole numbers, are at the same distance from 0 of 10 and -10 are... -3/7 are also integers terminating decimal, therefore it is not true regarding sets of numbers answers morganarmstrong380 Answer! Denominator that is not an integer as a ratio of two integers,. Addition, subtraction, and repeating decimals are rational numbers and their negative counterpart e.g See answers morganarmstrong380 Answer... Hence, every integer is 1, 2, 3, 4⦠why Net. Fractions, however, the divisor of an integer as well as all rational numbers are all numbers 1 2! Allow any conclusionâmeaning the implication is true Pretoria on 14 February 2013 how long will the footprints on the last. Floating-Point numbers can be written without a decimal component it belongs to the integers, is., since they are the numbers you usually count and they will continue on infinity... ; negative numbers, finite decimals, they should be a all integers are rational numbers can., we can write 1.5 as 3/2, 6/4, 9/6, and are... Are not integers Each integer can be written as a fraction, both are numbers. A fraction, for example itself over 1 sir and how false explain the of. Are the numbers you usually count and they will continue on into infinity long! The thing that sets integers apart from these first two groups is that negative... And numerator are integers of rational numbers are integers ; negative numbers and their counterpart! Where and are integers ; negative numbers and their negative counterpart e.g true Each can... Under the transportation of dangerous goodstdg regulations i hope this helps be in... Decimal, therefore it is a rational number can be expressed as fractions statement in the form x if. Above ( natural, whole, integers include all whole numbers are a subset of numbers! Net cash used 3, 4â¦, integers ) PLUS some types of fraction/decimal c and... Any decimals, then all integers, which is the longest reigning Champion! The divisor of an integer should not have any decimals, then your number is rational if we can 1.5... The Answer is always a rational number and all fractions that is not equal to.... Into infinity website for the most clear and detailed math lessons and 0 is called number. Be made by dividing two integers form where and are integers other words, it is equal... Premise of... ⢠0.278254â¦.. is a number x and 0 is called a number with a number No... To the integers are closed under addition, subtraction, and multiplication 35256 ) ( Show Source:! A whole number to 1, all integers, the Answer is a! Then all integers are irrational numbers '' Answer: D. Step-by-step explanation i! Since the divisor of an integer is a rational number can be written in a p/q.... With a long line of different decimals, and more and -10 they are fractions whose numerator denominator! ) all rational numbers much money do you start with in monopoly?. Are rational numbers includes all integers are rational numbers are integers, b, c and... All integers are not integers into infinity and they will continue on into infinity February?. Long will the footprints on the moon last logic allow any conclusionâmeaning the is. And they will continue on into infinity barton get the title sir and how zero are integers numbers... It is a fraction, for example can be equal to zero is not equal to.... As well as a fraction, where q is not equal to zero dividing two integers numbers! Statement is true or false ) all rational numbers '' Answer:.., are at a specific distance from 0 as in the p/q form and the No D.. Sets integers apart from these first two groups is that they also include negative.. N'T \ '' count\ '' zero ⦠Correct answers: 2 question: which the. The quotient of two integers ca n't \ '' count\ '' zero it to. Is represented as a fraction, for example can be expressed as a ratio of integers. In monopoly revolution hereâs a hint: if youâre working with a number can... Question 6: Tell whether the given statement is true or false word âratio.â in this case, divisor... Integer should not have any decimals, and more subset of rational numbers ( and so on.! Have any decimals, and more false explain | whose numerator denominator. Opposites have the same absolute value since they are both at the same distance from 0 not equal zero! Numbers true or false ratio 9/10 and so on ) are being under! Some fractions, p/q, where q is not equal to zero: 2 question: which of whole. You can put this solution on your website the rational numbers check out our for! ) all rational numbers can be made by dividing two integers numerator are integers 0... ¦-4, -3, -2, ⦠some integers are rational numbers is... Explanation: New questions in Mathematics is preferred to Net cash provided from investing activities is preferred to cash... An Answer to your question âThere are whole numbers and their negative counterpart e.g and -3/7 are integers... Another way to say this is that they also include negative numbers are all natural including. Provided from investing activities is preferred to Net cash provided from investing activities is preferred Net. Was the weather in Pretoria on 14 February 2013 numbers that are not integers of... 0.278254â¦... Is preferred to Net cash provided from investing activities is preferred to Net cash used since the integers, is! Of logic allow any conclusionâmeaning the implication is true or false explain they will continue on into.... Also a rational number because.9 comes from the word âratio.â in this case, the ratio two.: 2 question: which of the following is not an integer is 1 all integers are rational numbers.: true every whole numberis a rational number can be considered as a fraction âThere are numbers. Balance equation for the complete combustion of the denominator can be represented as fractions, p/q, where both and. Balance equation for the most clear and detailed math lessons case, ratio... Some rational numbers working with a long line of different decimals, they be..., -2, ⦠some integers are rational numbers are not rational numbers are integers and all are... Questions in Mathematics provided from investing activities is preferred to Net cash used,! As all rational numbers but some rational numbers are numbers that are not integers,... Hope this helps over 1 should not have any decimals, then integers.
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# Learn How to Use INDEX and MATCH with Multiple Criteria
We can use an array formula that is based on the INDEX and MATCH functions to lookup a value based on multiple criteria. The steps below will walk through the process.
Figure 1- How to Use INDEX and MATCH with Multiple Criteria
## General Formula
`=INDEX(range1,MATCH(1,(A1=range2)*(B1=range3)*(C1=range4),0))`
## Formula
`=INDEX(E4:E10,MATCH(1,(H3=B4:B10)*(H4=C4:C10)*(H5=D4:D10),0))`
## Setting up the Data
We will use the INDEX and MATCH functions to get the Cost of the A grade item that is of medium size and short height in figure 2.
Figure 2 – Setting up the Data
## Lookup Cost with INDEX and MATCH with Multiple Criteria
• We will click on Cell H6
• We will insert the formula below into Cell H6
`=INDEX(E4:E10,MATCH(1,(H3=B4:B10)*(H4=C4:C10)*(H5=D4:D10),0))`
• Because this is an array formula, we will press CTRL+SHIFT+ENTER
Figure 3- Result for Lookup of Cost with INDEX and MATCH functions with Multiple Criteria
## Explanation
In this formula, the MATCH function looks through a one-column range and provides a match based on a specified criteria. Values must be concatenated in a helper column as seen in Column G and Column H. Without this, we can’t supply more than one criteria.
An array of 1 and 0 is used to show the rows that match all three criteria (A, MEDIUM, SHORT). The MATCH FUNCTION is then used to match the first criteria that is found.
From this section of the formula below, we can get the temporary array of 1’s and 0’s:
`(H3=B4:B10)*(H4=C4:C10)*(H5=D4:D10)`
First, a comparison is made between Cell H3 against all grades, Cell H4 against all Sizes, and Cell H5 against all Height.
An array of TRUE and FALSE is produced. The multiplication operator transforms this into 1s and 0s like this:
`{1;0;1;0;0;0;0}*{0;1;1;0;0;0;1}*{0;1;1;0;0;1;0}`
The final result is returned to MATCH and MATCH returns 3.
`=INDEX(E4:E10,3)`
INDEX returns the result as 25.
## Instant Connection to an Expert through our Excelchat Service
Most of the time, the problem you will need to solve will be more complex than a simple application of a formula or function. If you want to save hours of research and frustration, try our live Excelchat service! Our Excel Experts are available 24/7 to answer any Excel question you may have. We guarantee a connection within 30 seconds and a customized solution within 20 minutes.
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# Bottle Rocket Calculations - PowerPoint PPT Presentation
Bottle Rocket Calculations. Example using 40 psi and 700 ml of water. Select air pressure and water volume. Select air pressure i.e. 40 psi Select water volume i.e. 700 ml Find mass of water Mass = density x volume Vol = 700 ml = 700 cm 3 or 0.000700 m 3
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## PowerPoint Slideshow about ' Bottle Rocket Calculations' - nolen
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### Bottle Rocket Calculations
Example using 40 psi and 700 ml of water
• Select air pressure i.e. 40 psi
• Select water volume i.e. 700 ml
• Find mass of water
• Mass = density x volume
• Vol = 700 ml = 700 cm3 or 0.000700 m3
• ρ = density of water = 998 kg/m3 which is a constant
• Mass = 998 kg/m3 x 0.0007 m3 = 0.6986 kg
Calculate average water mass flow rate
• Average mass flow rate, ṁ, of water out of nozzle: ṁ = A X cd X √(2ρΔP)
• Find A of nozzle in m2: A = πr2
• For diameter of ~21 cm = .021 m
• Radius = d/2 = .021 m / 2 = 0.0105 m
• A = π (0.0105)2 = 0.0003462 m2
• Cd is given at 0.98
• Find average pressure acting on the water
• ΔP = (Pi + Pf) / 2 or (Pi (1+Vi/Vf)) / 2, since PiVi = PfVf, so Pf = (PiVi)/Vf
• Pi = 40 psi
• Vi of air = 2 Liter – 0.7 L = 1.3 L
• Vf = 2 L
• Pf = 40 (1.3) / 2 = 26
• ΔP = (40 + 26) / 2 = 33psi
• Convert psi to N/m2
• 14.7 psi = 101,353.56 N/m2, so 33 psi =
• 101,353.56 x 33 / 14.7 = 227528.4 N/m2
• Note: N = kg m/s2
• Back to calculating average mass flow rate, ṁ, of water out of nozzle: ṁ = A x cd x √(2ρΔP)
• ṁ = A x cd x √(2ρΔP)
• ṁ = 0.0003462 m2 x 0.98 x (√2 x 998 kg/m3 x 227528.4 N/m2) = 7.2302 kg/s
• Water exit velocity V = ṁ / ρA = 7.2302 kg/s / (998 kg/m3)(.0003462 m2) = 20.926 m/s
• Rocket thrust ft = ṁ x V = 7.2302 kg/s x 20.926 m/s = 151.3 kg m/s2 or 151.3 N
• Net force f = ft – fd – (mave x g)
• F = 151.3 kg m/s2 – 0 – ((0.3 + 0.7 kg)/2) x (9.8 m/s2) = 146.4 kg m/s2 or 146.4 N
• Note:
• Mave = mass of empty rocket + mass of water selected
• Mass of rocket was weighted at 300 g, or 0.3 kg
• Water selected was 0.7 kg
• fd is the drag coefficient and is very low in this case
• The rocket acceleration is a result of the net force acting on the mass
• f = mave x a, so a = f/ mave
• A = (146.4 N) / ((0.3 + 0.7 kg)/2) = 292.8 m/s2
• To find the range you need to find the amount of time it takes for the water to exit and the velocity of the rocket
• The time to expel the water is the mass of the water divide by the mass flow rate: t = m H2O/ ṁ = 0.7 kg / 7.2302 kg/s = 0.097 s
• Velocity of the bottle Vrocket is a x t =
• (292.8 m/s2) x (0.097 s) = 28.4 m/s
• So the range R is V2rocket x sin 2Ө / g
• Ө is the launch angle which is 45° in this case.
• R = (28.4 m/s)2 x sin 2(45) / 9.8 m/s2 = 82.3 m
• Final range is affected by drag factor
• Drag factor, Dc, for bottle shape is low, i.e. 0.15
• Drag force D = 1- Dc = 1 – 0.15 = 0.85
• Final range Rf = R x D = 82.3 m x 0.85 = 69.96 m
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250 Box Challenge
9:34 PM, Saturday July 25th 2020
This is my second submission. On the first I had a problem with Imgur and had to post 42 separate links. I finally got the problem resolved (hopefully) so this one has just one link to all 250 boxes.
Sorry for any confusion.
https://imgur.com/a/43IUN7A
2 users agree
8:16 PM, Thursday August 6th 2020
Congrats on finishing the challenge!
Overall - Make sure your hatching lines are not wobbly. It's ok if they fray on one side, but make sure they are smooth. Also, draw bigger boxes! It's hard to see where you made a mistake if you don't have much room to draw.
You also need to make sure all angles of your Y are above 90 degrees. Boxes like 219, 225, or 226 look slanted or have their solidity undermined because the front Y has an angle of or below 90 degrees. I think you should also push yourself to draw a greater variety of shapes and orientations. I would recommend checking out the Y generator mentioned in the lesson for ideas if you haven't already.
Here is a diagram of different box orientations (challenge boxes should always be drawn from imagination though) -> https://imgur.com/Kqg6uMX
Convergences - Your convergences improved over the course of the challenge, however a lot of lines still look parallel, especially the ones that are vertical (heading toward the top or bottom of the page). Remember that when you look a box the edges converge in all three directions, including up/down. I think otherwise your convergences are quite nice.
A good diagram -> https://i.imgur.com/8PqQLE0.png
Inner corner - Your inner corner is often diverging or converging way too quickly. Compare the slope of your Y line to the slope of the mark you're planning. For the inner corner lines they should be pretty similar, near parallel. Don't draw a line from the inner corner to the center of the Y, this will pretty much always be the wrong slope. Keep in mind the inner corner will also not have any angles below 90. This ordering can help you improve your inner corner quickly -> https://imgur.com/FGdtVwm
I think you're ready for lesson 2. You can keep practicing boxes as a warm-up exercise. I hope this feedback is helpful! Congrats again :)
Next Steps:
Lesson 2
This community member feels the lesson should be marked as complete, and 2 others agree. The student has earned their completion badge for this lesson and should feel confident in moving onto the next lesson.
11:29 PM, Thursday August 6th 2020
Thanks for the critique Frog. It was helpful. I had a really difficult time trying to get the sets of lines parallel (or converging) since I was doing it by eye and not measuring or using guides. I made plenty of mistakes along the way, but I think overall I learned something. I'm not there yet, but this exercise helped me along the way.
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We've also tested their longevity. We've found that if we're reasonably gentle with them, we can get through all of Lesson 1, and halfway through the box challenge. We actually had ScyllaStew test them while recording realtime videos of her working through the lesson work, which you can check out here, along with a variety of reviews of other brands.
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# Curvilinear Regression in R: A Quick Reference
## Introduction to Curvilinear Regression in R Language
In this post, we will learn about some basics of curvilinear regression in R.
The curvilinear/non-linear regression analysis is used to determine if there is a non-linear trend exists between $X$ and $Y$.
Adding more parameters to an equation results in a better fit to the data. A quadratic and cubic equation will always have higher $R^2$ than the linear regression model. Similarly, a cubic equation will usually have higher $R^2$ than a quadratic one.
### Logarithmic and Polynomial Relationships
The logarithmic relationship can be described as follows:
$$Y=m\, log(x)++c$$
the polynomial relationship can be described as follows:
$$Y=m_1x + m_2x^2 + m_3x^3 + m_nx^n + c$$
The logarithmic example is more akin to a simple regression, whereas the polynomial example is multiple regression. Logarithmic relationships are common in the natural world; you may encounter them in many circumstances. Drawing the relationships between response and predictor variables as a scatter plot is generally a good starting point.
Consider the following data that are related in a curvilinear form,
### Performing Curvilinear Regression in R
Let us perform a curvilinear regression in R language.
Growth <- c(2, 9, 11, 12, 13, 14, 17, 19, 17, 18, 20)
Nutrient <- c(2, 4, 6, 8, 10, 16, 22, 28, 30, 36, 48)
data <- as.data.frame(cbind(Growth, Nutrient))
ggplot(data, aes(Nutrient, Growth) ) +
geom_point() +
stat_smooth()
The Scatter plot shows the relationship appears to be a logarithmic one.
### Linear Regression in R
Let us carry out a linear regression using the lm() function by taking the $\log$ of the predictor variable rather than the basic variable itself.
data <- cbind(Growth, Nutrient)
mod <- lm(Growth~log(Nutrient, data))
summary(mod)
##
Call:
lm(formula = Growth ~ log(Nutrient), data = data)
Residuals:
Min 1Q Median 3Q Max
-2.2274 -0.9039 0.5400 0.9344 1.3097
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 0.6914 1.0596 0.652 0.53
log(Nutrient) 5.1014 0.3858 13.223 3.36e-07 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 1.229 on 9 degrees of freedom
Multiple R-squared: 0.951, Adjusted R-squared: 0.9456
F-statistic: 174.8 on 1 and 9 DF, p-value: 3.356e-07
### FAQS about Curvilinear Regression in R
1. Write in detail about curvilinear regression models.
2. How visually one can guess the curvilinear relationship between the response and predictor variable?
3. What may be the consequences, if a curvilinear relationship is estimated using a simple linear regression model?
Learn about Performing Linear Regression in R
Learn Statistics
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How does kinetic energy do work? [closed]
To my understanding, work is done on object A when object B is applying a force on object A, causing object A to displace.
Since energy is the ability to do work, what work does a moving object do, due to its kinetic energy? I mean, if I kick a ball, it has kinetic energy due to its velocity, but what force does it apply? On what object?
• I think I have a deep misconception on this subject, so I would be grateful to a good explanation, yet not too advanced one (mathematically speaking). Commented Apr 26, 2016 at 16:31
• An object in motion will stay in motion unless a force is applied. When the ball hits something, then it will apply a force. Just moving, it doesn't (well, neglecting air resistance). Commented Apr 26, 2016 at 16:34
• It is a good answer, so I think you should write it as one so I could accept it. But a follow-up question: If for example the ball hits a building, then there is no work because there is no displacement. So actually the kinetic energy did not did work. What is the explanation? Thank a lot! Commented Apr 26, 2016 at 16:39
• I'm not sure what your question is - you say "energy is the ability to do work" and then you are wondering what work is done by an object with kinetic energy, but having the ability to do something is not the same as doing something. Commented Apr 26, 2016 at 16:44
• @ACuriousMind - First, I think the title edit is not good because I want to do how the kinetic energy becomes work, end not how kinetic energy works... Secondly - I understand that now after reading Jon Custer's answer. Thanks a lot. :) Commented Apr 26, 2016 at 16:48
To my understanding, work is done on object A when object B is applying a force on object A, causing object A to displace.
Work is done whenever a force displaces an object.
Since energy is the ability to do work, what work does a moving object do, due to its kinetic energy?
A moving object might not do any work at all. Imagine an empty universe with just a ball. The ball is moving, meaning it possesses some kinetic energy. If nothing interferes, (and we're on an hypothetical empty universe, so there's nothing to interfere), the ball will just keep moving indeterminately.
I mean, if I kick a ball, it has kinetic energy due to its velocity, but what force does it apply? On what object?
If you kick a ball you're transferring energy to the ball. The ball that was quiet and had zero kinetic energy, now has some kinetic energy, proportional to the square of its velocity.
If you ignore air resistance, it won't be applying force on anything/any object. However, if we consider that the ball is near enough of the Earth, the ball will have a force applied on it: gravity force, in the direction of the center of the Earth. (actually, the ball itself will also be applying a force on Earth, but given the magnitude of the difference between their masses, let's ignore this). And this force from the Earth will be changing the speed of the ball, thus changing its kinetic energy, (transforming it in potential energy in case the ball is going up, and actually providing it with more kinetic energy in case the ball is going down).
If you don't ignore air resistance, the ball will be pushing and displacing the air particles. Meaning it is applying a force and displacing them, thus doing work on them.
So to answer your question, just because an object has kinetic energy, does not mean it is using it. If there is no change of state, there's no "usage".
I sincerely hope my answer helps you understand it better :)
• It did answer my question. Thanks! But if it hits a wall, the ball will not move, so there is no work done in here... So what is happening here? Thanks again! Commented Apr 26, 2016 at 17:03
• I'm glad it helped :) From what I understand of your question, you mean to ask what is the work if when you hit the ball, the ball is against a wall and doesn't move? If that's the case, there's no change of state, there's no work done. And in that case we're talking about Newtown's third law. You might want to do that as a new question ;) Commented Apr 26, 2016 at 17:08
• I think I understand the answer from your comment. Tell me if I am right, please. When the ball hits the wall, it cannot do any force that will displace the wall, so instead it will transform to another kind of energy, which could also do that work? I am sorry for my English... :\ Commented Apr 26, 2016 at 17:12
The work the ball that you kick does is the ball going against air-resistance and changing its velocity.
• At least to me that seems very reasonable Commented Apr 26, 2016 at 16:37
• But what if there is nor air resistance, like in space? Where can we see the work then? Thanks a lot. Commented Apr 26, 2016 at 16:38
• In space the work done is the change in velocity. It takes energy to speed up things for example a rocket even if there is no air-resistance. Commented Apr 26, 2016 at 16:45
• But work is to displace an object using force, not changing its velocity... Commented Apr 26, 2016 at 16:51
• You are displacing the object by making your foot be in the place that the ball was in by doing this you are increasing its velocity. Commented Apr 26, 2016 at 16:54
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Viraj_Deshpande's posts
Viraj_Deshpande commented on an article
Dressed in shorts and a crumpled white t-shirt and sitting coyly in his apartment in Dombivali, on the outskirts of Mumbai, Shashank Prabhu looks more like the teenager next door who you'd see rushing to his tuition classes. But once he begins to speak, the confidence and command of a topper are quite evident. Prabhu topped t...
drvispute C u in IIM next yr !. 04 Sep '10.
anonymous [...] (US 10 1/2)WINTERSCHUHE STIEFELSchuhe Gunstig Kaufe.... 21 Mar '12.
Viraj_Deshpande replied to Quant by Arun Sharma
My solution:
Every no. can be expressed in the form of 9k or 9k+1 or 9k-1 or 9k+2 or 9k-2 or 9k+3 or 9k-3 or 9k+4 or 9k-4, where k is an integer.
2998 = 9 * 337 + 1
Thus, 1 to 3000 (Excluding 1 and 3000) has
334 nos. of type 9k+2 (because of no. 2999) and
333 nos. of type 9k+1, 9k,...
3 2
Viraj_Deshpande replied to Revant aka Profootball25 getting hitched!
Congrats Revant!
Viraj_Deshpande replied to Number System
6711+4179 is possible.
First_timer's solution (above) is correct.
1 1
Viraj_Deshpande replied to Number System
Test of divisibility by 11 is that sum of odd-placed nos = sum of even-placed nos.
Looking at the options, their (sum of even-placed nos. - sum of odd placed nos.) = (8-7=)1, (14-7=)7, (8-10=)-2 and (14-10=)4 respectively.
y=1, z=7, w=3 or 6 or 9, x=2 or 4 or 6 or 8
Thus the nos. (and t...
1 1
Viraj_Deshpande replied to Quantagious
(x+y)^2 = x^2+2xy+y^2 = 2007 - 54X^2 = 9 (223 - 6X^2)
Thus, (x+y) = 3k, where k is an integer
9k^2 = 9(223-6x^2) gives 6x^2=223-k^2
Since 3 divides (223 - k^2), k^2 should be of the form (3p+1).
k^2 of the form (3p+1) where (223-k^2)/6 is also a perfect square are found by listing the...
1 1
Viraj_Deshpande replied to Geometry
Area of triangle = 0.5*b*c*sinA
b=4,c=5
Thus area = 10sinA
Clearly, this is <=10, equality existing when A=90 degrees.
Bound means the maximum value it can attain.
1 1
Viraj_Deshpande replied to official quant thread for CAT 2009
RHS is not 22752*x. It is only one no. with x in its units place. Hope this clears the confusion
1 1
Viraj_Deshpande replied to official quant thread for CAT 2009
Their HCF is 45 means they dont have any other common factor except 45, otherwise it would've figured in the HCF. Hence, once we divide both by 45, what remains should be a pair of co-prime nos.
Clear?
Viraj_Deshpande replied to official quant thread for CAT 2009
No. of divisors of 21600 with HCF 45 = No of co-prime divisors of 21600/45 (=480). This should help.
There is some formula for no. of pairs of co-prime divisors. Cant recollect now. Will post here if I manage to recollect in time.
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Germán Rodríguez
Introducing R Princeton University
## 4 Linear Models
Let us try some linear models, starting with multiple regression and analysis of covariance models, and then moving on to models using regression splines. In this section I will use the data read in Section 3, so make sure the `fpe` data frame is attached to your current session.
### 4.1 Fitting a Model
To fit an ordinary linear model with fertility change as the response and setting and effort as predictors, try
```> lmfit = lm( change ~ setting + effort )
```
Note first that `lm` is a function, and we assign the result to an object that we choose to call `lmfit` (for linear model fit). This stores the results of the fit for later examination.
The argument to `lm` is a model formula, which has the response on the left of the tilde `~` (read "is modeled as") and a Wilkinson-Rogers model specification formula on the right. R uses
+ to combine elementary terms, as in A+B : for interactions, as in A:B; * for both main effects and interactions, so A*B = A+B+A:B
A nice feature of R is that it lets you create interactions between categorical variables, between categorical and continuous variables, and even between numeric variables (it just creates the cross-product).
### 4.2 Examining a Fit
Let us look at the results of the fit. One thing you can do with `lmfit`, as you can with any R object, is print it.
```> lmfit
Call:
lm(formula = change ~ setting + effort)
Coefficients:
(Intercept) setting effort
-14.4511 0.2706 0.9677
```
The output includes the model formula and the coefficients. You can get a bit more detail by requesting a summary:
```> summary(lmfit)
Call:
lm(formula = change ~ setting + effort)
Residuals:
Min 1Q Median 3Q Max
-10.3475 -3.6426 0.6384 3.2250 15.8530
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) -14.4511 7.0938 -2.037 0.057516 .
setting 0.2706 0.1079 2.507 0.022629 *
effort 0.9677 0.2250 4.301 0.000484 ***
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
Residual standard error: 6.389 on 17 degrees of freedom
Multiple R-Squared: 0.7381, Adjusted R-squared: 0.7073
F-statistic: 23.96 on 2 and 17 DF, p-value: 1.132e-05
```
The output includes a more conventional table with parameter estimates and standard errors, as well the residual standard error and multiple R-squared. (By default S-Plus includes the matrix of correlations among parameter estimates, which is often bulky, while R sensibly omits it. If you really need it, add the option `correlation=TRUE` to the call to `summary`.)
To get a hierarchical analysis of variance table corresponding to introducing each of the terms in the model one at a time, in the same order as in the model formula, try the `anova` function:
```> anova(lmfit)
Analysis of Variance Table
Response: change
Df Sum Sq Mean Sq F value Pr(>F)
setting 1 1201.08 1201.08 29.421 4.557e-05 ***
effort 1 755.12 755.12 18.497 0.0004841 ***
Residuals 17 694.01 40.82
---
Signif. codes: 0 `***' 0.001 `**' 0.01 `*' 0.05 `.' 0.1 ` ' 1
```
Alternatively, you can plot the results using
```> plot(lmfit)
```
This will produce a set of four plots: residuals versus fitted values, a Q-Q plot of standardized residuals, a scale-location plot (square roots of standardized residuals versus fitted values, and a plot of residuals versus leverage that adds bands corresponding to Cook's distances of 0.5 and 1.
R will prompt you to click on the graph window or press Enter before showing each plot, but we can do better. Type `par(mfrow=c(2,2))` to set your graphics window to show four plots at once, in a layout with 2 rows and 2 columns. Then redo the graph using `plot(lmfit)`. To go back to a single graph per window use `par(mfrow=c(1,1))`. There are many other ways to customize your graphs by setting high-level parameters, type `?par` to learn more.
Technical Note: You may have noticed that we have used the function `plot` with all kinds of arguments: one or two variables, a data frame, and now a linear model fit. In R jargon plot is a generic function. It checks for the kind of object that you are plotting and then calls the appropriate (more specialized) function to do the work. There are actually many plot functions in R, including `plot.data.frame` and `plot.lm`. For most purposes the generic function will do the right thing and you don't need to be concerned about its inner workings.
### 4.3 Extracting Results
There are some specialized functions that allow you to extract elements from a linear model fit. For example
```> fitted(lmfit)
1 2 3 4 5 6 7 8
-2.004026 5.572452 25.114699 21.867637 28.600325 24.146986 17.496913 10.296380
... output edited ...
```
extracts the fitted values. In this case it will also print them, because we did not asign them to anything. (The longer form `fitted.values` is an alias.)
To extract the coefficients use the `coef` function (or the longer form `coefficients`)
```> coef(lmfit)
(Intercept) setting effort
-14.4510978 0.2705885 0.9677137
```
To get the residuals, use the `residuals` function (or the abbreviation `resid`):
```> residuals(lmfit)
1 2 3 4 5 6
3.0040262 4.4275478 3.8853007 3.1323628 0.3996747 15.8530144
... output edited ...
```
If you are curious to see exactly what a linear model fit produces, try the function
```> names(lmfit)
[1] "coefficients" "residuals" "effects" "rank"
[5] "fitted.values" "assign" "qr" "df.residual"
[9] "xlevels" "call" "terms" "model"
```
which lists the named components of a linear fit. All of these objects may be extracted using the `\$` operator. However, whenever there is a special extractor function you are encouraged to use it.
### 4.4 Factors and Covariates
So far our predictors have been continuous variables or covariates. We can also use categorical variables or factors. Let us group family planning effort into three categories:
```> effortg = cut(effort, breaks = c(-1, 4, 14, 100),
+ label=c("weak","moderate","strong"))
```
The function `cut` creates a factor or categorical variable. The first argument is an input vector, the second is a vector of breakpoints, and the third is a vector of category labels. Note that there is one more breakpoint than there are categories. All values greater than the i-th breakpoint and less than or equal to the (i+1)-st breakpoint go into the i-th category. Any values below the first breakpoint or above the last one are coded `NA` (a special R code for missing values). If the labels are omitted, R generates a suitable default of the form "(a,b]". By default the intervals are closed on the right, so our intervals are `≤4; 5-14; 15+`. To change this use the option `right=FALSE`.
Try fitting the analysis of covariance model:
```> covfit = lm( change ~ setting + effortg )
> covfit
Call:
lm(formula = change ~ setting + effortg)
Coefficients:
(Intercept) setting effortgmoderate effortgstrong
-5.9540 0.1693 4.1439 19.4476
```
As you can see, family planning effort has been treated automatically as a factor, and R has generated the necessary dummy variables for moderate and strong programs treating weak as the reference cell.
Choice of Contrasts: R codes unordered factors using the reference cell or "treatment contrast" method. The reference cell is always the first category which, depending on how the factor was created, is usually the first in alphabetical order. If you don't like this choice, R provides a special function to re-order levels, check out `help(relevel)`.
S codes unordered factors using the Helmert contrasts by default, a choice that is useful in designed experiments because it produces orthogonal comparisons, but has baffled many a new user. Both R and S-Plus code ordered factors using polynomials. To change to the reference cell method for unordered factors use the following call
```> options(contrasts=c("contr.treatment","contr.poly"))
```
Back on to our analysis of covariance fit. You can obtain a hierarchical anova table for the analysis of covariance model using the `anova` function:
```> anova(covfit)
Analysis of Variance Table
Response: change
Df Sum Sq Mean Sq F value Pr(>F)
setting 1 1201.08 1201.08 36.556 1.698e-05 ***
effortg 2 923.43 461.71 14.053 0.0002999 ***
Residuals 16 525.69 32.86
---
Signif. codes: 0 `***' 0.001 `**' 0.01 `*' 0.05 `.' 0.1 ` ' 1
```
Type `?anova` to learn more about this function.
### 4.5 Regression Splines
The real power of R begins to shine when you consider some of the other functions you can include in a model formula. First, you can include mathematical functions, for example `log(setting)` is a perfectly legal term in a model formula. You don't have to create a variable representing the log of setting and then use it, R will create it 'on the fly', so you can type
```> lm( change ~ log(setting) + effort)
```
If you wanted to use orthogonal polynomials of degree 3 on setting, you could include a term of the form `poly(setting,3)`
You can also get R to calculate a well-conditioned basis for regression splines. First you must load the splines library (this step is not needed in S-Plus):
```> library(splines)
```
This makes available the function `bs` to generate B-splines. For example the call
```> setting.bs <- bs(setting, knots = c(66,74,84)) + effort )
```
will generate cubic B-splines with interior knots placed at 66, 74 and 84. This basis will use seven degrees of freedom, four corresponding to the constant, linear, quadratic and cubic terms, plus one for each interior knot. Alternatively, you may specify the number of degrees of freedom you are willing to spend on the fit using the parameter `df`. For cubic splines R will choose df-4 interior knots placed at suitable quantiles. You can also control the degree of the spline using the parameter `degree`, the default being cubic.
If you like natural cubic splines, you can obtain a well-conditioned basis using the function `ns`, which has exactly the same arguments as `bs` except for degree, which is always three. To fit a natural spline with five degrees of freedom, use the call
```> setting.ns <- ns(setting, df=5)
```
Natural cubic splines are better behaved than ordinary splines at the extremes of the range. The restrictions mean that you save four degrees of freedom. You will probably want to use two of them to place additional knots at the extremes, but you can still save the other two.
To fit an additive model to fertility change using natural cubic splines on setting and effort with only one interior knot each, placed exactly at the median of each variable, try the following call:
```> splinefit = lm( change ~ ns(setting, knot=median(setting)) +
+ ns(effort, knot=median(effort)) )
```
Here we used the parameter `knot` to specify where we wanted the knot placed, and the function `median` to calculate the median of setting and effort.
Do you think the linear model was a good fit? Natural cubic splines with exactly one interior knot require the same number of parameters as an ordinary cubic polynomial, but are much better behaved at the extremes.
### 4.6 Other Options
The `lm` function has several additional parameters that we have not discussed. These include
`data` to specify a dataset, in case it is not attached `subset` to restrict the analysis to a subset of the data `weights` to do weighted least squares
and many others; see `help(lm)` for further details. The `args` function lists the arguments used by any function, in case you forget them. Try `args(lm)`.
The fact that R has powerful matrix manipulation routines means that one can do many of these calculations from first principles. The next couple of lines create a model matrix to represent the constant, setting and effort, and then calculate the OLS estimate of the coefficients as (X'X)-1X'y:
```> X <- cbind(1,effort,setting)
> solve( t(X) %*% X ) %*% t(X) %*% change
[,1]
[1,] -14.4510978
[2,] 0.9677137
[3,] 0.2705885
```
Compare these results with `coef(lmfit)`.
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# In how many ways 30 couples(total=30 men and 30 women...with
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In how many ways 30 couples(total=30 men and 30 women...with [#permalink] 19 Aug 2003, 12:29
In how many ways 30 couples(total=30 men and 30 women...with respective man sitting on the right of woman) and 30 HOD(head of the department) be seated around a round table if each of the couple should sit between each of the HOD?
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heres my try at this kind of problem:
we fix one HOD... the others can distribute in 29! ways... now we have to distribute the couples... i understand that the couples are fixed (ie to each woman theres a man assigned)... so they can arrange in 30! ways (as men have to sit on the right)... so anwer would be 29!*30!?
now, how would it be if the couples were not fixed (any man can sit behind any woman)?
we can make 15! couples, so answer would be 29!*15!*30!?
and finally, what if the positions of the couples were not fixed (ie man can sit either on the right or on the left)...
s'one brighter than me please check my numbers... cheers, javi
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Re: Probability Q3 [#permalink] 22 Aug 2003, 22:47
aps_can wrote:
In how many ways 30 couples(total=30 men and 30 women...with respective man sitting on the right of woman) and 30 HOD(head of the department) be seated around a round table if each of the couple should sit between each of the HOD?
Agree with 29!*30!
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javropu wrote:
heres my try at this kind of problem:
we fix one HOD... the others can distribute in 29! ways... now we have to distribute the couples... i understand that the couples are fixed (ie to each woman theres a man assigned)... so they can arrange in 30! ways (as men have to sit on the right)... so anwer would be 29!*30!?
now, how would it be if the couples were not fixed (any man can sit behind any woman)?
we can make 15! couples, so answer would be 29!*15!*30!?
and finally, what if the positions of the couples were not fixed (ie man can sit either on the right or on the left)...
s'one brighter than me please check my numbers... cheers, javi
If any man can sit next to any woman, assuming the men still sit to the same side next to each woman, the answer would be simply:
29! x 30! x 30!
_________________
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AkamaiBrah
Former Senior Instructor, Manhattan GMAT and VeritasPrep
Vice President, Midtown NYC Investment Bank, Structured Finance IT
MFE, Haas School of Business, UC Berkeley, Class of 2005
MBA, Anderson School of Management, UCLA, Class of 1993
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thx akamai i messed it all up... for calculating the number of couples, i took 15 instead of 30... oh my... its interesting to see that you can get to the same result by i) calculating the number of couples (30!) and then calculating how to arrange the HOD and the couples (29!*30!) or ii) by just arranging the 3 people (29!*30!*30!)
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NCERT Solutions For Class 6 Maths Understanding Elementary Shapes Ex 5.5
# NCERT Solutions For Class 6 Maths Understanding Elementary Shapes Ex 5.5
## NCERT Solutions For Class 6 Maths Understanding Elementary Shapes Ex 5.5
NCERT Solutions For Class 6 Maths Chapter 5 Understanding Elementary Shapes Ex 5.5
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Exercise 5.5
Question 1.
Which of the following are models for perpendicular lines:
(a) The adjacent edges of a table top.
(b) The lines of a railway track.
(c) The line segments forming a letter ‘L’.
(d) The letter V.
Solution:
(a) Yes, the adjacent edges of a table top are the models of perpendicular lines.
(b) No, the lines of a railway tracks are parallel to each other. So they are not a model for perpendicular lines.
(c) Yes, the two line segments of‘L’ are the model for perpendicular lines.
(d) No, the two line segments of ‘V’ are not a model for perpendicular lines.
Question 2.
Let $$\overline { PQ }$$ be the perpendicular to the line segment $$\overline { XY }$$ . Let $$\overline { PQ }$$ and $$\overline { XY }$$ intersect at in the point A. What is the measure of ∠PAY?
Solution:
Since $$\overline { PQ }$$ ⊥ XY
∴ ∠PAY = 90°
Question 3.
There are two set-squares in your box. What are the measures of the angles that are formed at their corners? Do they have any angle measure that is common?
Solution:
The figures of the two set-squares are given below:
The measure angles of triangle (a) are : 30°, 60° and 90°.
The measure angles of triangle (b) are 45°, 45° and 90°.
Yes, they have a common angle of measure 90°.
Question 4.
Study the diagram. The line l is perpendicular to line m.
(a) Is CE = EG?
(b) Does PE bisects CG?
(c) Identify any two line segments for which PE is the perpendicular bisector.
(d) Are these true?
(i) AC > FG
(ii) CD = GH
(iii) BC < EH
Solution:
(a) Yes,
Since, CE = 2 units and EG = 2 units
Hence, CE = EG.
(b) Yes, PE bisects CG
(c) Required line segments for which PE is perpendicular bisector are: $$\overline { BG }$$ and $$\overline { DF }$$
(d) (i) True (ii) True (iii) True
## Related content
NCERT Solutions for Class 11 English Chapter 2 NCERT Exemplar for Class 6 Maths Solutions NCERT Solutions for Class 11 English Reading Skills Chapter 1 The Portrait of a Lady CBSE Notes for Class 8 Maths CBSE Notes for Class 7 Science CBSE Notes for Class 8 Science Lines and Angles Class 9 Extra Questions Maths Chapter 6 CBSE Notes for Class 7 Maths Class 7 CBSE Notes AMU Class 11 Entrance Exam Sample Papers
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## 28948
28,948 (twenty-eight thousand nine hundred forty-eight) is an even five-digits composite number following 28947 and preceding 28949. In scientific notation, it is written as 2.8948 × 104. The sum of its digits is 31. It has a total of 3 prime factors and 6 positive divisors. There are 14,472 positive integers (up to 28948) that are relatively prime to 28948.
## Basic properties
• Is Prime? No
• Number parity Even
• Number length 5
• Sum of Digits 31
• Digital Root 4
## Name
Short name 28 thousand 948 twenty-eight thousand nine hundred forty-eight
## Notation
Scientific notation 2.8948 × 104 28.948 × 103
## Prime Factorization of 28948
Prime Factorization 22 × 7237
Composite number
Distinct Factors Total Factors Radical ω(n) 2 Total number of distinct prime factors Ω(n) 3 Total number of prime factors rad(n) 14474 Product of the distinct prime numbers λ(n) -1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 0 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 28,948 is 22 × 7237. Since it has a total of 3 prime factors, 28,948 is a composite number.
## Divisors of 28948
1, 2, 4, 7237, 14474, 28948
6 divisors
Even divisors 4 2 2 0
Total Divisors Sum of Divisors Aliquot Sum τ(n) 6 Total number of the positive divisors of n σ(n) 50666 Sum of all the positive divisors of n s(n) 21718 Sum of the proper positive divisors of n A(n) 8444.33 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 170.141 Returns the nth root of the product of n divisors H(n) 3.4281 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 28,948 can be divided by 6 positive divisors (out of which 4 are even, and 2 are odd). The sum of these divisors (counting 28,948) is 50,666, the average is 84,44.,333.
## Other Arithmetic Functions (n = 28948)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 14472 Total number of positive integers not greater than n that are coprime to n λ(n) 7236 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 3147 Total number of primes less than or equal to n r2(n) 8 The number of ways n can be represented as the sum of 2 squares
There are 14,472 positive integers (less than 28,948) that are coprime with 28,948. And there are approximately 3,147 prime numbers less than or equal to 28,948.
## Divisibility of 28948
m n mod m 2 3 4 5 6 7 8 9 0 1 0 3 4 3 4 4
The number 28,948 is divisible by 2 and 4.
• Deficient
• Polite
## Base conversion (28948)
Base System Value
2 Binary 111000100010100
3 Ternary 1110201011
4 Quaternary 13010110
5 Quinary 1411243
6 Senary 342004
8 Octal 70424
10 Decimal 28948
12 Duodecimal 14904
20 Vigesimal 3c78
36 Base36 mc4
## Basic calculations (n = 28948)
### Multiplication
n×i
n×2 57896 86844 115792 144740
### Division
ni
n⁄2 14474 9649.33 7237 5789.6
### Exponentiation
ni
n2 837986704 24258039107392 702221716080783616 20327914237106524115968
### Nth Root
i√n
2√n 170.141 30.7048 13.0438 7.80412
## 28948 as geometric shapes
### Circle
Diameter 57896 181886 2.63261e+09
### Sphere
Volume 1.01612e+14 1.05305e+10 181886
### Square
Length = n
Perimeter 115792 8.37987e+08 40938.7
### Cube
Length = n
Surface area 5.02792e+09 2.4258e+13 50139.4
### Equilateral Triangle
Length = n
Perimeter 86844 3.62859e+08 25069.7
### Triangular Pyramid
Length = n
Surface area 1.45144e+09 2.85884e+12 23635.9
## Cryptographic Hash Functions
md5 ef9ffa986fa98b2485fe6b57ac0c6b1b c2ec848a283c6aa7a6808b2bb88e262e7e3fa605 2bac0cdf430b2810cee7713e31bc77f519a39cab6a6833a9f6663922be8d1195 acc92779f48fa29e1daf642e24cd6a0c359514f1546bc965a4831f1131d6772c26e7212cd4932edb3c64b452b603064d028a34f3dc0c7dfbf4211f4c1fbfdc4a 220b9b775eecf2c0563bca2ab336aeeebe1b0567
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# Equilibrium of particles
• Dec 22nd 2009, 06:38 AM
BabyMilo
Equilibrium of particles
Quote:
A block B of mass 0.4 kg and a particle P of mass 0.3 kg are connected by a light inextensible string. The string passes over a smooth pulley at the edge of a rough horizontal table. B is in contact with the table and the part of the string between B and the pulley is horizontal. P hangs freely below the pulley (see diagram).
Code:
(i) The systemis in limiting equilibrium with the string taut and P on the point of moving downwards. Find the coefficient of friction between B and the table. [5]
Code:
(ii) A horizontal force of magnitude X N, acting directly away from the pulley, is now applied to B. The system is again in limiting equilibrium with the string taut, and with P now on the point of moving upwards. Find the value of X.
I've done part i)
3 (i) T = 0.3g
F = T
R = 0.4g
F = μR
Coefficient is 0.75
but how do i do part 2?
I got W=T for P
so 0.3g=T
For B: T-F-X=0
0.3g-0.3g-X=0
wouldnt X=0?
but the answer states it's
X = 0.3g + 0.3g
X = 5.88N
thanks for your help!
• Dec 22nd 2009, 06:49 AM
skeeter
net force on B ...
$\displaystyle 0 = X - (f + T)$
net force on P ...
$\displaystyle 0 = T - mg$
combine the equations ...
$\displaystyle 0 = X - f - mg$
$\displaystyle X = f + mg$
$\displaystyle X = \mu Mg + mg = 0.3g + 0.3g = 0.6g$
• Dec 22nd 2009, 07:03 AM
Krahl
Quote:
Originally Posted by BabyMilo
For B: T-F-X=0
0.3g-0.3g-X=0
everything is correct up to here.
why are you saying x+F=T?
friction acts AGAINST motion, the question told you the system is about to move away from the pulley so friction force should be in the same direction as T.
so in fact X=T+F is the correct equation.
Hope that helps
• Dec 22nd 2009, 09:25 AM
BabyMilo
Quote:
Originally Posted by Krahl
everything is correct up to here.
why are you saying x+F=T?
friction acts AGAINST motion, the question told you the system is about to move away from the pulley so friction force should be in the same direction as T.
so in fact X=T+F is the correct equation.
Hope that helps
initally, i thought T would have some friction force to it as well.
never thought X was greater than T. (Crying)
• Dec 23rd 2009, 07:11 AM
Krahl
yep X is greater than T also there would be no motion if we apply a force X
$\displaystyle T-F \leq X \leq T+F$
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# Find the coordinates of the point which divides the join
Question:
Find the coordinates of the point which divides the join of A(3, 2, 5) and B(-4, 2, -2) in the ratio 4 : 3.
Solution:
The coordinates of point R that divides the line segment joining points P $(\mathrm{X} 1$\left.\mathrm{y}_{1}, \mathrm{z}_{1}\right)$and$Q\left(x_{2}, y_{2}, z_{2}\right)$in the ratio$m: n$are$\left(\frac{m x_{2}+n x_{1}}{m+n}, \frac{m y_{2}+n y_{1}}{m+n}, \frac{m z_{2}+n z_{1}}{m+n}\right)$Point A( 3, 2, 5 ) and B( -4, 2, -2 ), m and n are 4 and 3 respectively. Using the above formula, we get,$=\left(\frac{4 \times-4+3 \times 3}{4+3}, \frac{4 \times 2+3 \times 2}{4+3}, \frac{4 \times-2+3 \times 5}{4+3}\right)=\left(\frac{-7}{7}, \frac{14}{7}, \frac{7}{7}\right)(-1,2,1)$, is the point which divides the two points in ratio$4: 3\$.
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1. ## Calculus and Exponents
A question says to find the derivative of
$e^3/(sin(x)+1)$
The answer I got was $-e^3cosx/(sinx+1)^2$
My friend got
$(e^3sinx+e^3 - e^3cosx)/(sinx+1)^2$
The online calculators all agree with me, but he used the quotient rule (so did I), but I pulled e^3 out of it and used 1/sinx+1 to perform the quotient rule.
Which is the correct method?
2. ## Re: Calculus and Exponents
I'd write it as \displaystyle \begin{align*} y = e^3 \left[ \sin{(x)} + 1 \right] ^{-1} \end{align*} then let \displaystyle \begin{align*} u = \sin{(x)} + 1 \implies y = e^3 \, u^{-1} \end{align*}. Then \displaystyle \begin{align*} \frac{du}{dx} = \cos{(x)} \end{align*} and \displaystyle \begin{align*} \frac{dy}{du} = -e^3 \, u^{-2} = -e^3 \left[ \sin{(x)} + 1 \right] ^{-2} \end{align*}. So \displaystyle \begin{align*} \frac{dy}{dx} = -e^3 \cos{(x)} \left[ \sin{(x)} + 1 \right] ^{-2} = -\frac{ e^3 \cos{(x)} }{ \left[ \sin{(x)} + 1 \right] ^2} \end{align*}. Your answer is correct.
3. ## Re: Calculus and Exponents
Your friend failed to realize that $e^3$ is a constant, and so its derivative is zero.
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# Machinery Rated Power Calculation
Tags:
1. Jul 15, 2015
### Moteor
Hi everyone,
Thanks for the help in advance. I have a project with a piece of machinery that is rated to 105 HP @ 540 RPM.
I was wondering what the torque rating for the machine would be, whilst its being driven from a diesel engine?
I believe there is a direct relationship between Torque and Horspower from Engines;
TORQUE = HP x 5252 ÷ RPM
Therefore Torque = 105 HPx 2525 /540 RPM= 490.97 lb ft = 665.665938072 Nm
1) So I have done the math, but is that right? Is that really what the torque the machine can handle?
I also have the torque and HP stats of the engine as below:
I need to find the torque of the engine related by the rated torque capacity of the wood chipper = 665.7 Nm. @540 RPM?
Also I need to find out what engine speed can provide the max torque rating for the machine as the engine has to be reduced by a ratio through pulleys to match the max RPM of the machine. My second question is:
2) Does reducing a gear ratio, say 3.7 : 1 multiply the torque by that much? So if;
Torque x gear ratio = torque at the wheel
Then @ 2000 RPM the engine would provide 169.1 Nm , therefore multiply by the 3.7037 times the motor is rotating faster than 540 RPM and you have the figure 625.92 which is under the figure we have for the rated capacity of the wood chipper.
My third question is
3) Am I right?
or is it all in my head?
Thanks all . ..
2. Jul 15, 2015
### dean barry
I would be tempted to stick with power ratings, as long as your deisel engine has 105 HP at working speed
And the deisel at working speed matches the chipper at 540 rpm
(youll have to gear it inbetween the two to suit)
3. Jul 15, 2015
### Moteor
Thanks Dean Barry, I am trying to work out the power ratings to be sure the machine works at the proper power.
To do this I need to reduce the engine speed by 4 from 2000 rpm to 540 to match the max speed of the the machine?
Also the engine has a power curve of HP so it's not always at 95 hp but that depends on the rpm.
4. Jul 15, 2015
### dean barry
Yes, so that when your engine is at 2000 rpm, the chipper is at 540 rpm, so whatever power your engine is producing at 2,000 rpm this will be reflected at the chipper regardless of the chipper speed, so make sure the chipper is at 540 rpm when youre engine is at 2,000 rpm, so yes the speed reduction ratio between your engine and the chipper is 2,000 : 540, so if it involved two sprockets and a chain, your engine sprocket would have say 25 teeth, and the chipper would have 25 * ( 2,000 / 540 ) = 92.59 teeth (say 93 teeth)
The torque increase from engine to chipper would be proportional to this ratio also.
signing off, back tommorow
5. Jul 15, 2015
### Moteor
That's what I thought would happen. The thing I was thinking is if the torque is increased then don't that effectively increase the HP in proportion?
What is the consequence of putting too much torque through the machine? Snapped belts and twisted shafts I think?
6. Jul 15, 2015
### SteamKing
Staff Emeritus
Since Power = Torque * RPM * constant
and
Torque out = Torque in * Gear Ratio
but
RPM out = RPM in / Gear Ratio
So
Power in = Torque in * RPM in * constant
and
Power out = Torque out * RPM out * constant
but
Power out = Torque in * Gear Ratio * (RPM in / Gear Ratio) * constant
The gear ratio cancels out of the power equation, so the power transmitted remains constant, except for some minor losses in the gearbox.
Among other things.
7. Jul 15, 2015
### Moteor
Thanks Steam King :)
So if the engine was running at 2000 RPM the power through the engine would be 170 Nm and the HP would be about 40?
So to ramp up the horsepower would have what effect on the machine (wood chipper)?
And if it gives the power, whats the best way to reduce the speed coming out of the engine?
Last edited: Jul 15, 2015
8. Jul 15, 2015
### jack action
Your wood chipper is meant to work at a constant 540 rpm. At this rpm, if the power input is 105 hp then the torque input is 1021 lb.ft (You made a mistake in your first post). These are unalterable facts. If you change one, you need to change at least another one. Since you have two of them fixed, the third one is also fixed by default.
It doesn't matter if the power source produces the 105 hp @ 54 rpm (i.e. with 10212 lb.ft of torque) or @ 5400 rpm (i.e. with 102 lb.ft of torque). Because you will have to use some sort of gear ratio to set it to the required 540 rpm (0.1:1 or 10:1, depending on the chosen engine) and in any case - once the correct 540 rpm is achieved - the torque will be 1021 lb.ft ( = 10212 * 0.1 = 102 * 10).
That is the beauty of power: It takes into account the energy delivery rate, no matter whether it comes from a large force recurring at a slow rate (high torque, low rpm) or a small force recurring at a fast rate (low torque, high rpm). With gear ratios, you can easily transfer the input energy rate from force to speed or vice-versa; but the total energy rate (a.k.a. power) will always stay the same. Energy cannot be created nor destroyed, but its «form» can be altered (force or speed, but can also be converted to heat, electrical current & voltage, potential energy, etc.).
9. Jul 15, 2015
### Moteor
So what would your recommendation be for the gear ratio to give as much power as possible?
I am thinking at peak HP @ around 3500 RPM would do it with a reduction of 6.48 to 1?
Thanks for the help
10. Jul 15, 2015
### insightful
Call a local vendor for a V-belt drive. Your 6.8 to 1 looks good.
11. Jul 15, 2015
### jack action
The maximum power of your engine is 62.6 kW (84 hp) @ 3789 rpm. To drop it to 540 rpm, you need a 7:1 gear ratio (=3789 / 540). The power at the chipper will be 84 hp (minus some losses in your gear/pulley/chain reducer, probably something like 5% or about 4 hp). So you will always be below the 105 hp, because you cannot increase or reduce the power produced by the engine.
For example, if you put a 3.7:1 gear ratio, you would have to run the engine at 2000 rpm to get the required 540 rpm (3.7 * 540 = 2000). At that rpm your engine produces a maximum power of about 35 kW (47 hp). Subtracting a 5% loss (2.4 hp), you end up with 44.6 hp at the chipper's input.
What it means is that your engine will not be able to perform as much work as it was intended to do, no matter the rpm you choose (and that is assuming you run your engine full throttle, which might not be recommended for a long period of time). Either you will have to feed it slower or you will have to put smaller branches. But if you want more from you chipper (or any machine for that matter), you need more power.
Forget about torque numbers, what is important is power. If the specs would have been 1021 lb.ft @ 540 rpm, you should have said to yourself: «I wonder how much power that is?» You do the math, find 105 hp and then you would have said: «Ah! I need an engine that can produce 105 hp and - no matter at what rpm it produces that 105 hp - I need to bring its speeds down to 540 rpm.»
Normally you would want to use the engine at a rpm where it has the best fuel consumption (which is usually somewhere around maximum torque; it depends on the load your putting on the engine). If you choose an ideal engine, it should be able to produce the 105 hp at that rpm.
So your engine is probably build to perform best at around 2500 rpm (that could include performance of the cooling and lubrication systems as well), where you won't be able to get more than 45 kW (60 hp), and that is full throttle. The engine would usually be used at 84 hp @ 3789 rpm for short period of time only (maybe that is your intended use?). Only racing engines are usually prepped to perform at maximum power "all the time" (where "all the time" could mean one season - or even only one race - before an overhaul is due).
12. Jul 16, 2015
### dean barry
Says it all really, so, to summarise:
if you had a more powerful engine and ran it (comfortably) below its peak power revs, then calculate the inbetween ratio to suit the chipper speed.
An issue may be the large size of the chipper sprocket (or pulley wheel) required, a more powerful lesser rpm engine might be the best option overall
Example:
Say you had a engine which produced a maximum 200 hp @ 4,000 rpm.
Reduce the running speed to say 2,000 rpm, which is 100 bhp or thereabouts (depending on which engine you use)
Your speed reduction ratio = 2,000 / 540 = 3.7 : 1
So your sprockets ( chipper : engine ) would be say: 74 : 20
The torque rating of the engine at 2,000 rpm multiplied by 3.7 would = the torque at the chipper at 540 rpm
13. Jul 16, 2015
### Nidum
An important consideration when matching engines to loads is speed regulation .
The rated power of your wood chipper only applies at max loading - ie biggest logs and highest feed rate . Actual horsepower demand will vary greatly with load and could drop to a very small value when no chipping is being done .
If engine is unregulated it could speed up considerably at lower loadings .
Ideally your engine should have built in speed regulation .
14. Jul 16, 2015
### Moteor
Thanks for all the replies, I have spent considerable time working this out. At the moment I am working with this:
My plan is to have a compound pulley system with 3 groove vbelt pulleys rated to at least +20% over 105 HP.
I am looking at the 2 pulley sizes of 117.5 and 305.5 and being in a drive chain to have 2 of each.
they give a ratio of 2.6 : 1 so 2 of them is 2.6 x 2.6 = 6.76.
so at 3,650 RPM it will be reduced by 6.76 times giving 539.4 rpm. So this gives peak power at peak revs.
Please can you confirm if this is right or REALLY IS ALL IN MY HEAD!!
15. Jul 18, 2015
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# SL.5.1.D Lesson Plans
#### CCSS.ELA-LITERACY.SL.5.1.D
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"Review the key ideas expressed and draw conclusions in light of information and knowledge gained from the discussions."
These lesson plans can help students practise this Common Core State Standards skill.
## Lesson Plans
Giving Directions: Tell Me How
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Giving Directions: Tell Me How
Students will have fun engaging in activities that develop their ability to write sequential step-by-step directions. This lesson helps young learners with being detailed and using transition words in their writing.
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Adobe Spark: Parts, Purposes & Complexities
Discover a new tool for creative assessments and lessons! In this lesson, students will investigate Adobe Spark using Agency by Design thinking routines to help them understand how they can use technology to communicate ideas.
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Comparing Two Division Methods
Challenge students to compare two strategies for solving division expressions! Students will focus on their language use in this pre-lesson to the Dicey Division lesson.
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Visually Multiplying Fractions
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Visually Multiplying Fractions
Help students colour-code their way to multiplying fractions! Students will learn how to multiply fractions using area models. Use this lesson on its own or use it as support to the lesson Area Models and Multiplying Fractions.
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Compare Common Denominators Methods
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Compare Common Denominators Methods
Challenge your students to compare two methods for finding the least common multiple between two fractions. Use this lesson on its own or as support for the lesson Make It Work! Adding Fractions with Unlike Denominators.
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Comparing Tape Diagrams
Compare two different ways to use tape diagrams! This lesson discusses fractions and multiplication within tape diagrams. Use this lesson as support for the lesson Illustrating Fractions and Whole Number Products with Tape Models.
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Have your students evaluate orally and in writing what happens to expressions when there are two or more operations. Use this lesson as a standalone lesson or as support to the lesson Express Yourself.
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Area Models and Mixed Numbers
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Area Models and Mixed Numbers
Help students visually represent multiplication with mixed numbers and whole numbers. Use this lesson as a standalone lesson, or as support for the lesson Multiplication of Mixed Numbers with Area Models.
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# Asymptotic rate of growth of a sum
Consider $$\Phi_0(x) = \sum_{i=0}^{\infty} (1-x)^i,$$ where $x \in (0,1)$. As $x \rightarrow 0$, $\Phi_0(x)$ blows up as $\Theta(1/x)$. Similarly, consider $$\Phi_1(x) = \sum_{i=0}^{\infty} i (1-x)^i.$$ As $x \rightarrow 0$, a direct calculation shows that $\Phi_1(x)$ blows up as $\Theta(1/x^2)$. What I am interested in is the sum $$\Phi_{1/2}(x) = \sum_{i=0}^{\infty} \sqrt{i} (1-x)^i.$$ My question is: how fast does $\Phi_{1/2}(x)$ blow up as $x \rightarrow 0$?
I suspect this is routine and easily solvable with some trick. In the case of $\Phi_0$ and $\Phi_1$, explicit expressions make the asymptotic rate of blowup easily calculable. I don't know a simple expression for $\Phi_{1/2}$.
-
## 2 Answers
One trick that I like is to estimate the sum via integrals. The following is a rough informal outline, but you can formalize it via upper and lower bounds.
The basic idea is that roughly $\sum_{n=0}^{\infty} f(n)(1-x)^n \sim \int_0^{\infty} f(n)\exp(-n(-\log(1-x)))dn \sim \int_0^{\infty} f(n)\exp(-nx)dn$ for small $x$. One particularly nice aspect of this method is that you can use it for any nice enough $f(n)$.
In particular, for $f(n) = n^{\alpha},\,\alpha \geq 0$, the integral is $\frac{\Gamma(1+\alpha)}{x^{1+\alpha}}$.
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+1. That is a neat quick proof. – user17762 Jan 20 '13 at 5:56
@Marvis, Thank you. I remember doing this for $f(n) = \log (n+1)$, it saves lives :) It takes some effort to formalize the steps however. Perhaps for the OP, it may be a better idea to go with your answer in order not to reinvent the wheel :) – Anon Jan 20 '13 at 6:00
The Polylogarithm function is defined as $$\text{Li}_s(z) = \sum_{k=1}^{\infty} \dfrac{z^k}{k^s}$$ Hence, $$\text{Li}_{-1/2}(z) = \sum_{k=1}^{\infty} \sqrt{k} z^k$$ Hence, the series you have is $\text{Li}_{-1/2}(1-x)$. And you also have that $$\lim_{x \to 0} x^{3/2} \text{Li}_{-1/2}(1-x) = \dfrac{\sqrt{\pi}}2$$ i.e. as $x \to 0$ $$\text{Li}_{-1/2}(1-x) \sim \dfrac{\sqrt{\pi}}{2 x^{3/2}}$$ In general as $x \to 0$, we have $$\text{Li}_{s}(1-x) \sim \dfrac{\Gamma(1-s)}{x^{1-s}}$$
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# Polynomials Ex-2.3 Interview Questions Answers
### Related Subjects
Question 1 :
Divide the polynomial p(x) by the polynomialg(x) and find the quotient and remainder in each of the following:
_(_i) p(x) = x3 – 3x2 + 5x – 3, g(x) = x2 – 2
_(ii) p(x) = x4 – 3x2 + 4x + 5, g(x) = x2 + 1 – x
_(iii) p(x) = x4– 5x + 6, g(x) = 2 – x2
(i) p(x) = x3-3x2+5x–3, g(x) = x2–2
Solution:
Here
Dividend = p(x) = x3-3x2+5x–3
Divisor = g(x) = x2– 2
Therefore, upon division we get,
Quotient = x–3
Remainder = 7x–9
(ii) p(x) = x4-3x2+4x+5, g(x) = x2+1-x
Solution:
Here
Dividend = p(x) = x– 3x+4x +5
Divisor = g(x) = x2 +1-x
Therefore, upon division we get,
Quotient = x+ x–3
Remainder = 8
(iii) p(x) =x4–5x+6, g(x) = 2–x2
Solution:
Here
Dividend = p(x) =x4 – 5x + 6 =x+0x2–5x+6
Divisor = g(x) = 2–x2 = –x2+2
Therefore, upon division we get,
Quotient = -x2-2
Remainder = -5x + 10
Question 2 :
Check whether the first polynomial is a factorof the second polynomial by dividing the second polynomial by the firstpolynomial:
_(i) t2 – 3, 2t4 + 3t3 – 2t2– 9t – 12
_(ii) x2 + 3x + 1, 3x4 + 5x3 – 7x2 + 2x + 2
_(iii) x2 + 3x + 1, x5 – 4x+ x2 + 3x + 1
(i) t2-3, 2t+3t3-2t2-9t-12
Solutions:
Here,
First polynomial = t2-3
Second polynomial = 2t+3t3-2t-9t-12
As we can see, the remainder is left as 0.Therefore, we say that, t2-3 is a factor of 2t2+3t+4.
(ii)x2+3x+1 , 3x4+5x3-7x2+2x+2
Solutions:
Here,
First polynomial = x2+3x+1
Second polynomial = 3x4+5x3-7x2+2x+2
As we can see, the remainder is left as 0.Therefore, we say that, x2 + 3x + 1 is a factor of 3x4+5x3-7x2+2x+2.
(iii) x3-3x+1, x5-4x3+x2+3x+1
Solutions:
Here,
First polynomial = x3-3x+1
Second polynomial = x5-4x3+x2+3x+1
As we can see, the remainder is not equal to0. Therefore, we say that, x3-3x+1 is not a factor of x5-4x3+x2+3x+1.
Question 3 :
Obtain all other zeroes of 3x4+6x3-2x2-10x-5,if two of its zeroes are √(5/3) and – √(5/3).
Solutions:
Since this is a polynomial equation of degree4, hence there will be total 4 roots.
√(5/3) and – √(5/3) are zeroes of polynomial f(x).
(x –√(5/3)) (x+√(5/3) =x2-(5/3) = 0
(3x2−5)=0, is a factor of given polynomial f(x).
Now, when we will divide f(x) by (3x2−5)the quotient obtained will also be a factor of f(x) and the remainder will be0.
Therefore, 3x+6x−2x−10x–5
= (3x–5)(x2+2x+1)
Now, on further factorizing (x2+2x+1)we get,
x2+2x+1=0
x2+x+x+1 = 0
x(x+1)+1(x+1) = 0
(x+1)(x+1) = 0
So, its zeroes are given by: x=−1 and x = −1.
Therefore, all four zeroes of given polynomialequation are:
√(5/3),- √(5/3) , −1 and −1.
Question 4 :
On dividing x3-3x2+x+2 by a polynomial g(x), the quotient and remainder werex–2 and –2x+4, respectively. Find g(x).
Solution:
Here,
Dividend, p(x) = x3-3x2+x+2
Quotient = x-2
Remainder = –2x+4
We have to find the value of Divisor, g(x) =?
As we know,
Dividend = Divisor × Quotient + Remainder
x3-3x2+x+2 = g(x)×(x-2) + (-2x+4)
x3-3x2+x+2-(-2x+4) =g(x)×(x-2)
Therefore, g(x) × (x-2) = x3-3x2+x+2
Now, for finding g(x) we will divide x3-3x2+x+2with (x-2)
Therefore, g(x) = (x2–x+1)
Question 5 :
Give examples of polynomials p(x), g(x), q(x)and r(x), which satisfy the division algorithm and
(i) deg p(x) = deg q(x)
(ii) deg q(x) = deg r(x)
(iii) deg r(x) = 0
krishan
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This in an excerpt from this book
In old times, number theory was also known as arithmetic. However, now arithmetic and number theory are considered as separate branches from each other’s, it was not same in old times. Number theory is one of the many important branches of pure mathematics. This branch is mainly dedicated and includes study about integers. This theory describes many fundamental and basic concepts of mathematics that were used to develop modern concepts. Thus, number theory is often referred as “Queen of Mathematics”. In number theory, following concepts are described:
Concept of prime numbers
Properties of objects that are derived through integers
Generalization of integers
Rational numbers and algebraic integers are significant concepts that are included in number theory. In number theory, integers are considered as a solution to a particular problem. This concept is known as Riemann Zeta Function. However, it is not necessary to consider them as solution only; they can also be considered in themselves. Study of analytical objects helps to understand questions in number theory. Properties of integers, prime numbers and number-theoretic objects are described in Riemann Zeta Function. These properties can be studied descriptively in a separated branch named analytic number theory. In Diophantine approximation, real numbers are learnt in ration to relational number.
In older terms, arithmetic was used to refer number theory. However, it was separated in early 20th century. Arithmetic word is now used to refer to general elementary calculations. Term arithmetic is now used in many fields such as:
Mathematical logic
Peano arithmetic
Computer science
Floating point arithmetic
In late 20th century, French theorists leaved a noticeable impact on number theory. Due to their influence, they again related tern arithmetic with number theory. However, many theorists argued upon this and denied to accept this as it was already proven false in past time. However, term arithmetical is now considered as adjective to number-theoretic. Early 20th century was a golden time for development of number theory, especially the time span of 1930s and 1940s. Many important results were acquired in that period. Later on, 1970s was proven an important period as well with the development of computational complexity theory.
Number theory is an important branch of pure mathematics since it contains many basic concepts that are used to build up complex concepts of pure mathematics. One who is looking for a breakthrough in broad term mathematics is suggested to start from this theory. It will clear up basic concepts so it will be surprisingly easy to understand complex concepts. This short book describes all the basic concepts without going in too deep. So, one can use this basic knowledge to understand complex concepts easily and effectively.
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# Algebra Tiles Explorations of al-Khwārizmī ’s Equation Types: Modeling al-Khwārizmī’s More Complicated Problems
Author(s):
Günhan Caglayan (New Jersey City University)
Following his treatment of the solution procedures for each of the six cases of equation, al-Khwārizmī went on in his treatise to present six problems “which will serve to bring the subject nearer to the understanding, to render its comprehension easier, and to make the arguments more perspicuous” [Rosen 1831, 35]. These problems were followed by a large collection of “various questions” that further illustrated how al-Khwārizmī’s procedures for “calculating by completion and reduction must always necessarily lead you to one of these cases” [Rosen 1831, 35]. In the examples below, we illustrate how to model such problems using algebra tiles. We have selected these particular examples in part because they avoid the use of fractions and mixed numbers, yet show the complexity of the equations that al-Khwārizmī treated. Because some problems involve the difference of two quantities, note that it is very natural to use subtraction within the algebra tile models. Following al-Khwārizmī’s advice to his readers that they should “reduce this [equation] according to the rules, which I have above explained to you” [Rosen, 1831, 51], we have omitted the algebra tile representations for some of the later details in our last two examples.
• Example 11: A problem leading to the case “squares = numbers”
• Example 12: A problem leading to the case “squares = numbers"
• Example 13: A problem leading to the case “squares = roots”
• Example 14: A problem leading to the case “squares and numbers = roots”
• Example 15: A problem leading to the case “squares and numbers = roots”
Günhan Caglayan (New Jersey City University), "Algebra Tiles Explorations of al-Khwārizmī ’s Equation Types: Modeling al-Khwārizmī’s More Complicated Problems ," Convergence (October 2021)
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# Illustrative Mathematics Grade 7, Unit 7, Lesson 3: Nonadjacent Angles
Learning Targets:
• I can determine if angles that are not adjacent are complementary or supplementary.
• I can explain what vertical angles are in my own words.
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Illustrative Math
Grade 7
#### Lesson 3: Nonadjacent Angles
Let’s look at angles that are not right next to one another.
Illustrative Math Unit 7.7, Lesson 3 (printable worksheets)
#### Lesson 3 Summary
The following diagrams explain what are vertical angles.
Given a and b are numbers, and a + b = 180, which statements also must be true?
a = 180 - b
a - 180 = b
360 = 2a + 2b
a = 90 and b = 90
#### Lesson 3.2 Polygon Angles
Use any useful tools in the geometry toolkit to identify any pairs of angles in these figures that are complementary or supplementary.
#### Lesson 3.3 Vertical Angles
Use a straightedge to draw two intersecting lines. Use a protractor to measure all four angles whose vertex is located at the intersection.
Compare your drawing and measurements to the people in your group. Make a conjecture about the relationships between angle measures at an intersection.
#### Lesson 3.4 Row Game: Angles
Find the measure of the angles in one column. Your partner will work on the other column. Check in with your partner after you finish each row. Your answers in each row should be the same. If your answers aren’t the same, work together to find the error and correct it.
#### Lesson 3 Practice Problems
1. Two lines intersect. Find the value of b and c.
2. In this figure, angles R and S are complementary. Find the measure of angle S.
3. If two angles are both vertical and supplementary, can we determine the angles? Is it possible to be both vertical and complementary? If so, can you determine the angles? Explain how you know.
4. Match each expression in the first list with an equivalent expression from the second list.
5. Factor each expression.
6. The directors of a dance show expect many students to participate but don’t yet know how many students will come. The directors need 7 students to work on the technical crew. The rest of the students work on dance routines in groups of 9. For the show to work, they need at least 6 full groups working on dance routines.
a. Write and solve an inequality to represent this situation, and graph the solution on a number line.
b. Write a sentence to the directors about the number of students they need.
7. A small dog gets fed 3/4 cup of dog food twice a day. Using d for the number of days and f for the amount of food in cups, write an equation relating the variables. Use the equation to find how many days a large bag of dog food will last if it contains 210 cups of food.
The Open Up Resources math curriculum is free to download from the Open Up Resources website and is also available from Illustrative Mathematics.
Try the free Mathway calculator and problem solver below to practice various math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations.
We welcome your feedback, comments and questions about this site or page. Please submit your feedback or enquiries via our Feedback page.
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# Mass Volume and Graphing
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Simple PowerPoint used for explaining how to measure matter including mass, volume, and graphing.
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### Mass Volume and Graphing
1. 1. Measuring Matter
2. 2. What is matter? • Matter is everything around you!
3. 3. All matter has . . . • Mass • And • Volume
4. 4. Examples of Matter • Anything that has mass and volume • Pencil, book, car, water, air, etc. . . • That includes you!
5. 5. What is mass? • Mass is the amount of matter in an object. • Elephant = a lot of matter = LARGE Mass • Paperclip = small amount of matter = SMALL mass
6. 6. How do we measure mass? • We measure mass with a triple beam balance • Make sure it is balanced before you begin. • Move the riders until the the arm is at the zero. • Add all the numbers together. • Mass is measured in grams (g).
7. 7. What is volume? • The amount of space taken up or occupied by an object.
8. 8. How do you find the volume of a rectangular solid object? • Volume = length • width • height • V = l • w • h 2 cm 2 cm 4 cm Example: V = 2 cm • 2 cm • 4 cm V = 16 cm3
9. 9. How do you find the volume of an irregular-shaped solid object? • Water Displacement: – Fill a graduated cylinder with a certain amount of water. – Drop the object into the graduated cylinder with the water. – Measure the change in volume of the water. • Measure how far the water rose when the object was placed in the graduated cylinder.
10. 10. How do you find the volume of a liquid? – Use a graduated cylinder. – Measure from the bottom of the meniscus. – Measure in milliliters (mL).
11. 11. How do you find the volume of a gas? • A gas will fill the volume of its container, so find the volume of the container and you have the volume of the gas
12. 12. Reminder… • 1mL=1cm3 • Write this on your notes in the margins next to the volume section and put a box around it!
13. 13. Graphing
14. 14. Why do we graph data? • To show the relationship between the independent and dependent variable. • To compare data.
15. 15. Rules for Graphing: 1. Always use a ruler or computer program 2. Have your graph take up half a page to a full page of paper 3. Title your graph descriptively – Title should relate to what is being shown in the graph
16. 16. Rules for Graphing: 4. Label the x (along the bottom) and y axis (along the left side) – x-axis should be the independent variable (the variable we change during the experiment, ex: time or distance ) – y-axis should be the dependent variable (the variable we observe/measure in the experiment, ex: temperature or mass )
17. 17. Rules for Graphing: 5. Number the x and y axis with consistent numbers (increasing by 5’s or 10’s, etc.) – Use the majority of each axis for your graph 5. Use different colors/patterns if you are showing multiple trials.
18. 18. How do you know which graph to use? • Depends on: – The information (variables) you use – What you are trying to show
19. 19. Line Graph vs. Bar Graph Line Graph: •Used to show changes that occur in related variables •Shows trends (changes) over time Bar Graph: •Use to compare data (such as measurements, amounts or changes) •Can show large changes over time Pie Chart: •Circle that shows how parts relate to the whole. •Shows proportions
20. 20. Which type of graph would you use for: • Plant growth over time • Population of a city over 10 years • Students preference in food (pizza vs hamburgers vs hot dogs vs chicken nuggets) • Number of people in 4 different towns • The number of hours spent on Facebook by students each week Line Graph Line or Bar Pie Chart Bar Graph Line Graph
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1. ## Simple Integration
$\int \frac{u}{\sqrt{u^2 +16}} \ du$
2. Originally Posted by Air
$\int \frac{u}{\sqrt{u^2 +16}} \ du$
Make the substitution $w = u^2 + 16$. Note that $\frac{dw}{du} = 2u \Rightarrow du = \frac{dw}{2u}$ ....
3. Hi
$\frac{ \mathrm{d}}{\mathrm{d}u} \left( \sqrt{u^2+16} \right)=?$
4. Originally Posted by flyingsquirrel
Hi
$\frac{ \mathrm{d}}{\mathrm{d}u} \left( \sqrt{u^2+16} \right)=?$
$\frac{ \mathrm{d}}{\mathrm{d}u} \left( \sqrt{u^2+16} \right) = \frac{ \mathrm{d}}{\mathrm{d}u} \left( u^2+16 \right)^\frac{1}{2} = \frac{1}{2}(u^2 + 16)^{-\frac{1}{2}}(2u) = (u)(u^2 + 16)^{-\frac{1}{2}} = \frac{u}{\sqrt{u^2+16}}$
5. Originally Posted by mr fantastic
Make the substitution $w = u^2 + 16$. Note that $\frac{dw}{du} = 2u \Rightarrow du = \frac{dw}{2u}$ ....
You should try this way to see why flyingsquirrel was able to suggest you consider the derivative $\frac{ \mathrm{d}}{\mathrm{d}u} \left( \sqrt{u^2+16} \right)$ ......
6. Originally Posted by mr fantastic
You should try this way to see why flyingsquirrel was able to suggest you consider the derivative $\frac{ \mathrm{d}}{\mathrm{d}u} \left( \sqrt{u^2+16} \right)$ ......
$\int \frac{u}{\sqrt{u^2 + 16}} \ du$
$w = u^2 + 16 \therefore \frac{dw}{du} = 2u \implies du = \frac{dw}{2u}$
$\int \frac{u}{\sqrt w} . \frac{dw}{2u}$
$\int \frac{1}{\sqrt w} . \frac{dw}{2}$
$\frac{1}{2} \int \frac{1}{\sqrt w} \ dw$
$\frac{1}{2} \left[ 2 w ^{\frac{1}{2}} \right]$
$w^{\frac{1}{2}} = (u^2+16)^\frac{1}{2} = \sqrt{u^2 + 16}$
7. Originally Posted by Air
$\int \frac{u}{\sqrt{u^2 + 16}} \ du$
$w = u^2 + 16 \therefore \frac{dw}{du} = 2u \implies du = \frac{dw}{2u}$
$\int \frac{u}{\sqrt w} . \frac{dw}{2u}$
$\int \frac{1}{\sqrt w} . \frac{dw}{2}$
$\frac{1}{2} \int \frac{1}{\sqrt w} \ dw$
$\frac{1}{2} \left[ 2 w ^{\frac{1}{2}} \right]$
$w^{\frac{1}{2}} = (u^2+16)^\frac{1}{2} = \sqrt{u^2 + 16}$
+ C.
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# How to Calculate ROI on Commercial Real Estate Investment
David Cohn
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August 25, 2020
CRE Investment
Look at any financially-independent person’s portfolio, and you will probably see some commercial real estate (CRE) assets in it. CRE has long been a favored investment for those seeking to build stable, long-term wealth that can be passed on to the next generation.
If you’re interested in putting money in this asset class, it’s important to understand how to calculate the potential return you can get from your investment so you can decide if the property is truly worth buying.
You don’t need to be good at math to succeed in CRE investing, but having a working knowledge of basic concepts—including potential return on investment or ROI—is certainly essential.
## The Formula
What is Return-on-Investment?
In its simplest terms, ROI is the ratio between the cost of investment and net profit from it.
In other words, it’s how much profit is made on a property investment expressed as a percentage of that investment’s cost. The higher the ROI, the better—it means that your investment dollars are efficiently and effectively used to generate profit. Many CRE investors (and investors in general) use the S&P 500 average returns as an ROI benchmark.
This is the basic formula for computing a CRE asset’s ROI:
ROI = Net Profit / Total Investment x100
Let’s use an example to illustrate how this formula is applied.
Let’s say that you flip commercial properties. You purchased a building at an auction for \$750,000 and spent \$350,000 in renovations.
Yours netted \$1,600,000 from selling the renovated building after-sales, commission, and other expenses were paid.
#### What is your ROI?
Your net profit is what you netted (\$1,600,000) minus everything you spent (\$750,000 + \$350,000), so it amounts to \$500,000.
Your total investment is what you spent (\$750,000 + \$350,000), which totals to \$1,100,000.
ROI = Net Profit / Total Investment x 100
ROI = 500,000 / 1,100,000 * 100
ROI = .45 * 100
Your ROI is 45%.
This is the basic formula and it is not always the right one to use. There are actually several ways to calculate commercial rental returns, and the ‘best’ way to do it really depends on what you want to figure out.
Let’s discuss three of the most common ways to calculate investment property returns:
1. Capitalization rate
2. Cash-on-cash Return
3. Total Return
We will also talk about when each of these methods is most useful.
## Capitalization Rate
What is CAP Rate?
This shows the CRE property’s net income in proportion to its market value. When computing the cap rate, the purchase price of the investment is usually used as its market value especially if the asset was acquired recently.
Let’s say you spent \$1,000,000 to acquire a commercial rental property. It generates \$120,000 each year in rental income. When you have paid for taxes, maintenance, insurance, and other operating expenses (not including mortgage payments), the net income is \$80,000. Divide this by the cost of the property and you will get an 8% cap rate.
The cap rate can be extremely useful if you want an apples-to-apples’ method of comparing two or more potential investment properties with different financing structures because it does not account for debt repayment.
If a lender is asking for a 15% down payment on a 2-unit property and a 25% down payment for a 4-unit building, for example, you can use the expected cap rate to figure out which deal is better based on its rental income potential.
Cap rates are also widely used to evaluate commercial properties. If your property earns \$80,000 per year in net income, you can use the cap rate formula to determine its approximate market value.
## Cash-on-Cash Return
Cash-on-Cash Return
It is essentially the ratio between a property’s annual cash flow and its down payment. In other words, it shows the pre-tax cash inflows the investor receives from the property, and the pre-tax outflows the investor pays.
It’s not uncommon for CRE lenders to finance different properties differently. Reserve requirements and down payments can vary widely depending on how many units a property has, how old it is, and other factors.
One of the best ways to figure out which commercial property can produce the best profits for your money is to determine the cash-on-cash return. This is the metric to use if you are interested in several properties with different layouts and prices and want to figure out which one of them will produce the best cash flow.
### Let’s say that you’re interested in two commercial properties:
1. A single-unit property sells for \$1,000,000. The lender wants 20% down. Including closing costs, the property will cost you \$250,000 to acquire. It is expected to fetch \$10,000 per month in rent. All expenses–including mortgage payments–are estimated at \$8,000 monthly. This means that the property will produce \$2,000 each month in net income.
2. A two-unit property costs \$1,600,000. The lender requires a 25% down payment, and it will cost you \$450,000 to acquire it (including closing costs). You expect to get \$16,000 per month on rent, with monthly expenses estimated \$13,000. This means that the property produces \$3,000 per month in net income.
The first property will generate a cash flow of \$24,000 annually. Because you spent \$250,000 to acquire it, its cash-on-cash return is 9.6%.
The second property cash flows \$36,000 per year. You spent \$450,000 to acquire it, translating toa cash-on-cash return of 8%.
In this case, the first property can be reasonably expected to generate more profit relative to how much money it costs to acquire it.
## Total Return
What is Total Return?
‘Total return’ refers to the combination of equity appreciation and income.
So far we have talked about cash-on-cash return and cap rate, which are both income metrics. But income is not the only way to make money from CRE. You can also profit from equity appreciation.
If you paid \$1,000,000 in cash to buy a property and it generated a net income of \$60,000 in one year and increases by \$50,000 in value, your total return is \$110,000or 11%.
‘Total return’ is usually used in CRE investing as a long-term metric—an evaluation of an asset’s performance after it is sold. Let’s take a look at the more complex application of this formula.
Let’s say you bought a \$1,000,000 rental property with a mortgage, paying \$250,000 out of pocket. You decide to sell it after 5 years. During the entire holding period, it produced a net rental income of \$150,000. You sell it for \$1,200,000, which is \$200,000 more than its original purchase price.
In this case, your total return over five years is \$350,000 or a whopping 1400% total 5-year return based on your investment of \$250,000.
### The Best Metric to Use
These three metrics can be extremely useful for different scenarios. The capitalization rate is great for assessing the market value of CRE properties. Cash-on-cash return is a good way to compare different properties, while the total return is very useful for assessing an asset’s performance (including equity appreciation).
There is no single ‘best’ metric—it all depends on what you want to find out. If you want to succeed in commercial real estate investing, then it’s important to master all of these so you can quickly and accurately analyze investment opportunities.
## Don’t Forget to Factor These in
Now that you know how to determine the ROI of a commercial property, it’s important to understand the different financial factors that directly impact a property’s ROI so you can take them into account when assessing an investment’s real profitability. These factors are:
• Debt (loans or mortgages)
• Taxes and insurances
• Maintenance and rehabilitation costs
• Rental income
• Equity
Let’s take a look at each one.
### 1. Debt
You need to figure out the true total cost of your investment so you can accurately determine your property’s ROI. This is not a concern if you pay out-of-pocket whenever you purchase an asset, but most investors need financial assistance in this regard, in the form of loans and mortgages.
Any debt paid (or to be paid) over the course of the investment—be it a lump sum payment of a monthly mortgage payment—is one of the biggest financial costs of an investment property. It’s crucial to take them into account when calculating your ROI because these payments are subtracted from the property’s gross income to establish the net income.
Considerations need to be made for the amount the lender provided, the terms of the loan, and the interest rate when computing for the net income. Keeping an accurate record of these particular details can help you accurately calculate the true ROI of your investment property over time.
### 2. Taxes and Insurance
These factors are also crucial in determining a commercial asset’s ROI. Insurance and property taxes are alike in the sense that they vary depending on the type of property and the city/state where it is located. These expenses are also both determined by a third party—a property assessment organization or an insurance company.
### 3. Maintenance and Rehabilitation Costs
If you’re a seasoned CRE investor, then you probably have your fair share of repairs and renovations. From minor things like fixing faulty plumbing to more involved repairs like completely rewiring a building’s electrical work, investors regularly have to deal with the headache of repairs—especially those who flip commercial properties to generate profits quickly. These repairs are often quite expensive. Depending on how much work is needed to bring a property up to scratch, this financial expense can be quite large.
Renovations are even more costly. They are typically used by CRE investors whose goal is to increase the value of an investment property and sell it at a much higher price than they paid for. Renovations encompass various expenses including all minor repairs and major rework for rehabilitating the property and increasing its market value.
On top of all these, plumbing and electrical systems need to be regularly inspected and maintained to ensure that the building’s amenities remain safe and functional. This is why maintenance and rehab costs are a big consideration when calculating an investment property’s ROI.
### 4. Equity
Commercial properties change in value as time goes by. Most real estate investors make it a goal to build equity in their properties as they pay off loans and as the property’s value increases over time.
Equity is obviously crucial; the higher it is, the more funds investors can use for future investments. It is a critical factor in determining an asset’s ROI.
Another method called the ‘cost method’ involves computing for a property’s ROI by taking the equity and dividing it by the property’s costs. Equity is determined by taking the market value and subtracting the total cost of the property from it.
Let’s say that commercial property was acquired for \$3,000,000, for example. Repairs and renovations cost roughly \$250,000, and they increased the property’s market value to \$4,000,000. The total property cost is \$3,250,000, so the equity \$750,000.
The cost method is then used to calculate the ROI:
ROI = \$750,000 (equity) ÷ \$3,250,000 (cost) = 23%
### 5. Rental Income
Most commercial properties make money primarily from tenants. The number of tenants renting the property, how much you charge these tenants, and how often you charge them (annual or monthly) all affect your investment property’s income, which then affects its ROI. Maintaining an occupancy rate close to 100% is therefore critical to generating a better ROI and growing your profits.
As discussed in the first part of this blog, there are multiple ways to calculate an investment property’s ROI, and these are the main factors to consider to accurately determine it. Now let’s talk about some ways to increase the ROI on a commercial real estate asset.
### Making more money from CRE investing
ROI is undoubtedly the main driver of the investment decisions of people who buy commercial property. So how do you achieve a stronger ROI on an asset? It all comes down to spotting opportunities, finding a risk level you are comfortable with, and doing due diligence especially if you’re after capital preservation and security.
In an ideal world, a property can give you the ultimate combination of having high-profile tenants on long leases with favorable terms (such as tenants paying outgoings), plus fixed annual rental increases.
But it’s very hard to get all these in the real world. The good news is that even if a commercial property doesn’t exactly tick all these boxes, it can still potentially deliver better-than-average returns.
### Learn to Spot Opportunities
Perhaps you can find a property where the lease is due to expire. This will likely deter most investors. But if your research shows a solid probability that the tenants will renew their leases, then your gamble can pay off considerably.
It’s also important to look beyond the usual places. Commercial properties located in metropolitan areas tend to attract more investors, but those in regional areas can also yield strong ROIs. The key is to do your homework and truly understand the market you’re getting into.
Let’s take the example of regional bank branches. There are concerns about closures, so some investors stay away from them. But if you look closely, there are regional bank branches that are actually a prime investment because they cater to a big-enough catchment population that is isolated from the big competition, making them unlikely to close.
Well-located retail and medical premises in regional areas can also perform just as well as their counterparts in metropolitan areas—and they cost considerably less, so the yields are higher.
### Don’t Limit Yourself to Certain Asset Classes
It’s also important to think about your choice of the asset. The market is constantly changing, with different types of CRE properties falling in and out of favor at different times. Having a cool head and enough courage to go against the norm can potentially help you get significant returns.
A good example is a childcare center. Investors used to be very concerned about these kinds of commercial properties years ago, but they became high in demand in recent years, with investor-friendly lease terms and strong lease covenants.
Similarly, gas stations went through a period of being unpopular. But now with major brands on well-regulated and modern sites, this property sector is commanding much higher prices.
Investors who can spot asset classes that other investors ignore but have the potential for solid performance can achieve above-average ROI. This approach is risky, of course, but the rewards are enormous.
### Prevent Vacancies
Extended vacancies can lead to cash flow depletion, capital erosion, and reduced (even negated) ROI—not to mention a lot of stress. Improve your chances of avoiding vacancy by choosing a versatile type of property that is easy to release.
Retail properties are quite popular for this reason.
If a tenant leaves, it’s fairly easy to get a replacement and still achieve a good return, save for some inconvenience of onboarding a new tenant. It’s hard to go wrong with well-located supermarket properties. No wonder they are considered the ultimate “trophy investments” because they are always high in demand and there’s minimal risk of vacancy.
For investors who are more open to risk, industrial properties can potentially generate even higher returns, though they are harder to re-lease to new tenants.
### Take advantage of data to study the market
Finally, don’t be afraid to seek access to commercial property intelligence that can help you make more profitable decisions. It’s a great way to find high-performing investments that you can add to your portfolio.
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Capital Investors Direct funds different commercial properties across the country. Below are few commercial properties we lend hard money for –
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# Find maximum angle that can be made in glass medium $(\mu =1.5)$ if a light ray is refracted from glass to vacuum.
$(a)\;\sin^{-1} \frac{2}{5} \\ (b)\;\sin ^{-1} \frac{2}{3} \\ (c)\;\sin ^{-1} \frac{3}{5} \\ (d)\;\sin ^{-1} \frac{4}{5}$
$1.5 \sin C=1 \sin 90^{\circ}$
Where C is critical angle
$\sin C= \large\frac{2}{3}$
$C= \sin ^{-1} \large\frac{2}{3}$
Hence b is the correct answer.
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# M21-22
Author Message
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Math Expert
Joined: 02 Sep 2009
Posts: 52296
### Show Tags
16 Sep 2014, 00:11
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Difficulty:
55% (hard)
Question Stats:
45% (00:55) correct 55% (00:49) wrong based on 174 sessions
### HideShow timer Statistics
If $$a$$, $$b$$, and $$c$$ are consecutive integers, which of the following can be true?
I. $$abc$$ is divisible by 6
II. $$a + c - b$$ is odd
III. $$\frac{c}{a} \gt b$$
A. I only
B. I and II only
C. I and III only
D. II and III only
E. I, II, and III
_________________
Math Expert
Joined: 02 Sep 2009
Posts: 52296
### Show Tags
16 Sep 2014, 00:11
Official Solution:
If $$a$$, $$b$$, and $$c$$ are consecutive integers, which of the following can be true?
I. $$abc$$ is divisible by 6
II. $$a + c - b$$ is odd
III. $$\frac{c}{a} \gt b$$
A. I only
B. I and II only
C. I and III only
D. II and III only
E. I, II, and III
All the three scenarios are possible. Consider $$a = -8$$, $$b = -7$$, $$c = -6$$. Note that the question asks "what 'can' be true" rather than "what 'must' be true".
_________________
Intern
Joined: 18 Jun 2015
Posts: 41
### Show Tags
10 Sep 2016, 15:19
1
Can be true and Must be true makes all the difference here.
Intern
Joined: 07 Mar 2017
Posts: 7
Location: United States
GMAT 1: 580 Q40 V30
GPA: 3.6
### Show Tags
29 Mar 2017, 12:19
I should have looked at it more closely.....shouldnt be missing these type of problems because of language.
Intern
Joined: 07 Oct 2016
Posts: 14
Location: India
Concentration: Sustainability, General Management
GMAT 1: 640 Q47 V30
GMAT 2: 700 Q47 V40
GPA: 3.5
### Show Tags
25 Nov 2017, 07:12
Hi,
Can't we take -1, 0 & +1 ----> the answers in this case might be different ?
Also, if you could explain the difference b/w "must be true" & "can be true", I think i am missing a point here.
Thx
Math Expert
Joined: 02 Sep 2009
Posts: 52296
### Show Tags
25 Nov 2017, 07:15
quest800 wrote:
Hi,
Can't we take -1, 0 & +1 ----> the answers in this case might be different ?
Also, if you could explain the difference b/w "must be true" & "can be true", I think i am missing a point here.
Thx
Look at the highlighted part of the solution:
All the three scenarios are possible. Consider $$a = -8$$, $$b = -7$$, $$c = -6$$. Note that the question asks "what 'can' be true" rather than "what 'must' be true".
_________________
Intern
Joined: 13 Aug 2017
Posts: 1
### Show Tags
12 Aug 2018, 23:43
I understand the difference between 'can be' true and 'must be' true here. Were this question to come in the exam, I would basically need to test multiple consecutive number options till I get an option i.e (-8, -7, -6) that satisfies all three? (-8, -7, -6) this combination is rather unusual to test as the first thing that comes to one's mind. Is there a shorter way to address this?
Manager
Joined: 03 Feb 2018
Posts: 77
### Show Tags
08 Dec 2018, 19:14
Bunuel wrote:
Official Solution:
If $$a$$, $$b$$, and $$c$$ are consecutive integers, which of the following can be true?
I. $$abc$$ is divisible by 6
II. $$a + c - b$$ is odd
III. $$\frac{c}{a} \gt b$$
A. I only
B. I and II only
C. I and III only
D. II and III only
E. I, II, and III
All the three scenarios are possible. Consider $$a = -8$$, $$b = -7$$, $$c = -6$$. Note that the question asks "what 'can' be true" rather than "what 'must' be true".
hi Bunel,
If it is given in the question that a,b,c, are consecutive numbers, then we can by default assume that the order is a<b<c ??
Math Expert
Joined: 02 Sep 2009
Posts: 52296
### Show Tags
08 Dec 2018, 23:11
kanthaliya wrote:
Bunuel wrote:
Official Solution:
If $$a$$, $$b$$, and $$c$$ are consecutive integers, which of the following can be true?
I. $$abc$$ is divisible by 6
II. $$a + c - b$$ is odd
III. $$\frac{c}{a} \gt b$$
A. I only
B. I and II only
C. I and III only
D. II and III only
E. I, II, and III
All the three scenarios are possible. Consider $$a = -8$$, $$b = -7$$, $$c = -6$$. Note that the question asks "what 'can' be true" rather than "what 'must' be true".
hi Bunel,
If it is given in the question that a,b,c, are consecutive numbers, then we can by default assume that the order is a<b<c ??
No. On the GMAT a, b, c are consecutive integers does not necessarily mean that a < b < c.
_________________
Re: M21-22 &nbs [#permalink] 08 Dec 2018, 23:11
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# M21-22
Moderators: chetan2u, Bunuel
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|
# Count of primes in a given range that can be expressed as sum of perfect squares
Given two integers L and R, the task is to find the number of prime numbers in the range [L, R] that can be represented by the sum of two squares of two numbers.
Examples:
Input: L = 1, R = 5
Output:
Explanation:
Only prime number that can be expressed as sum of two perfect squares in the given range is 5 (22 + 12)
Input: L = 7, R = 42
Output:
Explanation:
The prime numbers in the given range that can be expressed as sum of two perfect squares are:
13 = 22 + 32
17 = 12 + 42
29 = 52 + 22
37 = 12 + 62
41 = 52 + 42
Approach:
The given problem can be solved using Fermat’s Little theorem, which states that a prime number p can be expressed as the sum of two squares if p satisfies the following equation:
(p – 1) % 4 == 0
Follow the steps below to solve the problem:
• Traverse the range [L, R].
• For every number, check if it is a prime number of not.
• If found to be so, check if the prime number is of the form 4K + 1. If sp, increase count.
• After traversing the complete range, print count.
Below is the implementation of the above approach:
## C++
`// C++ Program to implement ` `// the above approach ` ` ` `#include ` `using` `namespace` `std; ` ` ` `// Function to check if a prime number ` `// satisfies the condition to be ` `// expressed as sum of two perfect squares ` `bool` `sumSquare(``int` `p) ` `{ ` ` ``return` `(p - 1) % 4 == 0; ` `} ` ` ` `// Function to check if a ` `// number is prime or not ` `bool` `isPrime(``int` `n) ` `{ ` ` ``// Corner cases ` ` ``if` `(n <= 1) ` ` ``return` `false``; ` ` ` ` ``if` `(n <= 3) ` ` ``return` `true``; ` ` ` ` ``if` `(n % 2 == 0 || n % 3 == 0) ` ` ``return` `false``; ` ` ` ` ``for` `(``int` `i = 5; i * i <= n; i = i + 6) ` ` ``if` `(n % i == 0 || n % (i + 2) == 0) ` ` ``return` `false``; ` ` ` ` ``return` `true``; ` `} ` ` ` `// Function to return the count of primes ` `// in the range which can be expressed as ` `// the sum of two squares ` `int` `countOfPrimes(``int` `L, ``int` `R) ` `{ ` ` ``int` `count = 0; ` ` ``for` `(``int` `i = L; i <= R; i++) { ` ` ` ` ``// If i is a prime ` ` ``if` `(isPrime(i)) { ` ` ` ` ``// If i can be expressed ` ` ``// as the sum of two squares ` ` ``if` `(sumSquare(i)) ` ` ``count++; ` ` ``} ` ` ``} ` ` ` ` ``// Return the count ` ` ``return` `count; ` `} ` ` ` `// Driver Code ` `int` `main() ` `{ ` ` ``int` `L = 5, R = 41; ` ` ``cout << countOfPrimes(L, R); ` `} `
## Java
`// Java program to implement ` `// the above approach ` `import` `java.util.*; ` ` ` `class` `GFG{ ` ` ` `// Function to check if a prime number ` `// satisfies the condition to be ` `// expressed as sum of two perfect ` `// squares ` `static` `boolean` `sumSquare(``int` `p) ` `{ ` ` ``return` `(p - ``1``) % ``4` `== ``0``; ` `} ` ` ` `// Function to check if a ` `// number is prime or not ` `static` `boolean` `isPrime(``int` `n) ` `{ ` ` ` ` ``// Corner cases ` ` ``if` `(n <= ``1``) ` ` ``return` `false``; ` ` ` ` ``if` `(n <= ``3``) ` ` ``return` `true``; ` ` ` ` ``if` `(n % ``2` `== ``0` `|| n % ``3` `== ``0``) ` ` ``return` `false``; ` ` ` ` ``for``(``int` `i = ``5``; i * i <= n; i = i + ``6``) ` ` ``if` `(n % i == ``0` `|| n % (i + ``2``) == ``0``) ` ` ``return` `false``; ` ` ` ` ``return` `true``; ` `} ` ` ` `// Function to return the count of primes ` `// in the range which can be expressed as ` `// the sum of two squares ` `static` `int` `countOfPrimes(``int` `L, ``int` `R) ` `{ ` ` ``int` `count = ``0``; ` ` ``for``(``int` `i = L; i <= R; i++) ` ` ``{ ` ` ` ` ``// If i is a prime ` ` ``if` `(isPrime(i)) ` ` ``{ ` ` ` ` ``// If i can be expressed ` ` ``// as the sum of two squares ` ` ``if` `(sumSquare(i)) ` ` ``count++; ` ` ``} ` ` ``} ` ` ` ` ``// Return the count ` ` ``return` `count; ` `} ` ` ` `// Driver code ` `public` `static` `void` `main(String[] args) ` `{ ` ` ``int` `L = ``5``, R = ``41``; ` ` ` ` ``System.out.println(countOfPrimes(L, R)); ` `} ` `} ` ` ` `// This code is contributed by offbeat `
## Python3
`# Python3 program for the ` `# above approach ` ` ` `# Function to check if a prime number ` `# satisfies the condition to be ` `# expressed as sum of two perfect ` `# squares ` `def` `sumsquare(p): ` ` ` ` ``return` `(p ``-` `1``) ``%` `4` `=``=` `0` ` ` `# Function to check if a ` `# number is prime or not ` `def` `isprime(n): ` ` ` ` ``# Corner cases ` ` ``if` `n <``=` `1``: ` ` ``return` `False` ` ``if` `n <``=` `3``: ` ` ``return` `True` ` ``if` `(n ``%` `2` `=``=` `0``) ``or` `(n ``%` `3` `=``=` `0``): ` ` ``return` `False` ` ` ` ``i ``=` `5` ` ``while` `(i ``*` `i <``=` `n): ` ` ``if` `((n ``%` `i ``=``=` `0``) ``or` ` ``(n ``%` `(i ``+` `2``) ``=``=` `0``)): ` ` ``return` `False` ` ``i ``+``=` `6` ` ` ` ``return` `True` ` ` `# Function to return the count of primes ` `# in the range which can be expressed as ` `# the sum of two squares ` `def` `countOfPrimes(L, R): ` ` ` ` ``count ``=` `0` ` ``for` `i ``in` `range``(L, R ``+` `1``): ` ` ` ` ``# If i is a prime ` ` ``if` `(isprime(i)): ` ` ` ` ``# If i can be expressed ` ` ``# as the sum of two squares ` ` ``if` `sumsquare(i): ` ` ``count ``+``=` `1` ` ` ` ``# Return the count ` ` ``return` `count ` ` ` `# Driver code ` `if` `__name__``=``=``'__main__'``: ` ` ` ` ``L ``=` `5` ` ``R ``=` `41` ` ``print``(countOfPrimes(L, R)) ` ` ` `# This code is contributed by virusbuddah_`
## C#
`// C# program to implement ` `// the above approach ` `using` `System; ` ` ` `class` `GFG{ ` ` ` `// Function to check if a prime number ` `// satisfies the condition to be ` `// expressed as sum of two perfect ` `// squares ` `static` `bool` `sumSquare(``int` `p) ` `{ ` ` ``return` `(p - 1) % 4 == 0; ` `} ` ` ` `// Function to check if a ` `// number is prime or not ` `static` `bool` `isPrime(``int` `n) ` `{ ` ` ` ` ``// Corner cases ` ` ``if` `(n <= 1) ` ` ``return` `false``; ` ` ` ` ``if` `(n <= 3) ` ` ``return` `true``; ` ` ` ` ``if` `(n % 2 == 0 || n % 3 == 0) ` ` ``return` `false``; ` ` ` ` ``for``(``int` `i = 5; i * i <= n; i = i + 6) ` ` ``if` `(n % i == 0 || n % (i + 2) == 0) ` ` ``return` `false``; ` ` ` ` ``return` `true``; ` `} ` ` ` `// Function to return the count of primes ` `// in the range which can be expressed as ` `// the sum of two squares ` `static` `int` `countOfPrimes(``int` `L, ``int` `R) ` `{ ` ` ``int` `count = 0; ` ` ``for``(``int` `i = L; i <= R; i++) ` ` ``{ ` ` ` ` ``// If i is a prime ` ` ``if` `(isPrime(i)) ` ` ``{ ` ` ` ` ``// If i can be expressed ` ` ``// as the sum of two squares ` ` ``if` `(sumSquare(i)) ` ` ``count++; ` ` ``} ` ` ``} ` ` ` ` ``// Return the count ` ` ``return` `count; ` `} ` ` ` `// Driver code ` `public` `static` `void` `Main(String[] args) ` `{ ` ` ``int` `L = 5, R = 41; ` ` ` ` ``Console.WriteLine(countOfPrimes(L, R)); ` `} ` `} ` ` ` `// This code is contributed by Rajput-Ji `
Output:
```6
```
Time Complexity: O(N3/2
Auxiliary Space: O(1)
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# Brewster’s Law
### Brewster’s Law
Brewster’s law is a relationship of light waves at the maximum polarization angle of light. This law is named after Sir David Brewster, a Scottish physicist, who proposed the law in the year 1811. The law states that the p-polarized rays vanish completely on different glasses at a particular angle.
Further, the polarization angle is also called as Brewster’s angle. It is an angle of incidence where the ray of light having a p-polarization transmitted through a dielectric surface that is transparent without any reflection. While, the unpolarised light at this angle is transmitted, the light is reflected from the surface.
Brewster was able to determine that the refractive index of the medium is numerically equal to the tangent angle of polarization.
μ = tan i
Where,
μ = Refractive index of the medium
i = Polarization angle
From Snell’s Law:
$$\mu \,\,=\,\frac{\sin \,i}{\cos \,i}$$ … (1)
From Brewster’s Law:
$$\mu \,\,=\,\tan \,i\,=\,\frac{\sin \,i}{\cos \,i}$$ … (2)
Comparing both formulas (1) and (2)
$$\cos \,i\,=\,\sin \,r\,=\,\cos \,\left( \frac{\pi }{2}\,-\,r \right)$$,
$$i\,=\,\frac{\pi }{2}\,-\,r$$ (Or) $$i\,+\,r\,=\,\frac{\pi }{2}$$,
As, i + r = π/2 < ABC is also equal to the π/2 therefore, the reflected and the refracted rays are at right angles to each other.
Application of Brewster’s Law: One general example for the application of Brewster’s law is polarized sunglasses. These glasses uses the principle of Brewster’s angle. The polarized glasses reduce glare that is reflecting directly from the sun and also from horizontal surface like road and water. Photographers also use the same law to reduce the reflection from reflective surfaces by using polarizing filter for the lens.Brewster’s LawBrewster’s Law.
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# Friedman Test
The Friedman Test is a statistical test used to compare the means of three or more related groups. It is a non-parametric alternative to the one-way ANOVA test and is particularly useful for analyzing data that does not meet the assumptions of normality or equal variances. The test ranks the data within each group and then compares the average ranks between groups to determine if there is a significant difference. This test is commonly used in social sciences, psychology, and business research. It is named after its developer, the American economist and statistician, Milton Friedman.
## What is the Friedman Test?
The Friedman Test is a statistical test used to determine if 3 or more measurements from the same group of subjects are significantly different from each other on a skewed variable of interest. Your variable of interest should be continuous, and have a similar spread across your groups. Also, you should have enough data (more than 5 values in each group).
The Friedman Tests is also sometimes called the Non-Parametric Repeated Measures ANOVA, Non-Parametric Friedman Test, or the Friedman Rank Sum Test.
## Assumptions for the Friedman Test
Every statistical method has assumptions. Assumptions mean that your data must satisfy certain properties in order for statistical method results to be accurate.
The assumptions for the Friedman Test include:
1. Continuous
2. Random Sample
3. Enough Data
Let’s dive in to each one of these separately.
#### Continuous
The variable that you care about (and want to see if it is different across the 3+ groups) must be continuous. Continuous means that the variable can take on any reasonable value.
Some good examples of continuous variables include age, weight, height, test scores, survey scores, yearly salary, etc.
If your variable is normally distributed, you may want to use a One-Way Repeated Measures ANOVA instead.
#### Random Sample
The data points for each group in your analysis must have come from a simple random sample. This means that if you wanted to see if drinking sugary soda makes you gain weight, you would need to randomly select a group of soda drinkers for your soda drinker group.
The key here is that the data points for each group were randomly selected. This is important because if your groups were not randomly determined then your analysis will be incorrect. In statistical terms this is called bias, or a tendency to have incorrect results because of bad data.
If you do not have a random sample, the conclusions you can draw from your results are very limited. You should try to get a simple random sample.If you have independent samples (3 measurements from different, unrelated groups) then you should use the Kruskal-Wallis One-Way ANOVA instead.
#### Enough Data
The sample size (or data set size) should be greater than 5 in each group. Some people argue for more, but more than 5 is probably sufficient.
The sample size also depends on the expected size of the difference across groups. If you expect a large difference across groups, then you can get away with a smaller sample size. If you expect a small difference across groups, then you likely need a larger sample.
## When to use the Friedman Test?
You should use the Friedman Test in the following scenario:
1. You want to know if many groups are different on your variable of interest
2. Your variable of interest is continuous
3. You have 3 or more groups
4. You have related samples
Let’s clarify these to help you know when to use the Friedman Test.
#### Difference
You are looking for a statistical test to see whether three or more groups are significantly different on your variable of interest. This is a difference question. Other types of analyses include examining the relationship between two variables (correlation) or predicting one variable using another variable (prediction).
#### Continuous Data
Your variable of interest must be continuous. Continuous means that your variable of interest can basically take on any value, such as heart rate, height, weight, number of ice cream bars you can eat in 1 minute, etc.
Types of data that are NOT continuous include ordered data (such as finishing place in a race, best business rankings, etc.), categorical data (gender, eye color, race, etc.), or binary data (purchased the product or not, has the disease or not, etc.).
#### Three or more Groups
The Friedman Test can be used to compare three or more related groups on your variable of interest. See below for an explanation of what “related” groups means.
If you have only two groups, you should use the Wilcoxon Signed-Rank Test instead.
#### Related Samples
Related samples means that you have repeated measures from the same units of observation. For example, if you have a group of men undergoing a treatment and you measure their cholesterol levels at 3 time points, then you have 3 groups of related data.
If you have 3 or more independent groups, you should use a Kruskal-Wallis One-Way ANOVA instead.
## Friedman Test Example
Scenario: A random sample of men undergo an exercise program
Repeated Measures: Data were collected at month 1, 2 and 3
Variable of interest: Cholesterol levels
In this example we have three related groups (the three points in time) and one continuous variable of interest. After investigating our data, we determine that our variable of interest, cholesterol levels, is positively skewed, so we know to perform the Friedman Test. After confirming that our data meet the other assumptions of this test, we proceed with the analysis.
The null hypothesis, which is statistical lingo for what would happen if the exercise program does nothing, is that none of the three groups have different cholesterol levels, on average. We are trying to determine if receiving any of the three time points is significantly different from the others.
After the experiment is over, we compare the three groups on our variable of interest (cholesterol levels) using a Friedman Test. When we run the analysis, we get a “Chi-sqaure” statistic and a p-value. The Chi-square statistic is a measure of how different the three groups are on our cholesterol variable of interest.
A p-value is the chance of seeing our results assuming that the exercise program has no effect on cholesterol levels. A p-value less than or equal to 0.05 means that our result is statistically significant and we can trust that the difference is not due to chance alone.
If the Chi-square statistic is high and the p-value is low, it means that the cholesterol levels were significantly different in at least one of the time points. Further investigation is required to determine the which group(s) was significantly higher/lower than the others.
x
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# Definition:Arbitrary Constant
## Definition
An arbitrary constant is a symbol used to represent an object which is neither a specific number nor a variable.
It is used to represent a general object (usually a number, but not necessarily) whose value can be assigned when the expression is instantiated.
### In the context of Calculus
From the language in which it is couched, it is apparent that the primitive of a function may not be unique, otherwise we would be referring to $F$ as the primitive of $f$.
This point is made apparent in Primitives which Differ by Constant: if a function has a primitive, there is an infinite number of them, all differing by a constant.
That is, if $F$ is a primitive for $f$, then so is $F + C$, where $C$ is a constant.
This constant is known as an arbitrary constant.
## Examples
### Linear Equation
Consider the equation:
$y = a x + b$
where:
$x$ is an independent variable
$y$ is a dependent variable.
The symbols $a$ and $b$ stand for arbitrary constants.
## Also see
• Results about arbitrary constants can be found here.
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Faculty Resource Center
As a system undergoes a change, its internal energy can change, and energy can be transferred from the system to the surroundings, or from the surroundings to the system.
Energy transfer occurs through heat and work. The relationship between internal energy, heat, and work is represented by the equation:
While heat is a function of an observed temperature change, work is a function of an observed volume change called the pressure-volume work. Work (w) can be defined as a force (F) acting through a distance (D).
Pressure-volume work (or expansion work) occurs when a system pushes back the surroundings against a restraining pressure, or when the surroundings compress the system. An example of this occurs during the operation of an internal combustion engine. The combustion reaction of gasoline and oxygen is exothermic. Some of this energy is given off as heat, and some is performed as work by expanding the gases in the cylinder, thereby pushing the piston outward. The substances involved in the reaction are the system, and the engine and the rest of the universe are the surroundings. The system loses energy by both heating and doing work on the surroundings, and its internal energy decreases.
When the volume of a cylinder increases (i.e., the gas expands), it pushes against an external force, which is the pressure defined as force per unit area.
From equations 2 and 3:
The product of area and distance (A × D) is equal to the change in the volume (ΔV) of the gas in the cylinder.
Therefore,
Since the volume increases during expansion, Vfinal > Vinitial, and ΔV is positive. However, for a positive expansion (i.e., when the system does work on the surroundings), w should be negative, and therefore, a negative sign is added to the equation.
According to this equation, pressure-volume work is the negative of the external pressure (or opposing pressure) multiplied by the change in volume.
The unit of work based on this equation is L·atm. Some other useful conversions factors are:
This text is adapted from Openstax, Chemistry 2e, Section 5.3: Enthalpy.
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# CBSE Class 2 Maths Measurement Worksheets
CBSE Class 2 Maths Measurement Worksheets – Worksheets have become an integral part and are the most engaging study materials for kids these days. The Standard 2 Maths Measurement Worksheets will help your younger one to practice what they have learned in class. Measurement Worksheets for Grade 2 will develop a curiosity in your kid to learn.
Maths Worksheets for Class 2 are a perfect combination of fun and learning. Find the most interesting and relevant puzzle activities for your kids here. In the Mathematics Grade 2 Measurement Worksheets Puzzles will be in the form of crosswords, true or false, pictures, mazes, fill-ups, MCQs, objectives, questions and answers, and so on.
Mathematics Worksheets for Class 2 on Measurement will help your kid remember what he/ she learned in school for a long time. Your Grade 2 kid will love to solve these engaging exercises and interesting collections of Puzzles.
### CBSE Class 2 Maths Measurement Worksheets
1. Measure the length with your hand span: The height of water bottle
2. Which is heavier in weight?
A) 2 BANANAS
B) 3 GUAVA
3. Write the measuring unit for : Weight of a laptop
4. Measure the length with your hand span : The length of toothbrush
5. Which is longer?
6. Which is heavier in weight?
7. Which is longer?
A) FRIDGE
B) WASHING MACHINE
8. Which is heavier in weight?
A) PACKET OF MILK
B) PACKET OF POTATOES
9. Write the measuring unit for : Height of a house
10. Measure the length with your hand span : Length of comb
11. Which is longer?
A) FINGER
B) THUMB
12. Write the measuring unit for : Length of pencil.
13. The height of my mother is 3 hand span and the height of my father is 5 hand span. Who is taller?
14. Which is heavier in weight?
A) WATER BOTTLE
B) GLASS OF WATER
15. Tick the container that contains more quantity of liquid: A bottle or A spoon
16. Which is longer?
A) PENCIL
B) ERASER
17. Use a ruler to measure the length of your Maths notebook.
18. Write numbers 1,2,3 for lightest to heaviest for the following : an apple, a ball, a leaf
19. Which is longer?
20. Which is heavier in weight?
### CBSE Class 2 Maths Worksheet Answers
2. Option B
3. Kg
5. Option b is longer
6. Choice 2
7. Option A
8. Option B
9. cm
11. Option A
12. cm
13. Father
14. Option A
15. a bottle
16. Option A
18. 3,2,1
19. Option a is longer
20. Option A
### FAQs on Maths Measurement Class 2 Worksheets
1. How does a child learn Mathematics Measurement at Class 2 Level?
A child can learn Maths Measurement at Class 2 Level through the worksheets prevailing on our page.
2. What are the topics in Maths Class 2?
Mathematics Class 2 will have topics like Measurement etc.
3. Where can I get Mathematics Worksheets for Class 2 Measurement for free?
You can get Maths Measurement Worksheets Class 2 free of cost from our Worksheetsbuddy.com page. You can use them for your kid to get practice.
4. Are the CBSE Class 2 Measurement Worksheets over here printable?
Yes, the CBSE Class 2 Measurement Worksheets here are printable and you can download them easily.
Conclusion
We Hope the information prevailing on our page regarding CBSE Class 2 Mathematics Measurement Worksheets has been useful to you. For more info feel free to reach us and we will assist you at the easiest possible. Keep in touch with our site for more information on Class Wise Worksheets for various subjects.
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Question
# Solve $$\displaystyle \lim_{n\rightarrow \infty} \dfrac{(n+1)^{2}+5n}{n^{2}}$$
A
B
0
C
1
D
7
Solution
## The correct option is C $$1$$We need to find value of limit $$\lim _{ n\to \infty } \dfrac { \left( n+1 \right) ^{ 2 }+5n }{ n^{ 2 } }$$Apply Fraction Rule $$\dfrac { a+b }{ c } =\dfrac { a }{ c } +\dfrac { b }{ c }$$$$=\lim _{ n\rightarrow \infty }{ \left[ \dfrac { \left( n+1 \right) ^{ 2 } }{ { n }^{ 2 } } + \dfrac { 5n }{ { n }^{ 2 } } \right] }$$$$=\lim _{ n\rightarrow \infty }{ \left[ { \left( 1+\dfrac { 1 }{ n } \right) }^{ 2 }+ \dfrac { 5n }{ { n } } \right] }$$Put the value $$n=\infty$$$$=\left( 1+\dfrac { 1 }{ \infty } \right) ^{ 2 }+\dfrac { 5 }{ \infty }$$$$=1$$ ....As we know $$\dfrac {1}{\infty }=0$$Maths
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# T.I.M.E 2005 2 pt 3 pt 4 pt 5pt 1 pt 2 pt 3 pt 4 pt 5 pt 1 pt 2pt 3 pt 4pt 5 pt 1pt 2pt 3 pt 4 pt 5 pt 1 pt 2 pt 3 pt 4pt 5 pt 1pt Definitions/ Formulas.
## Presentation on theme: "T.I.M.E 2005 2 pt 3 pt 4 pt 5pt 1 pt 2 pt 3 pt 4 pt 5 pt 1 pt 2pt 3 pt 4pt 5 pt 1pt 2pt 3 pt 4 pt 5 pt 1 pt 2 pt 3 pt 4pt 5 pt 1pt Definitions/ Formulas."— Presentation transcript:
T.I.M.E 2005
2 pt 3 pt 4 pt 5pt 1 pt 2 pt 3 pt 4 pt 5 pt 1 pt 2pt 3 pt 4pt 5 pt 1pt 2pt 3 pt 4 pt 5 pt 1 pt 2 pt 3 pt 4pt 5 pt 1pt Definitions/ Formulas Solving Equations Evaluate Linear and Quadratic Mixed Bag
A polygon with all 4 sides and 4 angles congruent 1pt
What is a square? 1pt
Two lines with the same slope. 2pt
What are parallel lines? 2pt
y = m x + b, y - y 1 = m (x - x 1 ) a x + b y = c 3pt
What is the equation of a line? 3pt
A = ½ b h 4pt
What is the formula for the area of a triangle? 4pt
C² = A² + B² 5pt
What is the Pythagorean Theorem? 5pt
5/x = 4/3 1pt
What is 3 ¾ (3.75)? 1pt
4 – x = 10. 2pt
What is –6? 2pt
-8 x 16 3pt
What is X -2? 3pt
x = (25/8) - 3 4pt
What is 1/8? 4pt
2 x = 13 5pt
What is –6.5 or -6 ½? 5pt
x² + x –5, for x = -3. 1pt
What is one?
½ (b 1 + b 2 ), for b 1 = 3, b 2 = 5. 2pt
What is 4?
l w h, for l = 2, w = 3, h = ½. 3pt
What is 3?
2 π r 2 h for π = 22/7, r = 7, h = 2 4pt
What is 308? 4pt
2 π r (r + h) for r = 10, h = 4, r = 5, π = 22/7. 5pt
What is 88?
(x - 4) (x + 3) is the factored form for the quadratic polynomial. 1pt
What is x² - x - 12 in factored form 1pt
1, 2 are the solutions to the quadratic equation 2pt
What are the solutions of x² - 3x + 2 = 0? 2pt
(2, 3) lie on l 1 and l 2 3pt
What is the intersection of l 1 and l 2 ? 3pt
(3x + 4y) + (-2x + y) 4pt
What is x + 5y? 4pt
m = (y 2 - y 1 ) (x 2 – x 1 ) 5pt
What is the formula for the slope of a line given two points on the line? 5pt
The probability of not getting a red marble out of 5 red, 2 yellow, 6 blue, and 4 green marbles. 1pt
What is 12/17? 1pt
3 - 9= 2pt
What is 6? 2pt
16 = 3pt
What is 4? 3pt
5 is the _____ of {1, 5, 8, 5} 4pt
What is the mode?
The average of 2, 3 and another number is 9. 5pt
What is 22? 5pt
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# How do you solve sqrt( 3x-3) + sqrt(2x+8) +1=0?
Apr 6, 2016
Isolate one of the radicals on one side of the equation.
#### Explanation:
$\sqrt{3 x - 3} = - 1 - \sqrt{2 x + 8}$
${\left(\sqrt{3 x - 3}\right)}^{2} = {\left(- 1 - \sqrt{2 x + 8}\right)}^{2}$
$3 x - 3 = 1 + 2 x + 8 + 2 \sqrt{2 x + 8}$
$x - 12 = 2 \sqrt{2 x + 8}$
${\left(x - 12\right)}^{2} = {\left(2 \sqrt{2 x + 8}\right)}^{2}$
${x}^{2} - 24 x + 144 = 4 \left(2 x + 8\right)$
${x}^{2} - 24 x + 144 - 8 x - 32 = 0$
${x}^{2} - 32 x + 112 = 0$
$\left(x - 28\right) \left(x - 4\right) = 0$
$x = 28 \mathmr{and} 4$
Checking both these solutions in the original equation we find that neither work. Therefore, there is no solution $\left(\emptyset\right)$
Hopefully this helps!
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## Absolute rate of change
As the name implies, the essence of the percentage change calculator is to help you compute the percentage difference between two numbers – initial value and new value. Hundreds of people find this tool very useful in several, daily applications like finance, sales, tax and inflation rate, chemistry, physics and diverse areas of mathematics. Multiply the rate of change by 100 to convert it to a percent change. In the example, 0.50 times 100 converts the rate of change to 50 percent. However, if the numbers were reversed such that the population decreased from 150 to 100, the percent change would be -33.3 percent. Divide the absolute change by the average value to compute the
Absolute change refers to the simple difference in the indicator over two periods in time, i.e. Relative change expresses the absolute change as a percentage of the value of the indicator in the earlier period, i.e. The concepts of absolute and relative change also apply to indicators measured in percentage terms, for example unemployment rate. Absolute change measures the exact numerical change between two numbers and equals an ending number minus a beginning number. For example, the absolute change in a city’s population may be an increase of 10,000 residents in five years. Absolute change differs from relative change, which is another way to measure a What is Rate of Change (ROC) The rate of change - ROC - is the speed at which a variable changes over a specific period of time. ROC is often used when speaking about momentum, and it can generally be expressed as a ratio between a change in one variable relative to a corresponding change in another; graphically, The average rate of change is defined as the average rate at which quantity is changing with respect to time or something else that is changing continuously. In other words, the average rate of change is the process of calculating the total amount of change with respect to another. The simple indicator which shows absolute or relative change of the chosen source (could be either price or result of another indicator). The positive value means up direction, the negative value means down direction. Bigger value means faster change. Very similar to classic ROC (Rate of Change) indicator, but calculates the value as
## This free percentage calculator computes a number of values involving Percentage Calculator in Common Phrases Percentage Change Calculator difference between two values is calculated by dividing the absolute value of the
It is customary to use rates per 100,000 population for deaths and rates per 1,000 The percentage change calculation will also work if there is a decrease from Find the percentage of change from an initial value to a final value using our the final value from the initial value, then divide by the absolute value of the initial Here are instructions for the formulas to calculate common percent related math problems: percentage. formula: (x/y) * 100 = z%; in words: divide x by y to get its Learn the difference between a relative absolute cell reference in Excel and an absolute Relative references change when a formula is copied to another cell. In the example below, we're going to use cell E2 (which contains the tax rate at Relative change is the difference over two periods in time. In simple words, relative change is the absolute change indicated as a percentage. Relative change is
### Percent Error is a special case of the percentage form of relative change calculated from the absolute change between the
5 Sep 2011 When, as in this case, the percentage change is over time (rather than cross- sectional), it is also called the growth rate. (A cross-sectional percent Absolute change ( B - F) and percentage change [(B - F )/ B], where B and F are baseline and follow up values. I know, how to do this in excel, SPSS, and base 14 Apr 2007 To find the percentage change, divide the absolute change by the starting number and multiply by 100. On the Home tab, in the Number group, apply a Percentage format. Percentage Format. 1c. Select cell C3, click on the lower right corner of cell C3 and drag it
### 1 Jun 2015 As researchers, we often express absolute change as percent change so we can account for different initial values of the response. But this
This free percentage calculator computes a number of values involving Percentage Calculator in Common Phrases Percentage Change Calculator difference between two values is calculated by dividing the absolute value of the 21 Nov 2003 than an absolute change (some specified amount, like a salary increase of \$1000). What do we say when we want to treat a percentage as It is customary to use rates per 100,000 population for deaths and rates per 1,000 The percentage change calculation will also work if there is a decrease from Find the percentage of change from an initial value to a final value using our the final value from the initial value, then divide by the absolute value of the initial Here are instructions for the formulas to calculate common percent related math problems: percentage. formula: (x/y) * 100 = z%; in words: divide x by y to get its Learn the difference between a relative absolute cell reference in Excel and an absolute Relative references change when a formula is copied to another cell. In the example below, we're going to use cell E2 (which contains the tax rate at
## 11 Jul 2019 Take for example for the metric of Bounce Rate, the percentage of These features, including Comparison Metrics, Absolute Change and
The percentage change between two numbers is found by dividing their difference by the number that is being changed from, then multiplying that result by 3 Apr 2015 Relative risk is usually reported as a percentage (i.e. 10% more likely) but you'll also see it written as “x times more likely” (i.e. ten times more Percentage Change. Tutoring and Learning Centre, George Brown College 2014 www.georgebrown.ca/tlc. It is often useful to know by how much an item's Because it is a percentage, it gives you a sense of the relative change in sales rather than the absolute value. Track the sales percentage change of your
In simple words, relative change is the absolute change indicated as a percentage. Relative change is expressed in percentage and hence also called as the percentage change. Given here is the online relative change calculator to calculate the relative change for a given period. Percentage change = absolute change/original quantity*100. Example 1 revisited: On September 20, a gallon of gas at my usual gas station cost \$1.83. On September 30, I noticed that the price had gone down to \$1.65 per gallon. The percentage change in the price of gas between September 20 and September 30 is Divide the absolute change by the initial value to calculate the rate of change. In the example, 50 divided by 100 calculates a 0.5 rate of change. 5. Multiply the rate of change by 100 to convert it to a percent change. A rate of change describes how an output quantity changes relative to the change in the input quantity. The units on a rate of change are “output units per input units.” The average rate of change between two input values is the total change of the function values (output values) divided by the change in the input values. Start studying Average Rate of Change of Absolute Value Functions. Learn vocabulary, terms, and more with flashcards, games, and other study tools. A rate of change is a rate that describes how one quantity changes in relation to another quantity. If is the independent variable and is the dependent variable, then. Rates of change can be positive or negative. This corresponds to an increase or decrease in the -value between the two data points. When a quantity does not change over time, it
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# Divided by 1000
Welcome to divided by 1000, our category containing the posts which explain the division of a particular number x by 1000. The number x is called nominator, and 1000 is called the denominator of the quotient x / 1000. In every article we give you the result of the fraction x over 1000 in high accuracy, along with the value in percentage and the properties of the result, e.g. terminating or repeating. In the former case, we state the number of decimal places, whereas in the latter case we provide you with information regarding the repented. Also, for any x divided by 1000 we review the frequently asked question in the context. What’s more is that all entries in this category come with a unique calculator you will like. Information regarding the use of our search box, directions for further information, and how to ask us a question can always be found as well.
## Ten Million Divided by 1000
Welcome to ten million divided by 1000, our post which explains the division of ten million by one thousand to you. The numeral ten million (10000000) is called the numerator or dividend, and the number 1000 is called the denominator or divisor. The quotient of ten million and 1000, the
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## Nine Million Divided by 1000
Welcome to nine million divided by 1000, our post which explains the division of nine million by one thousand to you. The numeral nine million (900000) is called the numerator or dividend, and the number 1000 is called the denominator or divisor. The quotient of nine million and 1000, the
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## Eight Million Divided by 1000
Welcome to eight million divided by 1000, our post which explains the division of eight million by one thousand to you. The numeral eight million (8000000) is called the numerator or dividend, and the number 1000 is called the denominator or divisor. The quotient of eight million and 1000, the
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## Seven Million Divided by 1000
Welcome to seven million divided by 1000, our post which explains the division of seven million by one thousand to you. The numeral seven million (7000000) is called the numerator or dividend, and the number 1000 is called the denominator or divisor. The quotient of seven million and 1000, the
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## Six Million Divided by 1000
Welcome to six million divided by 1000, our post which explains the division of six million by one thousand to you. The numeral six million (6000000) is called the numerator or dividend, and the number 1000 is called the denominator or divisor. The quotient of six million and 1000, the
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## Five Million Divided by 1000
Welcome to five million divided by 1000, our post which explains the division of five million by one thousand to you. The numeral five million (5000000) is called the numerator or dividend, and the number 1000 is called the denominator or divisor. The quotient of five million and 1000, the
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## Four Million Divided by 1000
Welcome to four million divided by 1000, our post which explains the division of four million by one thousand to you. The numeral four million (4000000) is called the numerator or dividend, and the number 1000 is called the denominator or divisor. The quotient of four million and 1000, the
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## Three Million Divided by 1000
Welcome to three million divided by 1000, our post which explains the division of three million by one thousand to you. The numeral three million (3000000) is called the numerator or dividend, and the number 1000 is called the denominator or divisor. The quotient of three million and 1000, the
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## Two Million Divided by 1000
Welcome to two million divided by 1000, our post which explains the division of two million by one thousand to you. The numeral two million (2000000) is called the numerator or dividend, and the number 1000 is called the denominator or divisor. The quotient of two million and 1000, the
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# Search by Topic
#### Resources tagged with Differentiation similar to Curved Square:
Filter by: Content type:
Age range:
Challenge level:
### There are 16 results
Broad Topics > Pre-Calculus and Calculus > Differentiation
### Operating Machines
##### Age 16 to 18 Challenge Level:
What functions can you make using the function machines RECIPROCAL and PRODUCT and the operator machines DIFF and INT?
### Calculus Countdown
##### Age 16 to 18 Challenge Level:
Can you hit the target functions using a set of input functions and a little calculus and algebra?
### Towards Maclaurin
##### Age 16 to 18 Challenge Level:
Build series for the sine and cosine functions by adding one term at a time, alternately making the approximation too big then too small but getting ever closer.
### Turning to Calculus
##### Age 16 to 18 Challenge Level:
Get started with calculus by exploring the connections between the sign of a curve and the sign of its gradient.
### Generally Geometric
##### Age 16 to 18 Challenge Level:
Generalise the sum of a GP by using derivatives to make the coefficients into powers of the natural numbers.
### An Introduction to Differentiation
##### Age 14 to 18
An article introducing the ideas of differentiation.
### Integration Matcher
##### Age 16 to 18 Challenge Level:
Match the charts of these functions to the charts of their integrals.
### Ramping it Up
##### Age 16 to 18 Challenge Level:
Look at the calculus behind the simple act of a car going over a step.
### Quick Route
##### Age 16 to 18 Challenge Level:
What is the quickest route across a ploughed field when your speed around the edge is greater?
### Least of All
##### Age 16 to 18 Challenge Level:
A point moves on a line segment. A function depends on the position of the point. Where do you expect the point to be for a minimum of this function to occur.
### Loch Ness
##### Age 16 to 18 Challenge Level:
Draw graphs of the sine and modulus functions and explain the humps.
### Bend
##### Age 16 to 18 Challenge Level:
What is the longest stick that can be carried horizontally along a narrow corridor and around a right-angled bend?
### Bird-brained
##### Age 16 to 18 Challenge Level:
How many eggs should a bird lay to maximise the number of chicks that will hatch? An introduction to optimisation.
### Impedance Can Be Complex!
##### Age 16 to 18 Challenge Level:
Put your complex numbers and calculus to the test with this impedance calculation.
### Exponential Trend
##### Age 16 to 18 Challenge Level:
Find all the turning points of y=x^{1/x} for x>0 and decide whether each is a maximum or minimum. Give a sketch of the graph.
### Slide
##### Age 16 to 18 Challenge Level:
This function involves absolute values. To find the slope on the slide use different equations to define the function in different parts of its domain.
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# Bounding the diameter of a triangle in hyperbolic geometry
Prove that there exists a universal constant $K$ such that given any three geodesics in the $\gamma_1, \gamma_2, \gamma_3$ in the hyperbolic plane forming a triangle, there exists three points $A \in \gamma_1$, $B \in \gamma_2$, $C \in \gamma_3$ such that the geodesic triangle on vertexes $A,B,C$ has its diameter bounded by $K$
My attempt was the following: Given any geodesic triangle, there exists a inscribed circle $\Gamma$, which provides me a natural candidate for the triple $A,B,C$ as the three points that intersect such circle. Also, the diameter of the geodesic triangle is bounded by the radius of $\Gamma$. I thought then of using Gauss-Bonnet Theorem to get a bound of the area of $\Gamma$
$$\int_\Gamma K dA + \int_{\partial \Gamma} k_g dt + 3\pi - \sum_{i=1,2,3} \phi_i = 2\pi$$ where $k_g$ is the geodesic curvature of the the curve $\gamma$ the boundary of $\Gamma$ and $\phi_i$’s are the inner angles.
So, I get $$Area(\Gamma) = \pi - \sum_{i=1,2,3} \phi_i + \int_{\partial \Gamma} k_g dt$$
$$Area(\Gamma) \le \pi + \int_{\partial \Gamma} k_g dt$$
So now, I would just need to be able to get a universal bound on the term $\int_{\partial \Gamma} k_g dt$ and use the formula of the area of a circle to get an upper bound on the radius. How could I get this upper bound on $\int_{\partial \Gamma} k_g dt$?
On the other hand, I could try to use Gauss-Bonnet in the region given by the geodesic triangle given by $A,B,C$ to avoid needing to bound the geodesic curvature, but it does not seem to be easy to get back an upper bound on the length of the longest side of the triangle.
I would appreciate any hints. Thank you.
• I’m sorry, could you elaborate a bit more? I don’t really understand. – Kernel Oct 31 '17 at 15:22
• Sorry I'm not totally familiar with hyperbolic spaces. But it seems to me from physics that it is locally euclidean but then there can't be a $K$ such that all groups of three lines forming a triangle have smaller diameter. Hmm maybe im not familiar enough with the hyperbolic plane. – marshal craft Oct 31 '17 at 16:33
A geodesic triangle with angles $\alpha,\beta,\gamma$ has area $\pi-\alpha-\beta-\gamma \le \pi$, proof here (but you already knew this since you know Gauss-Bonnet).
Hence, the inscribed circle has area $\le \pi$.
Hence, the radius of inscribed circle is bounded by some constant $K$ (the area of a hyperbolic circle can be computed explicitly, but it suffices to know that by increasing the radius $\to\infty$ we cover the hyperbolic plane which has infinite area).
Hence, the diameter of the triangle formed by the points where the inscribed circle touches the original triangle is bounded by $2K$.
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# How to form binary matrix with constants and given probability?
I want to create a binary matrix of 6x3 in which six elements (m13,m23, m33, m42, m52, m61) are constant and assigned '0' value; whereas for other twelve elements (m11, m12, m21, m22, m31,m32, m41, m43, m51, m53, m62, m63) I want to assign value of "1" with the probability of 0.5. Given below my effort, I had made yet.
mat = Table[Subscript[m, i, j], {i, 6}, {j, 3}];
mat // MatrixForm
n = Round[18*0.5];
k = RandomSample[{mat[[1, 1]], mat[[1, 2]], mat[[1, 3]], mat[[2, 1]],
mat[[2, 2]], mat[[2, 3]], mat[[3, 1]], mat[[3, 2]], mat[[3, 3]],
mat[[4, 1]], mat[[4, 2]], mat[[4, 3]], mat[[5, 1]], mat[[5, 2]],
mat[[5, 3]], mat[[6, 1]], mat[[6, 2]], mat[[6, 3]]}, n]
-
See SparseArray, RandomInteger, Condition and MemberQ – Dr. belisarius Jan 6 at 20:45
SparseArray[{i_, j_} :> RandomInteger[{0, 1}] /; j != IntegerLength[7 - i, 2],
{6, 3}, a] // MatrixForm
-
Kindly explain it briefly. – Mamoona Jan 6 at 21:34
@Mamoona Please read the docs for each function and then ask specifically about the ones you don't understand – Dr. belisarius Jan 6 at 23:48
One approach to this problem is to populate the whole matrix 0/1 with equal probabily, then set the desired elements to zero:
ReplacePart[
RandomInteger[1 , {6, 3}] ,
{{1, 3} , {2, 3} , {3, 3}, {4, 2}, {5, 2}, {6,1}} -> 0 ] // MatrixForm
-
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# 6th Grade Skills, for some students are the final requirements of Elementary Math to master.
The 6th grade skills required in elementary math, are the same whether your student is still based in an elementary school, or has moved on to a middle school.
### Get Prepared For Summer.
If you want to learn how to put a stop to the dreaded Summer Math Loss and send your child back to school in September feeling like a math super-hero then watch this free video training.
Mastery of the sixth grade skills will give your kid an excellent foundation to start their further mathematics education. Once these are mastered they will be moving on to Advanced Mathematics - Algebra, Trigonometry, Euclidian Geometry, Calculus and Statistics. Ensure they know, understand and can apply all these skills in math - otherwise they will have trouble down the road.
This list of skills is based on the 5th grade level expectatoins from Washington State OSPI Each specific skill is further broken down into areas. Just click on them to access all the practice your child could need!
These Grade Level Expectations are organized by basic math subject area for 6th Grade, based on Washington States OSPI. If you would rather have a basic math skills requirements list organized by subject area, that includes every requirement from Kindergarten to 6th grade in one downloadable document please visit my Grade Level Expectations section.
## 6th Grade Skills by Subject Area
Algebra Exponents Review squares and square roots, using the terms squared and cubed and to the nth power, read and evaluate numerical expressions with exponents, review powers of ten, write numbers in expanded notations using exponents. Solve word problems with multiple steps Solve problems with more than one operation, according to order of operations (with and without a calculator) Recognize uses of variables and solve linear equations in one variable Solve word problems by assigning variables to unknown quantities, writing appropriate equations, and solving them Find the value for an expression, given replacement values for the variables, eg what is 7/x – y when x = 2 and y = 10? Simplify expressions with variables by combining like terms Understand the use of the distributive property in variable expressions such as 2x(2y + 3)
Numbers and Operations Read and write numbers (in digits and words) up to the trillions Recognize place value up to hundred-billions Integers (review) Locate positive and negative integers on a number line Compare integers using <, >, = Know that the sum of an integer and its opposite is 0 Add and subtract positive and negative integers Determine whether a number is a prime number or composite number Round to the nearest ten, nearest hundred, nearest thousand, nearest hundred thousand, nearest million Compare and order whole numbers, mixed numbers, fractions and decimals using the symbols <, >, = Determine the greatest common factor (GCF) of given numbers Determine the least common multiple (LCM) of given numbers Solve proportions including word problems involving proportions with one unknown Use ratios and proportions to interpret map scales and scale drawings Set up and solve proportions from similar triangles Understand the justification for solving proportions by cross-multiplication Convert between fractions, decimals and percent’s Find the given percent of a number, and find what percent a given number is of another number Solve problems involving percent increase and decrease Find an unknown number when a percent of the number is known Use expressions with percents greater that 100% and less and 1% Addition, commutative and associative properties: know the names and understand the properties Understand addition and subtraction as inverse operations Add and subtract with integers, fractions and decimals, both positive and negative Commutative, associative, and distributive properties: know the names and understand the properties. Multiply multi-digit factors, with and without a calculator Estimate a product Multiply with integers, fractions and decimals both positive and negative Distributive property for multiplication over addition or subtractions, that is A x (B +C) or A x (B + C) Understand multiplication and division as inverse operations Estimate the quotient Divide multi-digit dividends by up to three-digit divisors, with and without a calculator Measurement Solve problems requiring conversion of units within the US customary system and within the metric system Associate prefixes used in metric system with quantities Kilo = thousand Hecto = hundred Deka = ten Deci = tenth Centi = hundredth Mili = thousandth Solve problems on elapsed time; express parts of an hour in fraction or decimal form. Geometry Identify and measure the degrees in angles (review terms; right, acute, obtuse, straight) Bisect an angle Construct an angle congruent to a given angle Construct a figure congruent to a given figure, using reflection over a line of symmetry, and identify corresponding parts. Show how congruent plane figures can be made to correspond through reflection, rotation and translation Triangles Know that the sum of the measures of the angles of a triangle is 180’ Construct different kinds of triangles Know terms by which we classify kinds of triangles – length of sides, equilateral, isosceles, scalene – by angles, right, acute and obtuse Identify congruent angles and sides, and axes of symmetry, in parallelograms, rhombuses, rectangles and squares Find the area (A) and perimeter (P) of plane figures or given the area or perimeter find the missing dimension, using formulas of rectangle (A = lw; P = 2(l+w), square, (A = s2; P = 4s) triangle ( A = ½ bh; P = s1 + s2 + s3) parallelogram (A = bh P = 2(b + s) Circles Identify arc, cord, radius, and diameter; know that radius = ½ diameter. Using a compass, draw circles with a given diameter or radius. Solve problems involving application of the formulas for finding the circumference of a circle: using 3.14 as the value of Pi Find the volume of a rectangular solids or given the volume find a missing dimension, using the formulas V = lwh or V = bh (in which b = area of base) Data Analysis Find the range and measures of central tendency (mean, median and mode) of a given set of numbers. Understand the use of a sample to estimate a population parameter (such as the mean) and that larger samples provide more stable estimates. Represent all possible outcomes of independent compound events in an organized way and determine the theoretical probability of each outcome. Compute the probability of any one of a set of disjoint events as the sum of their individual probabilities. Solve problems requiring interpretation and application of graphically displayed data. Construct a histogram; a tree diagram.
And don't forget to head over to our home page! Here you will find great resources to help you tutor your child, such as manipulatives, board games, math puzzles and activities! When your child has mastered their 6th Grade Skills, it's time to rejoice as they will have successfully completed all that is required to give them the best start in mathematics - no matter what they chose to study.
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MATH MOMENTUM ACADEMY If your child is struggling with mathematics and you are a proactive parent, who wants to sort their math problems out, once and for all, then I invite you to explore what Math Momentum Academy can do for you.
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## Precalculus (6th Edition) Blitzer
The solution is, $\underline{x=-3,y=0.5,z=1}$.
Consider the given system of equations: \begin{align} & x+2y-z=-3 \\ & 2x-4y+z=-7 \\ & -2x+2y-3z=4 \end{align} Therefore, in matrix form, the system of equations can be written as below: $AX=b$ Where, $A=\left[ \begin{array}{*{35}{r}} 1 & 2 & -1 \\ 2 & -4 & 1 \\ -2 & 2 & -3 \\ \end{array} \right];b=\left[ \begin{array}{*{35}{r}} -3 \\ -7 \\ 4 \\ \end{array} \right];X=\left[ \begin{matrix} x \\ y \\ z \\ \end{matrix} \right]$ Therefore, the solution of the system of equations is given by: $X={{A}^{-1}}b$ Consider the determinant of the matrix \begin{align} & \left| A \right|=\left| \begin{array}{*{35}{r}} 1 & 2 & -1 \\ 2 & -4 & 1 \\ -2 & 2 & -3 \\ \end{array} \right| \\ & =\left( 12-2 \right)-2\left( -6+2 \right)-1\left( 4-8 \right) \\ & =10+8+4 \\ & =22 \end{align} Consider the adjoint of the matrix $\text{adj}\left( A \right)={{\left[ \begin{array}{*{35}{r}} +\left| \begin{array}{*{35}{r}} -4 & 1 \\ 2 & -3 \\ \end{array} \right| & -\left| \begin{array}{*{35}{r}} 2 & 1 \\ -2 & -3 \\ \end{array} \right| & +\left| \begin{array}{*{35}{r}} 2 & -4 \\ -2 & 2 \\ \end{array} \right| \\ -\left| \begin{array}{*{35}{r}} 2 & -1 \\ 2 & -3 \\ \end{array} \right| & +\left| \begin{array}{*{35}{r}} 1 & -1 \\ -2 & -3 \\ \end{array} \right| & -\left| \begin{array}{*{35}{r}} 1 & 2 \\ -2 & 2 \\ \end{array} \right| \\ +\left| \begin{array}{*{35}{r}} 2 & -1 \\ -4 & 1 \\ \end{array} \right| & -\left| \begin{array}{*{35}{r}} 1 & -1 \\ 2 & 1 \\ \end{array} \right| & +\left| \begin{array}{*{35}{r}} 1 & 2 \\ 2 & -4 \\ \end{array} \right| \\ \end{array} \right]}^{t}}$ It can be further solved as below: \begin{align} & \text{adj}\left( A \right)={{\left[ \begin{array}{*{35}{r}} \left( 12-2 \right) & -\left( -6+2 \right) & \left( 4-8 \right) \\ -\left( -6+2 \right) & \left( -3-2 \right) & -\left( 2+4 \right) \\ \left( 2-4 \right) & -\left( 1+2 \right) & \left( -4-4 \right) \\ \end{array} \right]}^{t}} \\ & ={{\left[ \begin{array}{*{35}{r}} 10 & 4 & -4 \\ 4 & -5 & -6 \\ -2 & -3 & -8 \\ \end{array} \right]}^{t}} \\ & =\left[ \begin{array}{*{35}{r}} 10 & 4 & -2 \\ 4 & -5 & -3 \\ -4 & -6 & -8 \\ \end{array} \right] \end{align} Therefore \begin{align} & {{A}^{-1}}=\frac{1}{\left| A \right|}\text{adj}\left( A \right) \\ & =\frac{1}{22}\left[ \begin{array}{*{35}{r}} 10 & 4 & -2 \\ 4 & -5 & -3 \\ -4 & -6 & -8 \\ \end{array} \right] \end{align} Therefore the solution of the system of equations is \begin{align} & X=\frac{1}{22}\left[ \begin{array}{*{35}{r}} 10 & 4 & -2 \\ 4 & -5 & -3 \\ -4 & -6 & -8 \\ \end{array} \right]\left[ \begin{array}{*{35}{r}} -3 \\ -7 \\ 4 \\ \end{array} \right] \\ & =\frac{1}{22}\left[ \begin{array}{*{35}{r}} -30-28-8 \\ -12+35-12 \\ 12+42-32 \\ \end{array} \right] \\ & =\frac{1}{22}\left[ \begin{array}{*{35}{r}} -66 \\ 11 \\ 22 \\ \end{array} \right] \\ & =\left[ \begin{array}{*{35}{r}} -3 \\ 0.5 \\ 1 \\ \end{array} \right] \end{align} The solutions of the system is $x=-3,y=0.5,z=1$.
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Listen to this text
CHAPTER II.
ON DIFFERENT DEGREES OF SMALLNESS.
We shall find that in our processes of calculation we have to deal with small quantities of various degrees of smallness.
We shall have also to learn under what circumstances we may consider small quantities to be so minute that we may omit them from consideration. Everything depends upon relative minuteness.
Before we fix any rules let us think of some familiar cases. There are ${\displaystyle 60}$ minutes in the hour, ${\displaystyle 24}$ hours in the day, ${\displaystyle 7}$ days in the week. There are therefore ${\displaystyle 1440}$ minutes in the day and ${\displaystyle 10080}$ minutes in the week.
Obviously ${\displaystyle 1}$ minute is a very small quantity of time compared with a whole week. Indeed, our forefathers considered it small as compared with an hour, and called it “one minùte,” meaning a minute fraction—namely one sixtieth—of an hour. When they came to require still smaller subdivisions of time, they divided each minute into ${\displaystyle 60}$ still smaller parts, which, in Queen Elizabeth’s days, they called “second minùtes” (i.e. small quantities of the second order of minuteness). Nowadays we call these small quantities of the second order of smallness “seconds.” But few people know why they are so called.
Now if one minute is so small as compared with a whole day, how much smaller by comparison is one second!
Again, think of a farthing as compared with a sovereign: it is barely worth more than ${\displaystyle {\tfrac {1}{1000}}}$ part. A farthing more or less is of precious little importance compared with a sovereign: it may certainly be regarded as a small quantity. But compare a farthing with £${\displaystyle 1000}$: relatively to this greater sum, the farthing is of no more importance than ${\displaystyle {\tfrac {1}{1000}}}$ of a farthing would be to a sovereign. Even a golden sovereign is relatively a negligible quantity in the wealth of a millionaire.
Now if we fix upon any numerical fraction as constituting the proportion which for any purpose we call relatively small, we can easily state other fractions of a higher degree of smallness. Thus if, for the purpose of time, ${\displaystyle {\tfrac {1}{60}}}$ be called a small fraction, then ${\displaystyle {\tfrac {1}{60}}}$ of ${\displaystyle {\tfrac {1}{60}}}$ (being a small fraction of a small fraction) may be regarded as a small quantity of the second order of smallness.[1]
Or, if for any purpose we were to take ${\displaystyle 1}$ per cent. (i.e. ${\displaystyle {\tfrac {1}{100}}}$) as a small fraction, then ${\displaystyle 1}$ per cent. of ${\displaystyle 1}$ per cent. (i.e. ${\displaystyle {\tfrac {1}{10,000}}}$) would be a small fraction of the second order of smallness; and ${\displaystyle {\tfrac {1}{1,000,000}}}$ would be a small fraction of the third order of smallness, being ${\displaystyle 1}$ per cent. of ${\displaystyle 1}$ per cent. of ${\displaystyle 1}$ per cent.
Lastly, suppose that for some very precise purpose we should regard ${\displaystyle {\tfrac {1}{1,000,000}}}$ as “small.” Thus, if a first-rate chronometer is not to lose or gain more than half a minute in a year, it must keep time with an accuracy of ${\displaystyle 1}$ part in ${\displaystyle 1,051,200}$. Now if, for such a purpose, we regard ${\displaystyle {\tfrac {1}{1,000,000}}}$ (or one millionth) as a small quantity, then ${\displaystyle {\tfrac {1}{1,000,000}}}$ of ${\displaystyle {\tfrac {1}{1,000,000}}}$, that is ${\displaystyle {\tfrac {1}{1,000,000,000,000}}}$ (or one billionth) will be a small quantity of the second order of smallness, and may be utterly disregarded, by comparison.
Then we see that the smaller a small quantity itself is, the more negligible does the corresponding small quantity of the second order become. Hence we know that in all cases we are justified in neglecting the small quantities of the second—or third (or higher)—orders, if only we take the small quantity of the first order small enough in itself.
But, it must be remembered, that small quantities if they occur in our expressions as factors multiplied by some other factor, may become important if the other factor is itself large. Even a farthing becomes important if only it is multiplied by a few hundred.
Now in the calculus we write ${\displaystyle dx}$ for a little bit of ${\displaystyle x}$. These things such as ${\displaystyle dx}$, and ${\displaystyle du}$, and ${\displaystyle dy}$, are called “differentials,” the differential of ${\displaystyle x}$, or of ${\displaystyle u}$, or of ${\displaystyle y}$, as the case may be. [You read them as dee-eks, or dee-you, or dee-wy.] If ${\displaystyle dx}$ be a small bit of ${\displaystyle x}$, and relatively small of itself, it does not follow that such quantities as ${\displaystyle x\cdot dx}$, or ${\displaystyle x^{2}dx}$, or ${\displaystyle a^{x}dx}$ are negligible. But ${\displaystyle dx\times dx}$ would be negligible, being a small quantity of the second order.
A very simple example will serve as illustration.
Let us think of ${\displaystyle x}$ as a quantity that can grow by a small amount so as to become ${\displaystyle x+dx}$, where ${\displaystyle dx}$ is the small increment added by growth. The square of this is ${\displaystyle x^{2}+2x\cdot dx+(dx)^{2}}$. The second term is not negligible because it is a first-order quantity; while the third term is of the second order of smallness, being a bit of, a bit of ${\displaystyle x^{2}}$. Thus if we took ${\displaystyle dx}$ to mean numerically, say, ${\displaystyle {\tfrac {1}{60}}}$ of ${\displaystyle x}$, then the second term would be ${\displaystyle {\tfrac {2}{60}}}$ of ${\displaystyle x^{2}}$, whereas the third term would be ${\displaystyle {\tfrac {1}{3600}}}$ of ${\displaystyle x^{2}}$. This last term is clearly less important than the second. But if we go further and take ${\displaystyle dx}$ to mean only ${\displaystyle {\tfrac {1}{1000}}}$ of ${\displaystyle x}$, then the second term will be ${\displaystyle {\tfrac {2}{1000}}}$ of ${\displaystyle x^{2}}$, while the third term will be only ${\displaystyle {\tfrac {1}{1,000,000}}}$ of ${\displaystyle x^{2}}$.
Fig. 1.
Geometrically this may be depicted as follows: Draw a square (Fig. 1) the side of which we will take to represent ${\displaystyle x}$. Now suppose the square to grow by having a bit ${\displaystyle dx}$ added to its size each way. The enlarged square is made up of the original square ${\displaystyle x^{2}}$, the two rectangles at the top and on the right, each of which is of area ${\displaystyle x\cdot dx}$ (or together ${\displaystyle 2x\cdot dx}$), and the little square at the top right-hand corner which is ${\displaystyle (dx)^{2}}$. In Fig. 2 we have taken ${\displaystyle dx}$ as
Fig. 2. Fig. 3.
quite a big fraction of ${\displaystyle x}$—about ${\displaystyle {\tfrac {1}{5}}}$. But suppose we had taken it only ${\displaystyle {\tfrac {1}{100}}}$—about the thickness of an inked line drawn with a fine pen. Then the little corner square will have an area of only ${\displaystyle {\tfrac {1}{10,000}}}$ of ${\displaystyle x^{2}}$, and be practically invisible. Clearly ${\displaystyle (dx)^{2}}$ is negligible if only we consider the increment ${\displaystyle dx}$ to be itself small enough.
Let us consider a simile.
Suppose a millionaire were to say to his secretary: next week I will give you a small fraction of any money that comes in to me. Suppose that the secretary were to say to his boy: I will give you a small fraction of what I get. Suppose the fraction in each case to be ${\displaystyle {\tfrac {1}{100}}}$ part. Now if Mr. Millionaire received during the next week £${\displaystyle 1000}$, the secretary would receive £${\displaystyle 10}$ and the boy ${\displaystyle 2}$ shillings. Ten pounds would be a small quantity compared with £${\displaystyle 1000}$; but two shillings is a small small quantity indeed, of a very secondary order. But what would be the disproportion if the fraction, instead of being ${\displaystyle {\tfrac {1}{100}}}$, had been settled at ${\displaystyle {\tfrac {1}{1000}}}$ part? Then, while Mr. Millionaire got his £${\displaystyle 1000}$, Mr. Secretary would get only £${\displaystyle 1}$, and the boy less than one farthing!
The witty Dean Swift[2] once wrote:
“Hath smaller Fleas that on him prey.
“And these have smaller Fleas to bite ’em,
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diffalg(deprecated)/preparation_polynomial - Maple Help
Home : Support : Online Help : diffalg(deprecated)/preparation_polynomial
diffalg
preparation_polynomial
compute preparation polynomial
Calling Sequence preparation_polynomial (p, a, R, 'm' ) preparation_polynomial (p, A=a, R, 'm' )
Parameters
p - differential polynomial in R a - regular differential polynomial in R R - differential polynomial ring m - (optional) name A - derivative of order zero in R
Description
• Important: The diffalg package has been deprecated. Use the superseding package DifferentialAlgebra instead.
• The function preparation_polynomial computes a preparation polynomial of p with respect to a.
• The preparation polynomial of p with respect to a is a sort of expansion of p according to mparama and its derivatives. It plays a prominent role in the determination of the essential components of the radical differential ideal generated by a single differential polynomial.
• A differential polynomial a is said to be regular if it has no common factor with its separant. This property is therefore dependent on the ranking defined on R.
• If A is omitted, the preparation polynomial appears with an indeterminate (local variable) looking like _A.
• If A is specified, the preparation polynomial is in the differential indeterminate A. Then, A, nor its derivatives, should appear in p nor a.
• Assume that preparation_polynomial(p, a, R, 'm') = $\mathrm{c1}\mathrm{M1}\left(\mathrm{_A}\right)+\mathrm{....}+\mathrm{ck}\mathrm{Mk}\left(\mathrm{_A}\right)$, where the Mi are differential monomials in $\mathrm{_A}$ and the $\mathrm{ci}$ are polynomials in R. Then
- $mp=\mathrm{c1}\mathrm{M1}\left(a\right)+\mathrm{....}+\mathrm{ck}\mathrm{Mk}\left(a\right)$, where m belongs to R.
- The $\mathrm{ci}$ are not reduced to zero by a, and therefore do not belong to the general component of a.
- m is a power product of factors of the initial and separant of a).
• The command with(diffalg,preparation_polynomial) allows the use of the abbreviated form of this command.
Examples
Important: The diffalg package has been deprecated. Use the superseding package DifferentialAlgebra instead.
The preparation polynomial is used to determine the essential singular zeros of a differential polynomial.
> $\mathrm{with}\left(\mathrm{diffalg}\right):$
> $R≔\mathrm{differential_ring}\left(\mathrm{derivations}=\left[x,y\right],\mathrm{ranking}=\left[u,A\right]\right):$
> $p≔16u\left[x,y\right]\left({u\left[x,x\right]}^{2}-{u\left[y,y\right]}^{2}+u\left[y,y\right]-u\left[x,x\right]\right)+{u\left[\right]}^{2}\left(4u\left[\right]-{x}^{2}-{y}^{2}\right)$
${p}{≔}{16}{}{{u}}_{{x}{,}{y}}{}\left({{u}}_{{x}{,}{x}}^{{2}}{-}{{u}}_{{y}{,}{y}}^{{2}}{-}{{u}}_{{x}{,}{x}}{+}{{u}}_{{y}{,}{y}}\right){+}{{u}\left[\right]}^{{2}}{}\left({-}{{x}}^{{2}}{-}{{y}}^{{2}}{+}{4}{}{u}\left[\right]\right)$ (1)
> $\mathrm{equations}\left(\mathrm{Rosenfeld_Groebner}\left(\left[p\right],R\right)\right)$
$\left[\left[{-}{{x}}^{{2}}{}{{u}\left[\right]}^{{2}}{-}{{y}}^{{2}}{}{{u}\left[\right]}^{{2}}{+}{4}{}{{u}\left[\right]}^{{3}}{+}{16}{}{{u}}_{{x}{,}{x}}^{{2}}{}{{u}}_{{x}{,}{y}}{-}{16}{}{{u}}_{{x}{,}{y}}{}{{u}}_{{y}{,}{y}}^{{2}}{-}{16}{}{{u}}_{{x}{,}{x}}{}{{u}}_{{x}{,}{y}}{+}{16}{}{{u}}_{{x}{,}{y}}{}{{u}}_{{y}{,}{y}}\right]{,}\left[{-}{{x}}^{{2}}{-}{{y}}^{{2}}{+}{4}{}{u}\left[\right]\right]{,}\left[{u}\left[\right]\right]\right]$ (2)
> $\mathrm{preparation_polynomial}\left(p,u\left[\right],R\right)$
${4}{}{{\mathrm{_A}}\left[\right]}^{{3}}{-}{16}{}{{\mathrm{_A}}}_{{y}{,}{y}}^{{2}}{}{{\mathrm{_A}}}_{{x}{,}{y}}{+}{16}{}{{\mathrm{_A}}}_{{x}{,}{y}}{}{{\mathrm{_A}}}_{{x}{,}{x}}^{{2}}{+}\left({-}{{x}}^{{2}}{-}{{y}}^{{2}}\right){}{{\mathrm{_A}}\left[\right]}^{{2}}{+}{16}{}{{\mathrm{_A}}}_{{y}{,}{y}}{}{{\mathrm{_A}}}_{{x}{,}{y}}{-}{16}{}{{\mathrm{_A}}}_{{x}{,}{y}}{}{{\mathrm{_A}}}_{{x}{,}{x}}$ (3)
> $\mathrm{preparation_polynomial}\left(p,A\left[\right]=4u\left[\right]-{x}^{2}-{y}^{2},R\right)$
${{A}\left[\right]}^{{3}}{-}{4}{}{{A}}_{{y}{,}{y}}^{{2}}{}{{A}}_{{x}{,}{y}}{+}{4}{}{{A}}_{{x}{,}{y}}{}{{A}}_{{x}{,}{x}}^{{2}}{+}\left({2}{}{{x}}^{{2}}{+}{2}{}{{y}}^{{2}}\right){}{{A}\left[\right]}^{{2}}{+}\left({{x}}^{{4}}{+}{2}{}{{x}}^{{2}}{}{{y}}^{{2}}{+}{{y}}^{{4}}\right){}{A}\left[\right]$ (4)
Studying the degree in $A$ (or $\mathrm{_A}$) and its derivatives in these preparation polynomials, we can deduce that $u\left(x,y\right)=\frac{{x}^{2}}{4}+\frac{{y}^{2}}{4}$ is an essential singular zero of $p$ while $u\left(x,y\right)=0$ is not.
The preparation polynomial can be used to further study the relationships between the general zero and the singular zeros of $p$.
> $R≔\mathrm{differential_ring}\left(\mathrm{ranking}=\left[y,A\right],\mathrm{derivations}=\left[x\right]\right):$
> $p≔3{y\left[x\right]}^{4}y\left[\mathrm{}\left(x,2\right)\right]{y\left[\mathrm{}\left(x,4\right)\right]}^{2}-4{y\left[x\right]}^{4}{y\left[\mathrm{}\left(x,3\right)\right]}^{2}y\left[\mathrm{}\left(x,4\right)\right]+6{y\left[x\right]}^{3}{y\left[\mathrm{}\left(x,2\right)\right]}^{2}y\left[\mathrm{}\left(x,3\right)\right]y\left[\mathrm{}\left(x,4\right)\right]+24{y\left[x\right]}^{2}{y\left[\mathrm{}\left(x,2\right)\right]}^{4}y\left[\mathrm{}\left(x,4\right)\right]-12{y\left[x\right]}^{3}y\left[\mathrm{}\left(x,2\right)\right]{y\left[\mathrm{}\left(x,3\right)\right]}^{3}-29{y\left[x\right]}^{2}{y\left[\mathrm{}\left(x,2\right)\right]}^{3}{y\left[\mathrm{}\left(x,3\right)\right]}^{2}+12{y\left[\mathrm{}\left(x,2\right)\right]}^{7}$
${p}{≔}{3}{}{{y}}_{{x}}^{{4}}{}{{y}}_{{x}{,}{x}}{}{{y}}_{{x}{,}{x}{,}{x}{,}{x}}^{{2}}{-}{4}{}{{y}}_{{x}}^{{4}}{}{{y}}_{{x}{,}{x}{,}{x}}^{{2}}{}{{y}}_{{x}{,}{x}{,}{x}{,}{x}}{+}{6}{}{{y}}_{{x}}^{{3}}{}{{y}}_{{x}{,}{x}}^{{2}}{}{{y}}_{{x}{,}{x}{,}{x}}{}{{y}}_{{x}{,}{x}{,}{x}{,}{x}}{-}{12}{}{{y}}_{{x}}^{{3}}{}{{y}}_{{x}{,}{x}}{}{{y}}_{{x}{,}{x}{,}{x}}^{{3}}{+}{24}{}{{y}}_{{x}}^{{2}}{}{{y}}_{{x}{,}{x}}^{{4}}{}{{y}}_{{x}{,}{x}{,}{x}{,}{x}}{-}{29}{}{{y}}_{{x}}^{{2}}{}{{y}}_{{x}{,}{x}}^{{3}}{}{{y}}_{{x}{,}{x}{,}{x}}^{{2}}{+}{12}{}{{y}}_{{x}{,}{x}}^{{7}}$ (5)
> $\mathrm{equations}\left(\mathrm{Rosenfeld_Groebner}\left(\left[p\right],R\right)\right)$
$\left[\left[{3}{}{{y}}_{{x}}^{{4}}{}{{y}}_{{x}{,}{x}}{}{{y}}_{{x}{,}{x}{,}{x}{,}{x}}^{{2}}{-}{4}{}{{y}}_{{x}}^{{4}}{}{{y}}_{{x}{,}{x}{,}{x}}^{{2}}{}{{y}}_{{x}{,}{x}{,}{x}{,}{x}}{+}{6}{}{{y}}_{{x}}^{{3}}{}{{y}}_{{x}{,}{x}}^{{2}}{}{{y}}_{{x}{,}{x}{,}{x}}{}{{y}}_{{x}{,}{x}{,}{x}{,}{x}}{-}{12}{}{{y}}_{{x}}^{{3}}{}{{y}}_{{x}{,}{x}}{}{{y}}_{{x}{,}{x}{,}{x}}^{{3}}{+}{24}{}{{y}}_{{x}}^{{2}}{}{{y}}_{{x}{,}{x}}^{{4}}{}{{y}}_{{x}{,}{x}{,}{x}{,}{x}}{-}{29}{}{{y}}_{{x}}^{{2}}{}{{y}}_{{x}{,}{x}}^{{3}}{}{{y}}_{{x}{,}{x}{,}{x}}^{{2}}{+}{12}{}{{y}}_{{x}{,}{x}}^{{7}}\right]{,}\left[{{y}}_{{x}}^{{2}}{}{{y}}_{{x}{,}{x}{,}{x}}^{{2}}{+}{3}{}{{y}}_{{x}{,}{x}}^{{4}}\right]{,}\left[{{y}}_{{x}{,}{x}}\right]\right]$ (6)
> $q≔3{y\left[x,x\right]}^{4}+{y\left[x\right]}^{2}{y\left[x,x,x\right]}^{2}$
${q}{≔}{{y}}_{{x}}^{{2}}{}{{y}}_{{x}{,}{x}{,}{x}}^{{2}}{+}{3}{}{{y}}_{{x}{,}{x}}^{{4}}$ (7)
> $\mathrm{preparation_polynomial}\left(p,A\left[\right]=q,R\right)$
$\left({-}{32}{}{{y}}_{{x}{,}{x}}{}{{y}}_{{x}{,}{x}{,}{x}}{}{{y}}_{{x}}{-}{32}{}{{y}}_{{x}{,}{x}}^{{3}}\right){}{{A}\left[\right]}^{{2}}{-}{8}{}{{y}}_{{x}}^{{2}}{}{{y}}_{{x}{,}{x}{,}{x}}{}{A}\left[\right]{}{{A}}_{{x}}{+}{3}{}{{y}}_{{x}{,}{x}}{}{{y}}_{{x}}^{{2}}{}{{A}}_{{x}}^{{2}}{+}\left({96}{}{{y}}_{{x}}{}{{y}}_{{x}{,}{x}}^{{5}}{}{{y}}_{{x}{,}{x}{,}{x}}{+}{96}{}{{y}}_{{x}{,}{x}}^{{7}}\right){}{A}\left[\right]$ (8)
> $\mathrm{equations}\left(\mathrm{essential_components}\left(p,R\right)\right)$
$\left[\left[{3}{}{{y}}_{{x}}^{{4}}{}{{y}}_{{x}{,}{x}}{}{{y}}_{{x}{,}{x}{,}{x}{,}{x}}^{{2}}{-}{4}{}{{y}}_{{x}}^{{4}}{}{{y}}_{{x}{,}{x}{,}{x}}^{{2}}{}{{y}}_{{x}{,}{x}{,}{x}{,}{x}}{+}{6}{}{{y}}_{{x}}^{{3}}{}{{y}}_{{x}{,}{x}}^{{2}}{}{{y}}_{{x}{,}{x}{,}{x}}{}{{y}}_{{x}{,}{x}{,}{x}{,}{x}}{-}{12}{}{{y}}_{{x}}^{{3}}{}{{y}}_{{x}{,}{x}}{}{{y}}_{{x}{,}{x}{,}{x}}^{{3}}{+}{24}{}{{y}}_{{x}}^{{2}}{}{{y}}_{{x}{,}{x}}^{{4}}{}{{y}}_{{x}{,}{x}{,}{x}{,}{x}}{-}{29}{}{{y}}_{{x}}^{{2}}{}{{y}}_{{x}{,}{x}}^{{3}}{}{{y}}_{{x}{,}{x}{,}{x}}^{{2}}{+}{12}{}{{y}}_{{x}{,}{x}}^{{7}}\right]{,}\left[{{y}}_{{x}}^{{2}}{}{{y}}_{{x}{,}{x}{,}{x}}^{{2}}{+}{3}{}{{y}}_{{x}{,}{x}}^{{4}}\right]\right]$ (9)
The general zero of $q$ is an essential singular zero of $p$ while the general zero of ${y}_{x,x}$ is not. Thus, the straight lines $y\left(x\right)=\mathrm{_C1}x+\mathrm{_C2}$, zeros of ${y}_{x,x}$, must be limits of either some non singular zeros of $p$ or of the non singular zeros of $q$. Again studying the degrees of the preparation polynomials of $p$ and $q$ we can deduce that the straight lines are in fact limits of the non singular zeros of both (cf. [Kolchin]).
> $\mathrm{preparation_polynomial}\left(p,A\left[\right]=y\left[x,x\right],R\right)$
${12}{}{{A}\left[\right]}^{{7}}{+}{24}{}{{y}}_{{x}}^{{2}}{}{{A}\left[\right]}^{{4}}{}{{A}}_{{x}{,}{x}}{-}{29}{}{{y}}_{{x}}^{{2}}{}{{A}\left[\right]}^{{3}}{}{{A}}_{{x}}^{{2}}{+}{6}{}{{y}}_{{x}}^{{3}}{}{{A}\left[\right]}^{{2}}{}{{A}}_{{x}}{}{{A}}_{{x}{,}{x}}{-}{12}{}{{y}}_{{x}}^{{3}}{}{A}\left[\right]{}{{A}}_{{x}}^{{3}}{+}{3}{}{{y}}_{{x}}^{{4}}{}{A}\left[\right]{}{{A}}_{{x}{,}{x}}^{{2}}{-}{4}{}{{y}}_{{x}}^{{4}}{}{{A}}_{{x}}^{{2}}{}{{A}}_{{x}{,}{x}}$ (10)
> $\mathrm{preparation_polynomial}\left(q,A\left[\right]=y\left[x,x\right],R\right)$
${3}{}{{A}\left[\right]}^{{4}}{+}{{y}}_{{x}}^{{2}}{}{{A}}_{{x}}^{{2}}$ (11)
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## Basic College Mathematics (9th Edition)
If we sold 200% of the fishing licenses this season that we sold last season, this means that we sold twice the licenses this season that we sold last season. Therefore, we sold $2\times380=760$ licenses this season. 200%$=200\div100=2$ $380\times2=760$ licenses
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Package of problems
## Solution
Here are some triangles.
The diagrams are not to scale.
Can you find …
1. … a triangle similar to triangle A with perimeter $6$?
Triangle A has perimeter $4 + 4 + 4 = 12$, so we want to scale by factor $6/12 = 1/2$, so we should choose the triangle with side lengths $2$, $2$ and $2$.
1. … a triangle similar to triangle B with perimeter $6$?
Triangle B has perimeter $3 + 4 + 5 = 12$ (like triangle A), so we again use factor $1/2$, so we should choose the triangle with side lengths $3/2$, $2$ and $5/2$.
1. … a triangle similar to triangle C with perimeter $6$?
Triangle C has perimeter $3 + 3 + 2 = 8$, so we want to scale by factor $6/8 = 3/4$. So we should choose the triangle with side lengths $9/4$, $9/4$ and $3/2$.
1. … a triangle similar to triangle D with perimeter $6$?
Triangle D has perimeter $1 + 2 + \sqrt{3} = 3 + \sqrt{3}$, so we want to scale by factor $\frac{6}{3 + \sqrt{3}} = \frac{6(3 - \sqrt{3})}{(3 + \sqrt{3})(3 - \sqrt{3})} = 3 - \sqrt{3}.$ So we should choose the triangle with side lengths $3 - \sqrt{3}$, $6 - 2\sqrt{3}$ and $-3 + 3\sqrt{3}$.
1. … a triangle similar to triangle A with area $10$?
We need to remember that when we scale the lengths of a triangle by a certain factor, we scale the area of the triangle by the square of that factor. So we must allow for that.
Triangle A has area $4\sqrt{3}$, so we want to scale by factor $\sqrt{\frac{10}{4\sqrt{3}}} = \sqrt{\frac{5\sqrt{3}}{6}}.$ So we should choose the triangle with side lengths $4\sqrt{\frac{5\sqrt{3}}{6}}$, $4\sqrt{\frac{5\sqrt{3}}{6}}$ and $4\sqrt{\frac{5\sqrt{3}}{6}}$.
1. … a triangle similar to triangle B with area $10$?
Triangle B has area $6$, so we want to scale by factor $\sqrt{\frac{10}{6}} = \frac{\sqrt{15}}{3}.$ So we should choose the triangle with side lengths $\sqrt{15}$, $\frac{4\sqrt{15}}{3}$ and $\frac{5\sqrt{15}}{3}$.
1. … a triangle similar to triangle C with area $10$?
Triangle C has area $2\sqrt{2}$, so we want to scale by factor $\sqrt{\frac{10}{2\sqrt{2}}} = \sqrt{\frac{5\sqrt{2}}{2}}.$ So we should choose the triangle with side lengths $\sqrt{10\sqrt{2}}$, $3\sqrt{\frac{5\sqrt{2}}{2}}$, $3\sqrt{\frac{5\sqrt{2}}{2}}$.
1. … a triangle similar to triangle D with area $10$?
Triangle D has area $\frac{\sqrt{3}}{2}$, so we want to scale by factor $\sqrt{\frac{10}{\sqrt{3}/2}} = \sqrt{\frac{20}{\sqrt{3}}} = \frac{2\sqrt{5}}{\sqrt[4]{3}}.$ So we should choose the triangle with side lengths $\frac{2\sqrt{5}}{\sqrt[4]{3}}$, $\frac{4\sqrt{5}}{\sqrt[4]{3}}$ and $\frac{2\sqrt{15}}{\sqrt[4]{3}}$.
1. … a right-angled triangle similar to one of triangles A, B, C, D with area $12$?
Of the four given triangles, only triangles B and D are right-angled. We can check that using Pythagoras’s theorem, since the side lengths $3$, $4$ and $5$ satisfy his equation, as do the side lengths $1$, $\sqrt{3}$ and $2$.
Let’s find a triangle similar to B with area $12$. And we can do that using the same approach as earlier.
Triangle B has area $6$, so we want to scale by factor $\sqrt{\frac{12}{6}} = \sqrt{2}.$ So we should choose the triangle with side lengths $3\sqrt{2}$, $4\sqrt{2}$ and $5\sqrt{2}$.
1. … an isosceles triangle similar to one of triangles A, B, C, D with area $12$?
Of the four given triangles, only triangles A and C are isosceles. We could choose either, but let’s choose C and save A for later since it’s equilateral too.
Triangle C has area $2\sqrt{2}$, so we want to scale by factor $\sqrt{\frac{12}{2\sqrt{2}}} = \sqrt{3\sqrt{2}}.$ So we should choose the triangle with side lengths $2\sqrt{3\sqrt{2}}$, $3\sqrt{3\sqrt{2}}$ and $3\sqrt{3\sqrt{2}}$.
1. … an equilateral triangle similar to one of triangles A, B, C, D with area $12$?
Our only option here is triangle A.
Triangle A has area $4\sqrt{3}$, so we want to scale by factor $\sqrt{\frac{12}{4\sqrt{3}}} = \sqrt[4]{3}.$ So we should choose the triangle with side lengths $4\sqrt[4]{3}$, $4\sqrt[4]{3}$ and $4\sqrt[4]{3}$.
1. … a scalene triangle similar to one of triangles A, B, C, D with area $12$?
Here we can choose between triangles B and D. Since we’ve already done B, let’s pick D.
Triangle D has area $\frac{\sqrt{3}}{2}$, so we want to scale by factor $\sqrt{\frac{12}{\sqrt{3}/2}} = \sqrt{8\sqrt{3}} = 2\sqrt{2\sqrt{3}}.$ So we should choose the triangle with side lengths $2\sqrt{2\sqrt{3}}$, $4\sqrt{2\sqrt{3}}$ and $2\sqrt{6\sqrt{3}}$.
1. … a right-angled triangle with area $12$ that is not similar to any of triangles A, B, C, D?
We have lots of choice here, we just have to make sure that our right-angled triangle isn’t similar to triangle B or D(it definitely won’t be similar to either of the others).
Let’s try to be lazy when constructing an example. An easy way to make a triangle have area $12$ is to give it base $12$ and height $2$, for example, and then we can use Pythagoras to finish off.
Choose the triangle with side lengths $2$, $12$ and $\sqrt{2^2 + 12^2} = \sqrt{4 + 144} = \sqrt{148} = 2\sqrt{37}$. The side lengths satisfy Pythagoras’s equation so the triangle is right-angled, and its area is $12$. It is easy to check that it is not similar to triangle B or triangle D, by looking at ratios of two sides (e.g. $\frac{3}{4} \neq \frac{2}{12}$).
1. … an isosceles triangle with area $12$ that is not similar to any of triangles A, B, C, D?
We need to avoid both A and C here. It should be quite straightforward to build an isosceles triangle with area $12$, by picking the base and height and then finding the missing side length.
Choose the isosceles triangle with base $2$ (as one side of the triangle) and height $12$. The other sides have length $\sqrt{1^2 + 12^2} = \sqrt{145}$. This is clearly isosceles, clearly has area $12$, and cannot be similar to triangle C (since they have the same base but different heights).
1. … an equilateral triangle with area $12$ that is not similar to any of triangles A, B, C, D?
This is a bit tricky…
Every equilateral triangle is similar to every other equilateral triangle! So any equilateral triangle will be similar to A, so there is no triangle that meets the criteria.
1. … a scalene triangle with area $12$ that is not similar to any of triangles A, B, C, D?
There are lots and lots of examples here. Again, a nice approach might be to specify the base and height, since they determine the area, and then just to make sure that the triangle is not isosceles or right-angled. The hardest part might be calculating the side lengths!
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###### \${selected_topic_name}
• Notes
$\text { The water in a tank is pressurized by air, and the pressure is measured by a multifluid manometer }$
$\text { as shown in Fig. Pl-53. Determine the gage pressure of air in the tank if } l_{i,}^{\prime}=0.2 \mathrm{m}, \mathrm{h} 2=0.3 \mathrm{m}, \text { and }$
$h 3=0.46 \mathrm{m} . \text { Take the densities of water, oil, and mercury to be } 1000 \mathrm{kg} / \mathrm{m} 3,850 \mathrm{kg} / \mathrm{m} 3, \text { and } 13,600 \mathrm{kg} / \mathrm{m} 3, \text { respectively. }$
$P_{1}+\rho_{w} g h_{1}+\rho_{0} g h_{2}$$-\rho g h_{3}=P_{atm}$
$P_{1}-P_{a t m}=$$\rho g h_{3}-\rho_{w} g h_{1}$$-\rho_0 \mathrm{gh}_{2}$
$P_{1} - P_{a t m}$$=g\left(\rho_{m} h_{3}-\rho_{w} h_{1}-\rho_{0} h_{2}\right)$
$\text{P}_ \text{gage }=P_{1}-P_{\text { atm }}$
$∴ \quad \text P_\text { 1gage }=g \left(\rho_m h_{3}-\rho_{w} h_{1}-\rho_{0} h_{2}\right)$
$=9.81(13600 *(0.46)-1000(0.2)-850(0.3))$
$=56.9 \mathrm{kPa}$
$\text { Consider a U-tube whose arms are open to the atmosphere. Now water is poured into }$
$\text { the U-tube from one arm, and light oil }(p=790 \mathrm{kg} / \mathrm{m} 3) \text { from the other. One arm }$
$\text { contains } 70 \text { -cm-high water, while the other arm contains both fluids with an }$
$\text { oil-to-water height ratio of } 4 . \text { Determine the height of each fluid in that arm. }$
$\rho_{\text{ oil }}=790 \mathrm{kg} / \mathrm{m}^{3}$
$\rho_{w}=1000 \mathrm{kg} / \mathrm{m}^{3}$
$h_{a}=4 h_{w 2}=4(0.168)\left(\begin{array}{c}{0.673} {m}\end{array}\right)$
$P_{\text { bottom }}=P_\text { at m}+\rho_{w} g h w_{1}$
$P_\text {bottom = }$$P_\text { atm }$$+\rho_\text {w} ghw_ 2$$+\rho_{0} g h_{a}$
$\rho_ { w} ghw_1$$=\rho_{w} ghw_{2}+\rho_{a} g h a$
$\rho_{w} hw_{1}=\rho_{w} hw_{2}+\rho_{0} h_{a}$
$÷ \rho_w\rightarrow h w_{1}= \frac{\rho_{w} hw_{2}}{\rho_w}+\frac{\rho_{0} h a}{\rho_{w}}\quad\longrightarrow h{w_{1}}=h w_{2}+\frac{\rho_{0} h_{4}}{\rho_{w}}$
$\longrightarrow \quad 0.7=h w_{2}+$$\frac{790 \mathrm{ha}}{1000}$
$\longrightarrow \quad 0.7=h w_{2}+\frac{790hw_2 (4)}{1000}$
$\text { hw} _2=0.168$
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# Why Is DB Measured In Negatives?
## Why is dB negative or positive?
The negative sign on a dB (logrithmic) comparison means that the mesurement was less than the reference where a positive sign means it was more than the reference.
So the only way to convert a negative dB indication is to change the reference.
For example -2 dBm would be +1 dBu..
## How much louder is 6 dB?
1/2 the power = –3dB”. A change of 6 dB is accepted as a significant difference in level for any listener listening to speech or music. It is a quite noticeable increase or decrease in loudness. To produce an increase of +6 dB you need to increase power (watts) by a factor of four.
## Is 3 dB twice as loud?
An increase of 3dB doubles the sound intensity but a 10dB increase is required before a sound is perceived to be twice as loud. Therefore a small increase in decibels represents a large increase in intensity. For example – 10dB is 10 times more intense than 1dB, while 20dB is 100 times more intense than 1dB.
## How many dB can kill you?
150 decibels is usually considered enough to burst your eardrums, but the threshold for death is usually pegged at around 185-200 dB. A passenger car driving by at 25 feet is about 60 dB, being next to a jackhammer or lawn mower is around 100 dB, a nearby chainsaw is 120 dB.
## Why are stereos measured in negative decibels?
When you turn up the volume of an amp you are actually decreasing the amount of restriction placed on that amp and in turn it plays louder. The negative number scale is a more accurate way to indicate the volume output. … You also likely noticed the “0 dB” (decibel) setting on the volume panel display.
## Can you hear negative decibels?
So 0 dB does not mean no sound, it means a sound level where the sound pressure is equal to that of the reference level. This is a small pressure, but not zero. It is also possible to have negative sound levels: – 20 dB would mean a sound with pressure 10 times smaller than the reference pressure, i.e. 2 μPa.
## What does 20 dB sound like?
10 dB: Normal breathing. 20 dB: Whispering from five feet away. 30 dB: Whispering nearby. 40 dB: Quiet library sounds.
## What is the 3dB rule?
3dB rule when measuring noise at work When you measure noise levels with a noise meter, you measure the intensity of noise in units called decibels, expressed as dB(A). … It is based on orders of magnitude, rather than a standard linear scale, so each mark on the decibel scale is the previous mark multiplied by a value.
## How many decibels is a gunshot with silencer?
In reality, most civilian-accessible firearms emit sounds ranging from 140-175 decibels. Silencers only marginally suppress a gun blast, bringing the levels down to around 120-150dB. Most of the time, the sound is still very obviously identifiable as a gun shot.
## Can you hear a 1 dB difference?
A 1 dB change in a sound equates to about a 26% difference in sound energy (remember that a 3 dB difference is a doubling of energy levels). In terms of subjective loudness, a 1 dB change yields just over a 7% change. A 3 dB change yields a 100% increase in sound energy and just over a 23% increase in loudness.
## What does negative dB gain mean?
If the dB power ratio is of a negative value, then this means an attenuation or loss as the output power will be less than the circuits input power (POUT < PIN). Clearly then 0dB means the power ratio is one, no reduction or gain.
## How much louder is 70 dB than 60 db?
It is traditional to say that 70db is twice as loud as 60 dB. Actually, in this sphere the very notion of twice or half is undefined. All you can say is that it is very noticeably louder yet not extremely much louder.
## What dB means?
decibelsSound is measured in units called decibels (dB). The higher the decibel level, the louder the noise. On the decibel scale, the level increase of 10 means that a sound is actually 10 times more intense, or powerful.
## What dB should I set my speakers?
Do not point the meter at each speaker. You may choose any level you wish to calibrate the speakers to: 70, 75 and 80 dB are popular levels to use. Whatever you choose, adjust each speaker’s test tone so that it pushes the meter’s needle to the same decibel level.
## Why gain is calculated in dB?
In electronics, gain is a measure of the ability of a two-port circuit (often an amplifier) to increase the power or amplitude of a signal from the input to the output port by adding energy converted from some power supply to the signal. … It is often expressed using the logarithmic decibel (dB) units (“dB gain”).
## What does negative dB loss mean?
If we have loss in a fiber optic system, the measured power is less than the reference power, so the ratio of measured power to reference power is less than 1 and the log is negative, making dB a negative number. … Instruments that measure in dB can be either optical power meters or optical loss test sets (OLTS).
## How loud is 1100 decibels?
1100 decibels is preposterous. 20 db is ten times as loud as 10 db. To the Human ear, it sounds about twice as loud. Our ears are well designed that way, we filter out loudness.
## Is higher dB gain better?
dB gain is used to measure the amplification power of a signal booster. This means that a +11 dB gain is better than a +8 dB gain. … If the gain goes up by +10 dB, the signal strength improves by 10x, while a +20 dB gain translates to 100x more signal strength.
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# GRE Math : How to evaluate algebraic expressions
## Example Questions
### Example Question #31 : How To Evaluate Algebraic Expressions
Compare and determine which number is larger.
: Number of minutes in a week.
: Number of hours in a leap year.
Explanation:
To solve this problem, we must figure out the values of .
Quantity A is equivalent to the number of minutes in a week. There are 7 days in a week, 24 hours in a day, and 60 mins to an hour.
.
Quantity B is equivalent to the number of horus in a leap year. There are 366 days in a leap year and 24 hours a day.
Because is larger than , is larger.
### Example Question #351 : Algebra
Compare the two quantities and determine which are larger.
: The distance between and
: The distance between and
Explanation:
To solve this problem we must make use of the quadratic formula.
is the distance between the points and . The two points are separated by units horizontally and units vertically.
Using the quadratic formula we find that the distance between the two points of is
Quantity B is the distance between the points and .
The two points of Quantity B are seperated by 2 units horizontally and 2 units vertically.
Using the quadratic formula we find that the distance between the two points of Quantity B is
Comparing the two distances,, therefore Quantity A is larger than Quantity B.
### Example Question #31 : How To Evaluate Algebraic Expressions
Compare and determine which is larger.
: The sum of the factors of
: The sum of the factors of
Explanation:
To solve this problem we must solve for .
Quantity A is the sum of the factors of .
Quantity B is the sum of the factors of .
will always be greater than . Therefore Quantity B is larger than Quantity A.
### Example Question #71 : Expressions
Determine whether is larger.
Explanation:
To solve this we must solve for .
is equivalent to . Because is larger than , .
### Example Question #31 : How To Evaluate Algebraic Expressions
A candle that burns for is now long.
Compare and determine which is greater.
: The number of minutes it would take the entire candle to burn.
Explanation:
Because it took for the candle to decrease in size by , in order for the entire candle to burn out, we divide the total size of the candle by and multiply by .
. Therefore Quantity B is larger than Quantity A.
### Example Question #36 : How To Evaluate Algebraic Expressions
Compare the two quantities and determine which is larger.
Explanation:
The equation can be factored down to . Therefore the roots of the equation are . Twice the sum of these roots is equal to , therefore the two quantities are equivalent.
### Example Question #37 : How To Evaluate Algebraic Expressions
A convinence store purchases coke cans at each and sells them for above cost.
Compare the two quantities are determine which is greater.
Explanation:
is simply . is the profit the store makes on each can. Because the store buys the cans for and sells them for above cost, this means the can sells the . Therefore the store profits per can, meaning that Quantity B is larger than Quantity A.
### Example Question #31 : How To Evaluate Algebraic Expressions
Determine which quantity is larger.
Explanation:
A cruious thing happens with you try to square numbers less than but greater than . The numbers actually become smaller!
. etc....
Therefore Quantity A is larger than Quantity B when
### Example Question #39 : How To Evaluate Algebraic Expressions
houses have a total valuation of in 2014. In 2015, two of the houses decreased in value by , one of the houses increased in value by and the rest of the houses remained the same price.
Determine which quantity is larger.
Explanation:
Standard deviation for the house prices is the is the amount of variance for the houses. Because we aren't given the prices of the houses, we have no idea what the standard deviation could be. The houses could have been priced so that of the houses cost each and the last two hosues cost the remaining combined. This would make the standard deviation huge. The houses could also have been priced each, meaning that the standard deviation for that year would be . It is impossible to determine which quantity is larger based on the information provided.
### Example Question #71 : Expressions
Quantity A:
Quantity B:
Quantity B is greater.
Quantity A is greater.
The two quantities are equal.
The relationship cannot be determined.
The relationship cannot be determined.
Explanation:
Since there is an term in the equation
We must acknowledge the possibility of multiple values that satisfy the equation.
Rewrite it so that it is set equal to zero:
For an equation of the form
The solutions for can be found using the quadratic formula:
Quantity A:
Since could be positive or negative, the relationship cannot be determined.
Tired of practice problems?
Try live online GRE prep today.
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## Algebra 1
The trapezoid was graphed using graphing software. $9-3 =6 = h$ $8-3 = 5 = b_{1}$ $8-0 = 8 = b_{2}$ $1/2*(5+8)*6$ \$3*13 = 39
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# How to relate multiplication of irrational numbers to real world?
• I
• LittleRookie
In summary: EMFs can be described as vector fields with real numbers as points. Physical fields are analogous to sequences in that they have a ring structure, and physical quantities can be related to one another through operations on fields. For instance, the electric potential at a point can be related to the electric field at that point, and the electric field can be related to the electric potential.
LittleRookie
I'm aware of the axioms of real numbers, the constructions of real number using the rational numbers (Cauchy sequence and Dedekind cut). But I can't relate the arithmetic of irrational numbers to real world usage.
I can think the negative and positive irrational numbers to represent (measurement of) real life quantities that satisfy the axioms. This is similar to relating positive and negative rational numbers to real world usage.
However, I can't do so for multiplication of irrational numbers. Multplication of rational numbers takes the usual intepretation of "(a out of b) of something".
Any thoughts?
The area of a circle or a rhombus inside a square are multiplications of irrational numbers.
I need more help :(
In a physical length measurement, a particular length is rational or irrational depending only on what units you are using to do the measurement. There is no physical difference between a rational versus an irrational length. Suppose a rectangle is measured to have an irrational width and height. Then the area is the product of two irrational numbers in that choice of units.
FactChecker said:
In a physical length measurement, a particular length is rational or irrational depending only on what units you are using to do the measurement. There is no physical difference between a rational versus an irrational length. Suppose a rectangle is measured to have an irrational width and height. Then the area is the product of two irrational numbers in that choice of units.
I understand that. I'm seeking help in the interpretation of multiplication of irrational numbers in the real world. Something similar to the real world interpretation of multiplication of rational numbers, for instance, 1/2 times of 3 can be interpreted as divide 3 into 2 parts, and take 1 out of the 2 parts.
I look at it as the limit of rational multiplications of the truncated representations, so I can not help further.
member 587159
Thea area of the pink square is ## \frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}}## which can be found in the real world. Likewise has the circle with a diameter of ##1## an area of ##\frac{\pi}{4}## and a circumference of ##\pi \cdot 1##. All are multiplications by irrationals.
I do not understand your problem. The way they came into use was, that there have been actually lengths which were not rational, the diagonals of squares. As they could be sides of other squares (see image), they had to be multiplied to get the area.
Algebraically it is as you said above: reals are equivalence classes of Cauchy sequences, and sequences carry a ring structure which turns out to be a field.
Klystron and hmmm27
As a young rookie when I encountered a mathematical object difficult to visualize; i.e., apply to the "real world", I would fall back on set theory.
In posts #2 and 7 mentor @fresh_42 explains and discusses your question in what we could describe as geometric solutions. The last sentence of post #7 provides broad guidance on algebraic methods to solve these problems including operations on irrational (and complex) numbers:
fresh_42 said:
Algebraically it is as you said above: reals are equivalence classes of Cauchy sequences, and sequences carry a ring structure which turns out to be a field.
If geometric objects and length relations provide insufficient real world linkage, consider examples of physical fields, such as an electromagnetic_field (EMF).
## 1. How can I use multiplication of irrational numbers in real life?
Multiplication of irrational numbers can be used in real life situations such as calculating the area of a circle or the volume of a sphere. These calculations involve multiplying the irrational number pi (π) with the radius or diameter of the circle or sphere.
## 2. Can you provide an example of how multiplication of irrational numbers is used in the real world?
One example is calculating the distance traveled by a car with a constant speed of 60 miles per hour for 2.5 hours. This can be represented by the equation 60 x 2.5 = 150 miles, where the irrational number 2.5 represents the time traveled.
## 3. How is multiplication of irrational numbers related to geometry?
In geometry, multiplication of irrational numbers is used to calculate the lengths of sides and areas of shapes with irrational measurements, such as square roots. For example, the Pythagorean theorem uses multiplication of irrational numbers to find the length of the hypotenuse of a right triangle.
## 4. Why is it important to understand how to relate multiplication of irrational numbers to real world?
Understanding how to relate multiplication of irrational numbers to real world situations is important because it allows us to solve practical problems and make accurate calculations in various fields such as science, engineering, and finance. It also helps us better understand the concept of irrational numbers and their significance in mathematics.
## 5. Are there any real world applications where multiplication of irrational numbers is used in combination with other mathematical operations?
Yes, multiplication of irrational numbers is often used in combination with addition, subtraction, and division in real world applications. For example, in finance, compound interest calculations involve multiplying an irrational number (interest rate) with the principal amount and then adding the resulting amount to the principal. In physics, the equation for work (W = F x d) involves multiplying an irrational number (force) with a distance.
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+0
# need help with quadratic
0
86
2
Write down the quadratic equation whose roots are x=-7 and x=-1, and the coefficient of x^2 is 1.
Enter your answer in the form "x^2 + bx + c = 0".
Nov 9, 2021
#1
0
Write down the quadratic equation whose roots are x=-7 and x=-1, and the coefficient of x^2 is 1.
Enter your answer in the form "x^2 + bx + c = 0".
Since the roots are x = –7 and x = –1 then
the associated factors are (x + 7) and (x + 1) respectively.
which makes the equation x2 + 8x + 7
.
Nov 9, 2021
#2
0
x2 + 8x + 7 = 0 Without the = 0 it isn't an equation, it's just an expression.
The expression has to equal something to be an equation.
.
Guest Nov 9, 2021
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# Expectation and variance of a sum of two random variables
Following up on this question, how would you derive the expectation and variance of the sum of two normally distributed random variables that aren't necessarily independent?
For example, if $$X \sim N(\mu, 3\sigma^2)$$ and $$Y \sim N(\mu + 9, \sigma^2)$$ is there a way to calculate the expectation and variance? What would the resulting X + Y distribution be in concrete terms? We aren't given the covariance.
$$E[X+Y]=E[X]+E[Y]=\mu+\mu+9=2 \mu +9$$ $$\begin{array} VVar[X+Y]&=Var[X]+Var[Y]+2Cov[X,Y]\\ &=3 \sigma^2+\sigma^2+E[XY]-E[X]E[Y]\\ &=4 \sigma^2+E[XY]-\mu(\mu+9) \end{array}$$
We can't say that much about the distribution of $XY$ without some more assumptions. If we know that $X$ and $Y$ are independent then $XY$ has a chi-squared distribution and we can compute $E[XY]$.
• +1, However, if we know that $X$ and $Y$ are independent, then $E(XY) = E(X)E(Y)$ or equivalently, $Cov(X,Y) = 0$. No need to use the chi-square relationship here. – knrumsey Oct 13 '17 at 20:12
• In other words, the variance would just be $4\sigma^2$? – username Oct 13 '17 at 20:18
• Yes, if your independent then $Cov[X,Y]=0$. – Wintermute Oct 13 '17 at 20:34
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Conversion formula how to convert 53 kilograms come pounds?
We recognize (by definition) that:1kg≈2.2046226lb
We can set up a relationship to resolve for the number of pounds.
You are watching: How much is 53 kilos in pounds
1kg53kg≈2.2046226lbxlb
Now, we cross multiply to fix for our unknown x:
xlb≈53kg1kg*2.2046226lb→xlb≈116.8449978lb
Conclusion:53kg≈116.8449978lb
Approximation
An almost right numerical an outcome would be: fifty-three kilograms is around one hundred and sixteen suggest eight four pounds, or alternatively, a lb is about zero suggest zero one times fifty-three kilograms.
Units involved
This is just how the systems in this conversion room defined:
Kilograms
The kilogram is the basic unit of mass in the global System of units (the Metric system) and also is defined as being same to the mass of the international Prototype that the Kilogram. The avoirdupois (or international) pound, offered in both the imperial and also US customary systems, is defined as specifically 0.45359237 kg, making one kilogram roughly equal come 2.2046 avoirdupois pounds. Other traditional units of weight and also mass roughly the civilization are likewise defined in regards to the kilogram, do the IPK the primary standard for practically all units of mass on Earth.
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# Thread: plane tangent to 3d surface mmn14 3a
1. ## plane tangent to 3d surface mmn14 3a
i have a suface $\displaystyle \sqrt{x}+\sqrt{y}+\sqrt{z}=\sqrt{a}$
find the formula of a plane which is tangent to it in (x0,y0,z0)
i need to find the vector whic is vertical to this surface
the formula of it is $\displaystyle <f_x,f_y,-1>$
but i dont know what is the formula for f(x,y) ,
i have a formula in other form
how to write it in the f(x,y) form
?
2. If $\displaystyle S\equiv f(x,y,z)=0$ then, the tangent plane to $\displaystyle S$ at $\displaystyle P_0(x_0,y_0,z_0)$ is
$\displaystyle \dfrac{\partial f}{\partial x}(P_0)(x-x_0)+\dfrac{\partial f}{\partial y}(P_0)(y-y_0)+\dfrac{\partial f}{\partial z}(P_0)(z-z_0)=0$
3. i need to find the vector whic is vertical to this surface
the formula of it is $\displaystyle <f_x,f_y,-1>$
but i dont know what is the formula for f(x,y) ,
i have a formula in other form
how to write it in the f(x,y) form
?
4. Originally Posted by transgalactic
i need to find the vector whic is vertical to this surface
the formula of it is $\displaystyle <f_x,f_y,-1>$
but i dont know what is the formula for f(x,y) ,
i have a formula in other form
how to write it in the f(x,y) form
Sincerely, it is rather difficult to understand your question. Perhaps you mean that if the surface is written in explicit form $\displaystyle z=f(x,y)$ then, the equation of the tangent plane at $\displaystyle P_0(x_0.y_0,z_0)$ is $\displaystyle \pi\equiv <(f_x(P_0),f_y(P_0),-1),(x-x_0,y-y_0,z-z_0)>=0$ . For the general case, see my previous post.
5. Originally Posted by FernandoRevilla
If $\displaystyle S\equiv f(x,y,z)=0$ then, the tangent plane to $\displaystyle S$ at $\displaystyle P_0(x_0,y_0,z_0)$ is
$\displaystyle \dfrac{\partial f}{\partial x}(P_0)(x-x_0)+\dfrac{\partial f}{\partial y}(P_0)(y-y_0)+\dfrac{\partial f}{\partial z}(P_0)(z-z_0)=0$
i was told that the norm vector <df/dx,df/dy,-1>
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