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1. ## simple college algebra
I have gone back to school after 30 yrs and need a little help on a couple of problems. I know they are simple but for some reason I am have problems with these two. I can find the answers in the back of the text book but that does me no good. If someone could show me how to solve the it woould be greatly apprieciated. thanx. Here they are. I dont know how to show something is squared so I will just put the exp. after it.
1) 5/x2-43/x=18 ( the first x is squared)
2) 2=3/2x-1+-1/(2x-1)2 ( the 2x-1 is squared)
2. 1) $\displaystyle \frac {5}{x^2} -\frac {43}{x}=18$
2) $\displaystyle 2 =\frac {3}{2x-1} -\frac {1}{(2x-1)^2}$
For problem 1 multiply through by a factor of $\displaystyle x^2$. This gets $\displaystyle 5 - 43x=18x^2$ Then apply the quadratic formula (a=18, b=43, c=-5)to obtain your values for x. The second problem is similar yet you multiply through by a factor of $\displaystyle (2x-1)^2$, expand and then simplify.
3. Thanx for the help
4. Originally Posted by dinNA89
1) $\displaystyle \frac {5}{x^2} -\frac {43}{x}=18$
For problem 1 multiply through by a factor of $\displaystyle x^2$. This gets $\displaystyle 5 - 43x=18x^2$ Then apply the quadratic formula (a=18, b=43, c=-5)to obtain your values for x.
Or, you could factor. Just for fun.
$\displaystyle 18x^2+43x-5=0$
$\displaystyle 18x^2+45x-2x-5=0$
$\displaystyle (18x^2+45x)-1(2x+5)=0$
$\displaystyle 9x(2x+5)-1(2x+5)=0$
$\displaystyle (9x-1)(2x+5)=0$
$\displaystyle 9x-1=0 \; \;\Longrightarrow x=\frac{1}{9}$
$\displaystyle 2x+5=0 \; \; \Longrightarrow x=\frac{-5}{2}$
5. I know you said just for fun it could be factored, but at this point for me that would have been anything BUT fun. I do thank you for the help though and will more than likely be needing it again.
6. Originally Posted by whiteowl
... If someone could show me how to solve the it woould be greatly apprieciated. thanx. Here they are. I dont know how to show something is squared so I will just put the exp. after it.
...
2) 2=3/2x-1+-1/(2x-1)2 ( the 2x-1 is squared)
$\displaystyle 2=\frac3{2x-1}+\frac{-1}{(2x-1)^2}$
Use the substitution $\displaystyle y = \frac1{2x-1}$ . Your equation becomes:
$\displaystyle 2=3y-y^2~\iff~y^2-3y+2=0$
Now use the formula to solve quadratic equations or factor the LHS. You should get:
y = 1 or y = 2
Plug in these values into the substitution and solve for x:
$\displaystyle 1 = \frac1{2x-1}~\implies~x = 1$
or
$\displaystyle 2 = \frac1{2x-1}~\implies~x = \frac34$
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# RS Aggarwal Class 7 Solutions Chapter 15 Properties of Triangles Ex 15B
## RS Aggarwal Class 7 Solutions Chapter 15 Properties of Triangles Ex 15B
These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 15 Properties of Triangles Ex 15B.
Other Exercises
Question 1.
Solution:
In ∆ABC,
∠A = 75°, ∠B = 45°
side BC is produced to D
forming exterior ∠ ACD
Exterior ∠ACD = ∠A + ∠B (Exterior angle is equal to sum of its interior opposite angles)
= 75° + 45° = 120°
Question 2.
Solution:
In ∆ABC, BC is produced to D forming an exterior angle ACD
∠ B = 68°, ∠ A = x°, ∠ ACB = y° and ∠ACD = 130°
In triangle,
Exterior angles is equal to sum of its interior opposite angles
∠ACD = ∠A + ∠B
⇒ 130° = x + 68°
⇒ x = 130° – 68° = 62°
But ∠ACB + ∠ACD = 180° (Linear pair)
⇒ y + 130° = 180°
⇒ y = 180° – 130° = 50°
Hence x = 62° and y = 50°
Question 3.
Solution:
In ∆ABC, side BC is produced to D forming exterior angle ACD.
∠ACD = 65°, ∠A = 32°
∠B = x, ∠ACB = y
In a triangle, the exterior angles is equal to the sum of its interior opposite angles
∠ACD = ∠A + ∠B
⇒ 65° = 32° + x
⇒ x = 65° – 32° = 33°
But ∠ ACD + ∠ ACB = 180° (Linear pair)
⇒ 65° + y = 180°
⇒ y = 180°- 65° = 115°
x = 33° and y = 115°
Question 4.
Solution:
In ∆ABC, side BC is produced to D forming exterior ∠ ACD
∠ACD = 110°, and ∠A : ∠B = 2 : 3
In a triangle, exterior angles is equal to the sum of its interior opposite angles
⇒ ∠ACD = ∠A + ∠B
⇒ ∠A + ∠B = 110°
But ∠A : ∠B = 2 : 3
But ∠A + ∠B + ∠C = 180° (sum of angles of a triangle)
⇒ 44° + 66° + ∠C = 180°
⇒ 110° + ∠C = 180°
⇒ ∠C = 180° – 110° = 70°
Hence ∠ A = 44°, ∠ B = 66° and ∠ C = 70°
Question 5.
Solution:
In ∆ABC, side BC is produced to forming exterior angle ACD.
∠ACD = 100° and ∠A = ∠B
Exterior angle of a triangle is equal to the sum of its interior opposite angles.
∠ACD = ∠A + ∠B But ∠A = ∠B
∠A + ∠A = ∠ACD = 100°
⇒ 2 ∠A = 100°
⇒ ∠A = 50°
∠B = ∠A = 50°
But ∠A + ∠B + ∠ ACB = 180° (sum of angles of a triangle)
⇒ 50° + 50° + ∠ ACB = 180°
⇒ 100° + ∠ ACB = 180°
⇒ ∠ ACB = 180° – 100° = 80°
Hence ∠ A = 50°, ∠ B = 50° and ∠ C = 80°
Question 6.
Solution:
In ∆ABC, side BC is produced to D From D, draw a line meeting AC at E so that ∠D = 40°
∠A = 25°, ∠B = 45°
In ∆ABC,
Exterior ∠ACD = ∠A + ∠B = 25° + 45° = 70°
Again, in ∆CDE,
Exterior ∠ AED = ∠ ECD + ∠ D = ∠ACD + ∠D = 70° + 40° = 110°
Hence ∠ACD = 70° and ∠AED = 110°
Question 7.
Solution:
In ∆ABC, sides BC is produced to D and BA to E
∠CAD = 50°, ∠B = 40° and ∠ACB = 100°
∠ ACB + ∠ ACD = 180° (Linear pair)
⇒ 100° + ∠ ACD = 180°
⇒ ∠ ACD = 180° – 100° = 80°
In ∆ACD,
∠ CAD + ∠ ACD + ∠ ADC = 180° (sum of angles of a triangle)
⇒ 50° + 80° + ∠ ADC = 180°
⇒ 130° + ∠ ADC = 180°
⇒ ∠ ADC = 180° – 130° = 50°
Now, in ∆ABD, BA is produced to E
Exterior ∠DAE = ∠ACD + ∠ADC = 80° + 50° = 130°
Hence ∠ ACD = 80°, ∠ ADC = 50° and ∠DAE = 130°
Question 8.
Solution:
In ∆ABC, BC is produced to D forming exterior ∠ ACD
∠ACD = 130°, ∠A = y°, ∠B = x° and ∠ACB = z°.
x : y = 2 : 3
Now, in ∆ABC,
Exterior ∠ACD = ∠A + ∠B
But ∠A + ∠B + ∠ACB = 180° (sum of angles of a triangle)
⇒ 78° + 52° + ∠ACB = 180°
⇒ 130° + ∠ACB = 180°
⇒ ∠ACB = 180° – 130°
⇒ ∠ACB = 50°
⇒ ∠ = 50°
Hence x = 52°, y = 78° and z = 50°
Hope given RS Aggarwal Solutions Class 7 Chapter 15 Properties of Triangles Ex 15B are helpful to complete your math homework.
If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.
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Joined: 13 Jan 2011
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24 Jan 2011, 08:37
Gayathari's SC notes page 6 indicate the following:
Quote:
When in doubt pick the more/most form.
Incorrect: Could you drive slower?
Correct: Could you please drive more slowly?
I don't understand why example 1 is incorrect. Can someone explain please? I would really appreciate it.
Enkie
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24 Jan 2011, 11:26
enkie wrote:
Gayathari's SC notes page 6 indicate the following:
Quote:
When in doubt pick the more/most form.
Incorrect: Could you drive slower?
Correct: Could you please drive more slowly?
I don't understand why example 1 is incorrect. Can someone explain please? I would really appreciate it.
Enkie
Hi,
I'm not sure what the source of the general advice is, but it's not necessarily correct.
Both forms are grammatically correct, so it comes down to style - i.e. what sounds best. The style difference between the two is small, so it's unlikely that you'll be forced to choose between two answers solely on that basis. If the original advice giver has some OG questions to back up the rule, I'd love to see them.
In fact, in some cases the advice is clearly incorrect. When we normally speak/write, we're far more likely to say:
1) Out of all my friends, Bob is the smartest;
than
2) Out of all of my friends, Bob is the most smart,
although the advice tells us to go with door number (2).
Similarly, we would say:
1) Bob is smarter than Fred;
2) Bob is more smart than Fred.
So, it really comes down to a case-by-case analysis of which version sounds more natural.
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# Showing that $a^5$ and $a$ have the same last digit using Euler's Theorem.
To show that, in base 10, $a^5$ and $a$ have the same final digit it is sufficient to prove that $$a^5 \equiv a \bmod 10$$
Provided that $a$ and $10$ are coprime then Euler's Theorem proves the result: $$a^{\phi(10)} \equiv 1 \bmod 10$$ where $\phi(10) = \phi(2\cdot 5) = \phi(2)\cdot \phi(5) = 1\cdot 4 = 4$, and hence $$a^4 \equiv 1 \bmod 10 \implies a^5 \equiv a \bmod 10$$
The problem I have is: What if $a$ an $10$ are not coprime?
I have proven the result by going through all of the remainders one-by-one, e.g. if $a \equiv 5 \bmod 10$ then $a^5 \equiv 3125 \bmod 10 \equiv 5 \bmod 10$, etc.
I would like a more general proof, using Euler's Theorem if possible. Checking the remainders isn't practical for larger numbers, e.g. $\bmod 1000$.
• Have you heard about the Chinese remainder theorem? Aug 29 '16 at 16:14
• @Arthur Yes: $\mathbb Z/10\mathbb Z \cong \mathbb Z/2\mathbb Z \times \mathbb Z/5\mathbb Z$ where $[a]_{10} \longmapsto ([a]_2,[a]_5)$. Aug 29 '16 at 16:15
• Another solution is done using induction. Aug 29 '16 at 16:17
It's special case $\ p^i = 2,\ q^j = 5,\ \phi = 4,\ k=1\$ of the following generalization of Euler $\!\rm\color{blue}{(E)}$.
Theorem $\ \ n^{\large k+\phi}\equiv n^{\large k}\pmod{p^i q^j}\ \$ assuming that $\ \color{#0a0}{\phi(p^i),\phi(q^j)\mid \phi},\,$ $\, i,j \le k,\,\ p\ne q.\ \ \$
Proof $\ \ p\nmid n\,\Rightarrow\, {\rm mod\ }p^i\!:\ n^{ \phi}\equiv 1\,\Rightarrow\, n^{k + \phi}\equiv n^k,\,$ by $\ n^{\Large \color{#0a0}\phi} = (n^{\color{#0a0}{\Large \phi(p^{ i})}})^{\large \color{#0a0}m}\overset{\color{blue}{\rm (E)}}\equiv 1^{\large m}\equiv 1.$
$\qquad\quad\ \color{#c00}{p\mid n}\,\Rightarrow\, {\rm mod\ }p^i\!:\ n^k\equiv 0\,\equiv\, n^{k + \phi}\$ by $\ n^k = n^{k-i} \color{#c00}n^i = n^{k-i} (\color{#c00}{p\ell})^i$ and $\,k\ge i.$
So $\ p^i\mid n^{k+\phi}\!-n^k.\,$ By symmetry $\,q^j$ divides it too, so their lcm $= p^iq^j\,$ divides it too. $\$ QED
Remark $\$ Obviously the proof immediately extends to an arbitrary number of primes. This leads the way to Carmichael's Lambda function, a generalization of Euler's phi function.
• the coloring is cool. Aug 29 '16 at 16:24
What if $a$ and $10$ are not co-prime. then $(2|a)$ or $(5|a)$ or $(10|a).$
$2$ divides $a$ and $10$ does not divide $a.$
Then we have these $4$ elements $\{2,4,6,8\}$
And these form a group. It is cyclical of order $4.$
$a^4\equiv e \pmod{10}\\ a^5\equiv a\pmod{10}$
The identity element of this group turns out to be $6.$
$5$ divides $a$ and $10$ does not divide $a.$
$\forall n, a^n \equiv 5 \pmod {10}$
$10$ divides $a$
$\forall n, a^n \equiv 0 \pmod {10}$
More generally, for any modulo, the integers will fall into to cyclical groups, and the order of those groups will divide $\phi$
• I don't see that in what you have posted. So, then, what is your question? Aug 29 '16 at 16:26
• So in the more general case, any integer falls into a cyclic group who's order divides $\phi.$ Another way to look at it is to look at the cosets of $z \cdot G$ where $G$ is the group formed by the integers co-prime to $G.$ and $z$ is any integer, and what groups these cosets form. Aug 29 '16 at 16:35
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10prob_10
# 10prob_10 - ENGG1007 Foundations of Computer Science...
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1 ENGG1007 Foundations of Computer Science Probability Probability Professor Francis Chin, Dr SM Yiu Text book - Chapter 6
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2 ENGG1007 FCS An expensive sport car is hiding behind one of the doors 1. Player chooses a door. 1. MC then opens one of the other two doors which does not have the car behind 1. Player will be asked if he wants to switch his choice? If the player changes his choice, will he get a better chance to get the sport car? What do you think?
3 ENGG1007 FCS Probability Probability Sample space (S) – a set of possible outcomes; e.g. dice = {1, 2, 3, 4, 5, 6} Event (E) – a subset of sample space; e.g., small numbers = {1, 2, 3} If all outcomes are equally likely , probability of an event E , P ( E ) = | E | / | S | Sample space S = { ♠, ♥, ♦ , ♣ } x {A, 2, 3, 4, 5, 6 7, 8, 9, T, J, Q, K} Event = { 9 }; P ( { 9} ) = | { 9} | / |S| = 1 / 52 Event = a card of heart = { A, 1, 2, 3, 4, 5, 6, 7, 8, 9, T, J, Q, K } = P ( a card of heart ) = 13 / 52 Event = a red card ; P ( a red card ) = 26 / 52
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4 ENGG1007 FCS Probability Probability Example: Probability when the sum of 2 rolled fair dice is 6 Solution1: | S | = 6 2 = 36 outcomes E = {(1,5 ) , (2,4), (3,3), (4,2), (5,1)} P ( E ) = | E | / | S | = 5/36 Solution2: S be the set of 2-combinations with repetition, | S | = C (6+2-1,2) = 21 E = {{1,5}, {2,4}, {3,3}} So P ( E ) = | E | / | S | = 3/21 = 1/7 P ( E ) = | E | / | S | with the assumption that all outcomes are equally likely. Solution 1 is correct because all its outcomes are equally likely
5 ENGG1007 FCS Probability on Cards Probability on Cards Example 1: Probability (a 5-card hand without a spade ace) | S | = C(52,5) = 52x51x50x49x48 / 5! | E 1 | = C(51,5) = 51x50x49x48x47 / 5! P ( E 1 ) = | E 1 | / | S | = 47 / 52 Why counting the combinations work for this example? Are all “combinations” equally probable? “Yes” - because each of the combinations can be obtained in equal number of ways = 5! ways .
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6 ENGG1007 FCS Probability (a 5-card hand without a spade ace) Another approach (counting the number of permutations using the rule of product) # of permutations of a 5-card hand without a ♠A = 51 x 50 x 49 x 48 x 47 Total number of permutations of a 5-card hand = 52 x 51 x 50 x 49 x 48 Since all possible permutations are equally likely, the probability that all 5 cards are not ♠A = (51 x 50 x 49 x 48 x 47)/ (52 x 51 x 50 x 49 x 48) = 47 / 52
7 ENGG1007 FCS Probability on cards Probability on cards Ex 2: Probability (a 5-card hand without any ace) | E 2 | = C(48,5) = 48x47x46x45x44 / 5! | S | = C(52,5) = 52x51x50x49x48 / 5! P ( E 2 ) = | E 2 | / | S | = C(48,5) / C(52,5) = (48/52) (47/51) (46/50) (45/49) (44/48) = 0.658842 Ex 3: Probability (a 5-card hand with two pairs) | E 3 | = C(13,2) C(4,2) C(4,2) C(44,1) Ex 4: Probability (a 5-card hand with one pair) | E 4 | = C(13,1) C(4,2) C(12,3) C(4,1) C(4,1) C(4,1)
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8 ENGG1007 FCS Ex 5: Probability (a 5-card hand at least one ace)
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10prob_10 - ENGG1007 Foundations of Computer Science...
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# Math
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The sides of parallelogram 2x+y,3y-2x,5y-8,4x-3 are given in cm.find x and y and the perimeter of the parallelogram
• Math -
I assume you have listed the sides in order going around the parallogram.
Opposite sides are equal, so ...
2x+y = 5y-8 ----> 2x-4y = -8
3y-2x = 4x-3 ---> 6x - 3y = 3
divide the 2nd by 3
2x - y = 1
2x-y = 1
2x-4y = -8
subtract them
3y = 9
y = 3
then 2x-3=1
x = 2
for perimeter, evaluate each of the sides using the values for x and y, then add up the length of the sides.
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# Triangle Congruence Proof Worksheet | Education Template With Triangle Congruence Proofs Worksheet
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Triangle Congruence Proofs Worksheet : Triangle Congruence Proof Worksheet | Education Template With Triangle Congruence Proofs Worksheet
Triangle Congruence Proofs Worksheet. All we have to do is to plug in our number of sides into the equation. Write an equation for the response that happened. All you need is easy internet connection and a tool to work on. If you didn’t get right here by way of My EP Assignments, I suggest you go there and create an account.
Read the lesson and through the examples. Reflexive property is of various varieties such as the reflexive property of equality, reflexive property of congruence, and reflexive property of relations. A binary relation R defined on a set A is said to be reflexive if, for each factor a ∈ A, we now have aRa, that is, ∈ R. In different words, we can say that a relation outlined on a set is a reflexive relation if and only if every element of the set is said to itself.
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# Thread: Factors Involving Negative Exponents
1. ## Factors Involving Negative Exponents
EDIT: Yeah, I posted this in the urgent forum. Really sorry mods! Please delete!
So I've been able to do all my Precalc homework using prior knowledge with Algebra 2 until I came across this here:
If someone could hit up number 59 with a detailed procedure (The instruction is to simplify the expression,) I'd appreciate it SO much. Many apologies for the bluriness, if you can't make out what the exponents are, this is the problem typed out:
-2(x^2-3)^-3(2x)(x+1)^3)-3(x+1)^2(x^2-3)^-2/ [(x+1)^3]^2
2. Would you be able to do this type of question with only positive powers? (Just getting an idea of how much of an explaination you need).
To approach this problem, you want to expand each term then simplify.
First step is to expand each set of brackets. You'll need to be very good at expanding (a+b)^3 to do this. Can you do this ok?
Then multiply by the coefficient. eg -2 for the first set of brackets.
Then simplify by adding like terms.
Then see what can be cancelled out via the denominator.
A negaitve power is the same as 1/ the positive power.
eg. (3x+1)^-2 = 1/(3x+1)^+2
There are often shortcuts to this type of problem which you'll become more familiar with, but for now it's probably best to expand everything then simplify.
Let me know if this makes sense and when I have a few minutes I'll write out a worked answer for you (I'm having some issues with my own work at the moment).
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HW6sol
# HW6sol - Stat 512 1 Solutions to Homework#6 Dr Levine 1 In...
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Stat 512 – 1 Solutions to Homework #6 Dr. Levine 1. In this exercise you will illustrate some of the ideas described in Chapter 7 of the text related to the extra sums of squares. (a) Create a new variable called SAT which equals SATM + SATV and run the following two regressions: (i) predict GPA using HSM , HSS , and HSE ; Analysis of Variance Sum of Mean Source DF Squares Square F Value Pr > F Model 3 27.71233 9.23744 18.86 <.0001 Error 220 107.75046 0.48977 Corrected Total 223 135.46279 (ii) predict GPA using SAT , HSM , HSS and HSE . Analysis of Variance Sum of Mean Source DF Squares Square F Value Pr > F Model 4 27.88746 6.97187 14.19 <.0001 Error 219 107.57533 0.49121 Corrected Total 223 135.46279 Calculate the extra sum of squares for the comparison of these two analyses. Use it to construct the F statistic (i.e., general linear test statistic) for testing the null hypothesis that the coefficient of the SAT variable is zero in the model with all four predictors. What are the degrees of freedom for this test statistic? Using the SSE definition, ( ) ( ) ( ) ( ) | , , , , , , , 107.75046 107.57533 0.17513 SSM sat hsm hss hse SSE hsm hss hse SSE sat hsm hss hse = = = or, using the SSM definition, ( ) ( ) ( ) ( ) | , , , , , , , 27.88746 27.71223 0.17513 SSM sat hsm hss hse SSM sat hsm hss hse SSM hsm hss hse = = = The numerator df is (220 – 219) = (4 – 3) = 1. ( ) ( ) ( ) | , , / 1 0.17513/1 0.35653 0.49121 SSM sat hsm hss hse F MSE full = = = , df = (1,219)
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The calculator performs straightforward and advanced operations through fractions, expressions through fractions an unified with integers, decimals, and mixed numbers. It also shows comprehensive step-by-step information about the fraction calculation procedure. Solve difficulties with two, three, or more fractions and numbers in one expression.
You are watching: 1/4 divided by 6
Divide: 6 : 1/4 = 6/1 · 4/1 = 6 · 4/1 · 1 = 24/1 = 24 splitting two fountain is the very same as multiply the very first fraction by the reciprocal worth of the 2nd fraction. The first sub-step is to find the reciprocal (reverse the numerator and also denominator, mutual of 1/4 is 4/1) the the 2nd fraction. Next, multiply the 2 numerators. Then, multiply the two denominators.In other words - six separated by one 4 minutes 1 = twenty-four.
Rules for expressions through fractions: Fractions - merely use a forward slash between the numerator and denominator, i.e., because that five-hundredths, go into 5/100. If you space using mixed numbers, be sure to leave a solitary space in between the whole and portion part.The slash separates the numerator (number over a portion line) and also denominator (number below).Mixed numerals (mixed fountain or combined numbers) compose as creature separated by one room and fraction i.e., 12/3 (having the very same sign). An example of a an adverse mixed fraction: -5 1/2.Because cut is both indications for fraction line and division, us recommended use colon (:) together the operator of division fractions i.e., 1/2 : 3.Decimals (decimal numbers) get in with a decimal suggest . and also they are automatically converted to fountain - i.e. 1.45.The colon : and slash / is the symbol of division. Have the right to be offered to divide combined numbers 12/3 : 43/8 or can be offered for write complex fractions i.e. 1/2 : 1/3.An asterisk * or × is the symbol for multiplication.Plus + is addition, minus sign - is subtraction and ()<> is math parentheses.The exponentiation/power symbol is ^ - because that example: (7/8-4/5)^2 = (7/8-4/5)2
Examples: • adding fractions: 2/4 + 3/4• subtracting fractions: 2/3 - 1/2• multiplying fractions: 7/8 * 3/9• dividing Fractions: 1/2 : 3/4• exponentiation of fraction: 3/5^3• spring exponents: 16 ^ 1/2• adding fractions and also mixed numbers: 8/5 + 6 2/7• splitting integer and fraction: 5 ÷ 1/2• complicated fractions: 5/8 : 2 2/3• decimal to fraction: 0.625• fraction to Decimal: 1/4• fraction to Percent: 1/8 %• to compare fractions: 1/4 2/3• multiplying a fraction by a totality number: 6 * 3/4• square source of a fraction: sqrt(1/16)• reducing or simple the fraction (simplification) - separating the numerator and denominator the a portion by the very same non-zero number - tantamount fraction: 4/22• expression through brackets: 1/3 * (1/2 - 3 3/8)• link fraction: 3/4 of 5/7• fountain multiple: 2/3 of 3/5• division to uncover the quotient: 3/5 ÷ 2/3The calculator follows popular rules for order that operations. The most typical mnemonics because that remembering this bespeak of operations are: PEMDAS - Parentheses, Exponents, Multiplication, Division, Addition, Subtraction. BEDMAS - Brackets, Exponents, Division, Multiplication, Addition, Subtraction BODMAS - Brackets, the or Order, Division, Multiplication, Addition, Subtraction.
See more: When Might The Inter-Quartile Range Be Better For Describing A Data Set Than The Range?
GEMDAS - Grouping signs - brackets (), Exponents, Multiplication, Division, Addition, Subtraction. be careful, constantly do multiplication and also division prior to addition and subtraction. Some operators (+ and -) and also (* and /) has the very same priority and also then need to evaluate from left come right.
## Fractions in indigenous problems:
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Difference Between Algorithm, Pseudocode, and Program
Algorithm Vs. Pseudocode Vs. Program: Find the Difference Between Algorithm, Pseudocode, and Program
All three of these are procedures used in a computer system, but there is a significant difference between algorithm, pseudocode, and program. In this article, we will discuss the same. But let us first know a bit more about each of these.
Algorithm – It is a well-defined, systematic logical approach that comes with a step-by-step procedure for computers to solve any given program.
Program – It refers to the code (written by programmers) for any program that follows the basic rules of the concerned programming language.
Pseudocode – A pseudocode is basically a simplified version of the programming codes. These codes exist in the plain English language, and it makes use of various short phrases for writing a program code before implementing it in any programming language.
What is an Algorithm?
We use an algorithm to generate a solution to any given problem in the form of steps. When we use a computer for solving any given problem (in well-defined multiple steps), we need to communicate these steps to the solution properly to the computer.
When we execute any algorithm on a computer, we need to combine various operations like subtraction and addition for performing various complex operations of mathematics. We can express algorithms using flowcharts, natural language, and many more.
What is a Pseudocode?
Pseudocode refers to a method using which we represent any algorithm for any program. Pseudocode does not consist of any specific syntax like a programming language. Thus, one cannot execute it on a computer. We can use several formats for writing pseudocode. A majority of them take down the structures from the available languages, like FORTRAN, Lisp, C, etc.
Various times, we can present an algorithm using the pseudocode because any programmer can easily read as well as understand them (those who are familiar with multiple programming languages). You get to include various control structures using pseudocode, such as repeat-until, if-then-else, while, for and case. It is present in various languages that are high-level.
Note– A pseudocode is not equivalent to a real programming language.
What is a Program?
It refers to a set of various instructions that a computer follows. No machine is actually capable of reading a program directly. It is because it only understands the available machine code. Instead, you can write anything in a computer language and then use a compiler or interpreter (for compiling and interpreting) so that it becomes understandable for any computer system.
Difference Between Algorithm, Pseudocode, and Program
Here is a list of the differences between Algorithm, Pseudocode, and Program.
Program
Meaning and Definition An algorithm is a well-defined, systematic logical approach that comes with a step-by-step procedure for computers to solve any given program. It refers to the code (written by programmers) for any program that follows the basic rules of the concerned programming language. It is basically a simplified version of the programming codes. These codes exist in the plain English language, and it makes use of various short phrases for writing a program code before implementing it in any programming language.
Expression and Use We can express algorithms using flowcharts, natural language, and many more. You get to include various control structures using pseudocode, such as repeat-until, if-then-else, while, for and case. No device can directly read a program. Instead, you can write anything in a computer language and then use a compiler or interpreter (for compiling and interpreting) so that it becomes understandable for any computer system.
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How do you solve using the completing the square method x^2-2x+2?
Apr 27, 2016
The solutions are:
color(green)(x = sqrt (-1) + 1 or, color(green)(x = -sqrt (-1) + 1
Explanation:
${x}^{2} - 2 x + 2$
${x}^{2} - 2 x = - 2$
To write the Left Hand Side as a Perfect Square, we add 1 to both sides:
${x}^{2} - 2 x + 1 = - 2 + 1$
${x}^{2} - 2 \cdot 1 \cdot x + {1}^{2} = - 1$
Using the Identity color(blue)((a-b)^2 = a^2 - 2ab + b^2, we get
${\left(x - 1\right)}^{2} = - 1$
$x - 1 = \sqrt{- 1}$ or $x - 1 = - \sqrt{- 1}$
color(green)(x = sqrt (-1) + 1 or color(green)(x = -sqrt (-1) + 1
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# Coefficient of Friction
Topics: Friction, Force, Mass Pages: 4 (1076 words) Published: November 16, 2012
Coefficient of Friction Lab
Abstract
The resisted force that acted along the tangent of two surfaces that were in contact was called friction. Friction was opposed to motion, and it acted in the opposite direction, where the surface of the object slid against the surface of the other object. The two types of friction that exist were called static friction and kinetic friction. When two surfaces are at rest with each other, but a push is caused to convey one of the surfaces to slide over the other was called static friction. However, the friction that was used in the lab was kinetic friction. Kinetic friction occurred when two surface were moving with contact to each other. The coefficient of kinetic friction is a constant shown as μk. The kinetic frictional force (fk) was given by the following equation: (fk= μkN), where N represented the normal force, which was the force that each body exerts on the other body, and acts perpendicular to each surface. The way that friction force is calculated is by the followed calculation: Ff=μFN, where (μ) was the coefficient of friction and (FN) was the normal force. Now in order to pinpoint the force of friction, the coefficient of friction should be figured out first. Now the way that the coefficient of friction was retrieved, the tension force (FT) was divided by the weight (Fg). An inclined plane that has an angle of θ was adjusted as shown in the following diagram:
2
If the block was placed on the plane, and the angle was slowly increased, the block would have began to slip at some angle. Now the normal force (N) acted perpendicularly to the plane, and a component of the weight of the block, acted in the opposite direction. Though when the angle is increased, the more force it took for the block to slide against another surface. So as the angle was increased, the friction cultivates, but when the angles decreased, so did the friction.1 Now some of the factors that can affect friction are the body surfaces....
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# Is 108 An Even Number?
• Yes, number 108 is an even number.
• One hundred and eight is an even number, because it can be divided by 2 without leaving a comma spot.
## Is 108 Odd Or Even?
• Number 108 is an even number.
## How To Calculate Even Numbers
• To find the set of even natural numbers, we use 2N where N is any natural number. Any number multiplied with an even number will result in an even number. For example: 0N, 2N, 4N, 6N, ... (where N is any natural number) the result is always an even number.
• The oddness of a number is called its parity, so an odd number has parity 1, while an even number has parity 0.
## Mathematical Information About Numbers 1 0 8
• About Number 1. The number 1 is not a prime number, but a divider for every natural number. It is often taken as the smallest natural number (however, some authors include the natural numbers from zero). Your prime factorization is the empty product with 0 factors, which is defined as having a value of 1. The one is often referred to as one of the five most important constants of analysis (besides 0, p, e, and i). Number one is also used in other meanings in mathematics, such as a neutral element for multiplication in a ring, called the identity element. In these systems, other rules can apply, so does 1 + 1 different meanings and can give different results. With 1 are in linear algebra and vectors and one Einsmatrizen whose elements are all equal to the identity element, and refers to the identity map.
• About Number 0. The number zero is the number of elements in an empty collection of objects, mathematically speaking, the cardinality of the empty set. Zero in mathematics by depending on the context variously defined objects, but often can be identified with each other, that is considered to be the same object, which combines several properties compatible with each other. As cardinal numbers (number of elements in a set) are identified with special ordinals, and the zero is just the smallest cardinal number is zero - elected as the first ordinal - in contrast to common parlance. As finite cardinal and ordinal it is depending on the definition often counted among the natural numbers. The zero is the identity element for addition in many bodies, such as the rational numbers, real numbers and complex numbers, and a common name for a neutral element in many algebraic structures, even if other elements are not identified with common numbers. Zero is the only real number that is neither positive nor negative.
• About Number 8. The octahedron is one of the five platonic bodies. A polygon with eight sides is an octagon. In computer technology we use a number system on the basis of eight, the octal system. Eight is the first real cubic number, if one disregards 1 cube. It is also the smallest composed of three prime number. Every odd number greater than one, raised to the square, resulting in a multiple of eight with a remainder of one. The Eight is the smallest Leyland number.
## What Is An Even Number?
An even number is an integer which is evenly divisible by two. This means that if the integer is divided by 2, it yields no remainder. Zero is an even number because zero divided by two equals zero. Even numbers can be either positive or negative. You can tell if any decimal number is an even number if its final digit is an even number. An integer that is not an even number is an odd number.
Parity is a mathematical term that describes the property of an integer's inclusion in one of two categories: even or odd. An integer is even if it is evenly divisible by two and odd if it is not even. For example, 6 is even because there is no remainder when dividing it by 2. By contrast, 3, 5, 7, 21 leave a remainder of 1 when divided by 2. A formal definition of an even number is that it is an integer of the form n = 2k, where k is an integer. It can then be shown that an odd number is an integer of the form n = 2k + 1.
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## Conversion formula
The conversion factor from ounces to kilograms is 0.028349523125, which means that 1 ounce is equal to 0.028349523125 kilograms:
1 oz = 0.028349523125 kg
To convert 460.6 ounces into kilograms we have to multiply 460.6 by the conversion factor in order to get the mass amount from ounces to kilograms. We can also form a simple proportion to calculate the result:
1 oz → 0.028349523125 kg
460.6 oz → M(kg)
Solve the above proportion to obtain the mass M in kilograms:
M(kg) = 460.6 oz × 0.028349523125 kg
M(kg) = 13.057790351375 kg
The final result is:
460.6 oz → 13.057790351375 kg
We conclude that 460.6 ounces is equivalent to 13.057790351375 kilograms:
460.6 ounces = 13.057790351375 kilograms
## Alternative conversion
We can also convert by utilizing the inverse value of the conversion factor. In this case 1 kilogram is equal to 0.07658263558311 × 460.6 ounces.
Another way is saying that 460.6 ounces is equal to 1 ÷ 0.07658263558311 kilograms.
## Approximate result
For practical purposes we can round our final result to an approximate numerical value. We can say that four hundred sixty point six ounces is approximately thirteen point zero five eight kilograms:
460.6 oz ≅ 13.058 kg
An alternative is also that one kilogram is approximately zero point zero seven seven times four hundred sixty point six ounces.
## Conversion table
### ounces to kilograms chart
For quick reference purposes, below is the conversion table you can use to convert from ounces to kilograms
ounces (oz) kilograms (kg)
461.6 ounces 13.086 kilograms
462.6 ounces 13.114 kilograms
463.6 ounces 13.143 kilograms
464.6 ounces 13.171 kilograms
465.6 ounces 13.2 kilograms
466.6 ounces 13.228 kilograms
467.6 ounces 13.256 kilograms
468.6 ounces 13.285 kilograms
469.6 ounces 13.313 kilograms
470.6 ounces 13.341 kilograms
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# 4th Grade Math Word Problems Area And Perimeter
Laurel Street kids tend to do better on math because it’s a kind of transitional language for students still learning to read and speak English fluently, said fourth-grade math teacher. emphasizes.
Students’ lack of fractions understanding has been cited as second only to word-problem difficulty. as the common-core math standards push more of the early work on fractions into 3rd grade, as.
“Let’s go over our math learning target for today,” said Christina Spontaneo, Love Elementary School teacher, to her fourth-grade class. They are using perimeter, but the word is never stated in.
My students will probably forget how to find the surface area of a pyramid two weeks after I teach it, but working in teams to solve complex problems is a skill they. Raymond Steinmetz teaches.
Fourth Grade Math Worksheets Activity Workbook – No Prep and Not Boring Fourth Grade Math Book New Fourth Grade Math Workbook Created Each book contains word problems, mixed math, and other fourth grade review materials. Use it each week for daily practice in your classroom. Area and Perimeter
Think of this new iron from Titleist as the answer to a first-grade math problem: 1 + 2 = 3. The all-new 718 AP3 long and mid-irons (3-7) were designed with a cast 17-4 stainless steel body. the.
The stress impedes their working memory—the area of. as early as 1st grade, 50 math problems to solve in three minutes. For many students, it is not an exaggeration to describe this experience as.
(Beaudin and Bowers, 1997 and extended by Zbiek and Hollebrands, 2008): Creative Commons — Attribution 4.0 International. word problem that addressed math standards at that grade level.
Some of the youngest students are learning a foreign language in Baltimore area schools. in the fourth grade, students are immersed in the Spanish language several times each week. The format this.
CCSS.Math.Content.4.MD.B.3 – Apply the area and perimeter formulas for rectangles in real world and mathematical problems. For example, find the width of a rectangular room given the area of the flooring and the length, by viewing the area formula as a multiplication equation with an unknown factor.
In one fourth. to the area in their classroom surrounded by yellow caution tape. Hiring and maintaining an adequate teaching staff also poses a problem for the district, according to the suit. Last.
Welcome to the Math Salamanders Area worksheets page. Here you will find a range of free printable area sheets, which will help your child to learn to work out the areas of a range of rectangles and rectilinear shapes. These sheets are aimed at children who are 3rd or 4th grade level.
Each worksheet has 10 problems solving word problems involving area and perimeter. Create New Sheet One atta Time Flash Cards Share Select a Worksheet Version 1 Version 2 Version 3 Version 4 Version 5 Version 6 Version 7 Version 8 Version 9 Version 10.
Apply the area and perimeter formulas for rectangles in real world and mathematical problems.For example, find the width of a rectangular room given the area of the flooring and the length, by viewing the area formula as a multiplication equation with an unknown factor.
In math, students will be required to solve problems using. what does the word terrain mean as it is used in the passage? "In his travels, he learned a great deal about the terrain and wildlife of.
Fixes looks at solutions to social problems and why they work. In response to Tuesday’s column about Jump Math, an approach to math education. Take for example a fifth-grade lesson on perimeter.
Area and Perimeter Math Problems and Solutions. In third grade, students typically learn to measure and calculate the area and perimeter of various shapes. Because it can be hard to remember the difference between these mathematical concepts, your child may need additional help at home.
The common problem they address is one. the alphabet with foam letters and a girl in fourth grade reading a song about diversity in India. At another table, a boy in first grade helped a girl learn.
Anna Grace Wise demonstrated drawing hand turkeys before stopping a 4-year-old from drinking directly from the sink. At Lawhon Elementary, Analiese Kent talked third-grade math students through word.
Focusing on finding the perimeter of squares, the worksheets here provide adequate practice in finding the perimeter of squares with integer, decimal and fraction dimensions, learn to find the diagonal and the side length using the perimeter and much more. Assess conceptual knowledge with the word problems. Perimeter of Rectangles Worksheets
Teachers’ math phobia, which faculties of education across North America view as a “huge problem,” are seen as one factor in Ontario’s falling student math scores, especially in grade. with the.
Common Core For Grade 3 New York State Common Core Math Module 7, Grade 3, Lesson 15 Application Problem Clara and Pedro each use four 3-inch by 5-inch cards to make the rectangles below. Whose rectangle has a greater perimeter? Concept Development Problem 1: Solve perimeter word problems with rectangles.
Area and Perimeter Word Problems. It is so important for students understand how to solve word problems. Sometimes it can be be the simple things that confuse students on how to solve a problem.
Seven-year-olds may be able to complete problems with answers in the teens. During their sixth year, some children will begin to solve basic word problems. one of four is one fourth, and one of.
Depth of understanding was hailed by its architects as a cornerstone of the Common Core, a set of educational guidelines for what students need to know in each grade in English and math. the area.
Astronomers Working On The Size Of The Universe Welcome. Please log in to Minneapolis Community & Technical College’s D2L Brightspace to view your courses. Assembling data gathered by eight radio telescopes around the world, astronomers created the picture showing the violent. Astronomers have discovered 83 supermassive black holes birthed by the universe in its infancy. But the Matsuoka team’s work, published Feb. 6
Apr 04, 2016 · 3_M_2_F: Partition a rectangle into rows and columns of the same-size squares and measure area by counting the unit squares (square cm, square m, square in,
In a Stanford University study, students who scored higher on an assessment of positive mindset have more brain activity throughout several areas associated with math problem-solving, as well as more.
Isotope 217 The Unstable Molecule Figure 3: SEM images in the SE mode of structures formed in sulfide gradient tubes under variable concentrations of dissolved organics. Scanning electron microscopy (SEM) imaging in the backscattered. In physical organic chemistry, a kinetic isotope effect (KIE) is the change in the reaction rate of a chemical reaction when one of the atoms in
Nov 14, 2017 · free 3rd grade math worksheets area and perimeter area & perimeter from an awesome mon core worksheet site this 58 best area and perimeter images on pinterest area & perimeter worksheets school pinterest 60 best area and perimeter images on pinterest 4th grade word problems worksheets perimeter worksheet 2 rounding free 3rd grade math worksheets area and perimeter area and perimeter word.
In one fourth. to the area in their classroom surrounded by yellow caution tape. Hiring and maintaining an adequate teaching staff also poses a problem for the district, according to the suit. Last.
Chemistry Extra Credit Ideas Credit: Courtesy of Science/AAAS Mercury is not just hellishly. told SPACE.com. Altogether, this surface chemistry suggests the planet formed from material now seen in certain stony chondritic. EXTRA CREDIT (29) Lots of homework (28) Respected (19) Skip class?. Difficult course w/ no chem background-she makes sure to show ideas in different. In addition, he’s layering
The next period I headed to the library commons area where some students were working. It has trouble with word problems, but if you can write down a word problem in math notation it shouldn’t be.
Directions: Read each question below. Click once in an ANSWER BOX and type in your answer; then click ENTER. Your answers should be given as whole numbers greater than zero. After you click ENTER, a message will appear in the RESULTS BOX to indicate whether your answer is correct or incorrect. To start over, click CLEAR.
Math Worksheets for Area and Perimeter of Rectangles. Basic Geometry: Area and Perimeter of Rectangles. The problems on these geometry worksheets require students to calculate the area and perimeter of rectangles given their dimensions.
5th Grade > CCSS 5th Grade Math > Word problems. Word problems – area and perimeter. Word problems – area and perimeter 5.NF.B.5a – Comparing the size of a product to the size of one factor on the basis of the size of the other factor, without performing the indicated multiplication. To link to this.
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# ln make me feel stu...
• Mar 1st 2008, 04:37 PM
eniav
ln make me feel stu...
Find the derivative:
y = 5^(7x)
• Mar 1st 2008, 04:40 PM
bobak
$\displaystyle y = 5^{7x}$
$\displaystyle \ln y = 7x \ln 5$
$\displaystyle \frac{1}{y} \frac{dy}{dx} = 7 \ln 5$
$\displaystyle \frac{dy}{dx} = 7 \ln 5 \ y$
$\displaystyle \frac{dy}{dx} = 7 \ln 5 \ 5^{7x}$
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It so happens that our nation is last among 1st world countries in math and science. Apparently we do quite well until about 4th grade and then we start to loose the edge by repeating the same stuff. While math is definitely incremental (meaning that it builds upon previous lessons), a word to the wise: take short times to review basic concepts daily. You don’t have to do it via rote memorization–play games, do a little here and a little there and before you know it you’ll have an expert mathematician.
* Counting, reading, writing to 1,000
Cool Teaching Idea: You can of course, practice reading and writing the numbers all the way up to 1,000. But why not add a little fun to it? Consider taking a year to collect 1,000 of something. Yes, I’m serious. It doesn’t have to be a big something, just something. Every time you add to your collection have your child label it with the number and the word for the number. We have done this twice so far: the first time we collected 1,000 photographs (due to my daughter’s budding interest in photography. The second time, we sent a Flat Traveler 1,000 miles. A well written letter to friends and family explaining your purpose and your goal will land you with 1,000 of something before you know it!
* Counting by 2’s, 3’s, 4’s, 5’s, and 10’s
Teaching Tip: I learned to count by 2’s and 5’s by playing jump rope. Our math teacher would take us outside and every time we jumped. . .we had to count. Unfortunately, my daughter hates to jump rope so we play catch instead. In any case, the rhythm while practicing makes for effortless memorization.
* Ordinal numbers to 10
* Zero as a place holder
* Using sets and number facts
Teaching Tip: This is where you teach your child the idea that 2+3 and 3+2 equal the same thing. The more you work on number facts, the better your child will grasp this concept. Also, using manipulative helps drive the idea home.
* Addition and subtraction facts to 20
* Basic multiplication and division facts
* Multiplication and division facts as inverse operations
* Multiplication properties of 0 and 1
Cool websites: Here http://cte.jhu.edu/techacademy/web/2000/heal/siteslist.htm is a fairly comprehensive list of websites on the internet that have math games to practice all these new facts.
* Place value through 100’s
* Fractions in daily life
* Decimal numeration system
* Basic concept of ratio
* Geometry puzzles and activities
* Estimation
* Common customary and metric measures of time, weight, length, volume, shape, temperature
* Telling time and using the calendar
Teaching Idea: The best way to teach these concepts and really have them stick is to put kids “in charge” of their own schedule. Give them a calendar to keep track of classes on, and consider giving them a schedule to keep. For example, our oldest daughter has a list to accomplish before 9am.
* Handling money (coins)
Teaching Idea: Again, the best way to practice handling money is to handle money! Let your kids pay and then help them count the change to make sure it was correct.
* Problem solving
* Charts and graphs
Related Articles:
What Your Kindergartner Should Know: Math
Cuisenaire Rods
Learning Wrap Ups
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# Fun with Fundamentals: Problem 219
Sept. 1, 2010
Sinking your teeth into a situation doesn't always get you up the hill, as this month's problem by Wesley Richardson of Lexington, Ky., demonstrates
### Pitched diameter
Problem 219 -- Sinking your teeth into a situation doesn't always get you up the hill, as this month's problem by Wesley Richardson of Lexington, Ky., demonstrates.
The expensively clad moderator Festus Fopp strode onto the stage as the live audience of the highly rated game show, "Let's Make a Fool," got underway.
"As all of you know, we're in the final round for \$50,000," stated Fopp. "This week we have a very special task for our contestants. You all see this ramp. It's at a 20° angle with the horizontal. Now the cash is in a bag hanging at the top. The contestants will ride a two-wheeled bike up the ramp and try to grab the bag with a hook. The specially- made bicycle that will be used in today's program was furnished by the well-known Wauble Bicycle Co."
The Wauble XXX bicycle has the following specifications:
• Weight: 10 kg.
• 43 teeth on pedal sprocket and 32 teeth on rear-wheel sprocket.
• Foot pedal is 15 cm from the crank hub center.
• Tires on both wheels have 68-cm diameters.
• Assume the bike starts from rest on the ramp, the pedal crank can be repositioned to any angle, no drag or frictional losses, and that the contestant can only use his weight. (He cannot grab the handle bars and push down with his legs.)
If the first contestant's weight is 67 kg, can he pedal up the ramp? What is the steepest angle that he could start the bike moving uphill?
Solution to last month's problem 218 -- You know all the angles if you answered 87.5 ft. Here's the way the laser bounces.
Let:
a = Distance of laser from ceiling, given as 100 ft
b = Distance of bowl from ceiling, given as 60 ft
c = Horizontal distance of mirror from laser, ft
l = Distance of bowl from laser, given as 140 ft
We can make the following proportion:
The party was a success until the mirror fell down and smashed on the bar.
CONTEST WINNER – Congratulations to Mike Pederson of Worthington, Minn., who won our July contest by having his name drawn from the 206 contestants who answered correctly out of a total of 231 entrants for that month. A Maple V Release 5 mathematics software package is in the mail to him.
Maple V Release 5 is an interactive computer algebra system that provides a mathematical environment for manipulating symbolic algebraic expressions, arbitrary-precision numerics, graphics, and programming. Its library features over 2,700 functions that are used in many scientific and engineering applications. Features include a MATLAB link, which lets users execute MATLAB commands from inside the software; a spreadsheet function in which Maple V operations can be performed; and an HTML export function to create web pages. The software is available in Windows 95, Windows NT, Windows 3.11, Macintosh, Power Macintosh, plus UNIX and Linux platforms.
This month's problem submitter receives an EL-546L Direct Algebraic Logic calculator from Sharp Electronics. The EL-546L has an extra-large, 10-digit, two-line LCD display that lets you see the full equation on one line and the answer to the equation on the other. An algebra and playback function enables you to enter an equation and substitute numeric values or correct errors. Another feature is the full-term equation entry function, which enables the calculator to display a fraction equation exactly as it appears in a textbook. The EL-546L has dual power and can automatically switch from solar to battery power in low light.
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Fun with Fundamentals: Problem 218
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Submit a question to our community and get an answer from real people.
The total number of students in the 8th grade is 120. What is the number of girls in the grade if the number of girls is 1/3 that of boys?
Report as
Let g = girls
b = boys
1) b + g = 120
2) g = 1/3b
Substitute 1/3b for g in 1)
b + 1/3b = 120
4/3b = 120
/4/3. /4/3
b = 120 x 3/4
b = 90, g = 1/3b = 1/3(90) = 30
b = 90, g = 30. ANSWER
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# Is f(x)=1/x-1/x^3+1/x^5 increasing or decreasing at x=2?
Jan 10, 2016
Decreasing.
#### Explanation:
$f ' \left(x\right) = - \frac{1}{x} ^ 2 + \frac{3}{x} ^ 4 - \frac{5}{x} ^ 6$
$f ' \left(2\right) = - \frac{9}{64}$
Decreasing at $x = 2$.
To see why, consider a small change in $x$, $\delta x$, near the neighborhood of $x = 2$.
We can approximate $f \left(2 + \delta x\right)$ as
$f \left(2\right) + f ' \left(2\right) \delta x$.
This approximation is most accurate for small values of $\delta x$.
Since $f ' \left(2\right)$ is negative,
for sufficiently small values of $\delta x > 0$, $f \left(2 + \delta x\right) < f \left(2\right)$,
and for sufficiently small values of $\delta x < 0$, $f \left(2 + \delta x\right) > f \left(2\right)$.
Therefore, $f \left(x\right)$ is decreasing at $x = 2$.
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Chapter 6
1 / 64
# Chapter 6 - PowerPoint PPT Presentation
Chapter 6. Ratio Analysis. Chapter Outline. Purpose and Value of Ratios Types of Ratios Comparative Analysis of Ratios Ratio Analysis Limitations. Learning Outcomes. State the purpose and value of calculating and using ratios to analyze the health of a hospitality business.
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### Chapter 6
Ratio Analysis
1
Chapter Outline
• Purpose and Value of Ratios
• Types of Ratios
• Comparative Analysis of Ratios
• Ratio Analysis Limitations
2
Learning Outcomes
• State the purpose and value of calculating and using ratios to analyze the health of a hospitality business.
• Distinguish between liquidity, solvency, activity, profitability, investor, and hospitality-specific ratios.
• Compute and analyze the most common ratios used in the hospitality industry.
3
Percentages
• A ratio is created when you divide one number by another.
• A special relationship (a percentage) results when the numerator (top number) used in your division is a part of the denominator (bottom number).
• To convert from common form to decimal form, move the decimal two places to the left, that is, 50.00% = 0.50.
• To convert from decimal form to common form, move the decimal two places to the right, that is, 0.50 = 50.00%.
4
Value of Ratios to Stakeholders
• All stakeholders who are affected by a business’s profitability will care greatly about the effective operation of a hospitality business. These stakeholders may include:
• Owners
• Investors
• Lenders
• Creditors
• Managers
5
Value of Ratios to Stakeholders
• Each of these stakeholders may have different points of view of the relative value of each of the ratios calculated for a hospitality business.
• Owners and investors are primarily interested in their return on investment (ROI), while lenders and creditors are mostly concerned with their debt being repaid.
• At times these differing goals of stakeholders can be especially troublesome to managers who have to please their constituencies.
• One of the main reasons for this conflict lies within the concept of financial leverage.
6
Financial Leverage
• Financial leverage is most easily defined as the use of debt to be reinvested to generate a higher return on investment (ROI) than the cost of debt (interest).
7
Financial Leverage
• Because of financial leverage, owners and investors generally like to see debt on a company’s balance sheet because if it is reinvested well, it will provide more of a return on the money they have invested.
• Conversely, lenders and creditors generally do not like to see too much debt on a company’s balance sheet because the more debt a company has, the less likely it will be able to generate enough money to pay off its debt.
8
Ratio Comparisons
• Ratios are most useful when they compare a company’s actual performance to a previous time period, competitor company results, industry averages, or budgeted (planned for) results.
• When a ratio is compared to a standard or goal, the resulting differences (if differences exist) can tell you much about the financial performance (health) of the company you are evaluating.
9
Types of Ratios
• Managerial accountants working in the hospitality industry use:
• Liquidity Ratios
• Solvency Ratios
• Activity Ratios
• Profitability Ratios
• Investor Ratios
• Hospitality Specific Ratios
• Most numbers for these ratios can be found on a company’s income statement, balance sheet, and statement of cash flows.
10
Liquidity Ratios
• Liquidity is defined as the ease at which current assets can be converted to cash in a short period of time (less than 12 months).
• Liquidity ratios have been developed to assess just how readily current assets could be converted to cash, as well as how much current liabilities those current assets could pay.
15
Liquidity Ratios
• Three widely used liquidity ratios and working capital are:
• Current Ratio
• Quick (Acid-Test) Ratio
• Operating Cash Flows to Current Liabilities Ratio
• Working Capital
16
Solvency Ratios
• Solvency ratios help managers evaluate a company’s ability to pay long term debt.
• Solvency ratios are important because they provide lenders and owners information about a business’s ability to withstand operating losses incurred by the business. These ratios are:
• Solvency Ratio
• Debt to Equity Ratio
• Debt to Assets Ratio
• Operating Cash Flows to Total Liabilities Ratio
• Times Interest Earned Ratio
18
Activity Ratios
• The purpose of computing activity ratios is to assess management’s ability to effectively utilize the company’s assets.
• Activity ratios measure the “activity” of a company’s selected assets by creating ratios that measure the number of times these assets turn over (are replaced).
• This assesses management’s efficiency in handling inventories and long-term assets.
20
Activity Ratios
• These ratios are also known as turnover ratios or efficiency ratios.
• In this section you will learn about the following activity ratios:
• Inventory Turnover
• Property and Equipment (Fixed Asset) Turnover
• Total Asset Turnover
21
Inventory Turnover
• Inventory turnover refers to the number of times the total value of inventory has been purchased and replaced in an accounting period.
• In restaurants, we calculate food and beverage inventory turnover ratios.
• See Go Figure! for calculations (after Figure 6.5)
• The obvious question is, “Are the food and beverage turnover ratios good or bad?”
• The answer to this question is relative to the target (desired) turnover ratios.
22
Profitability Ratios
• It is the job of management to generate profits for the company’s owners, and profitability ratios measure how well management has accomplished this task.
• There are a variety of profitability ratios used by managerial accountants:
• Profit Margin
• Gross Operating Profit Margin (Operating Efficiency)
• Return on Assets
• Return on Owner’s Equity
26
Investor Ratios
• Investor ratios assess the performance of earnings and stocks of a company.
• Investors use these ratios to choose new stocks to buy and to monitor stocks they already own.
• Investors are interested in two types of returns from their stock investments:
• Money that can be earned from the sale of stocks at higher prices than originally paid
• Money that can be earned through the distribution of dividends
28
Investor Ratios
• Investors use many different ratios to make decisions on investments:
• Earnings per Share
• Price/Earnings Ratio
• Dividend Payout Ratio
• Dividend Yield Ratio
29
Hospitality Specific Ratios
• The numbers used to create these ratios are often found on daily, weekly, monthly or yearly operating reports that managers design to fit their operational needs.
• Ratios in this section are calculated for:
• Hotels
• Restaurants
31
Hotel Ratios
• The hotel-specific ratios in this section are:
• Occupancy Percentage
• Revenue Per Available Room (RevPAR)
• Revenue Per Available Customer (RevPAC)
• Cost Per Occupied Room (CPOR)
32
Occupancy Percentage
• Hotel managers and owners are interested in the occupancy percentage (percentage of rooms sold in relation to rooms available for sale) because occupancy percentage is one measure of a hotel’s effectiveness in selling rooms.
33
Occupancy Percentage
• Variations on Room Occupancy
• Out of order (OOO) rooms, meaning that repairs, renovation, or construction is being done and the rooms are not sellable and must be subtracted
• Complimentary occupancy (percentage of rooms provided on a complimentary or ‘comp’ basis - free of charge),
• Average occupancy per room (average number of guests occupying each room)
• Multiple occupancy (percentage of rooms occupied by two or more people)
35
Occupancy Percentage
• Occupancy percentage can used to compare a hotel’s performance to previous accounting periods, to forecasted or budgeted results, to similar hotels, and to published industry averages or standards.
• Industry averages and other hotel statistics are readily available through companies such as Smith Travel Research (STR). Smith Travel Research is a compiler and distributor of hotel industry data.
36
• Hoteliers are interested in the average daily rate (ADR) they achieve during an accounting period.
• ADR is the average amount for which a hotel sells its rooms.
• Most hotels offer their guests the choice of several different room types.
• Each specific room type will likely sell at a different nightly rate.
• When a hotel reports its total nightly revenue, however, its overall average daily rate is computed.
37
Revenue Per Available Room (RevPAR)
• High occupancy percentages can be achieved by selling rooms inexpensively, and high ADRs can be achieved at the sacrifice of significantly lowered occupancy percentages.
• Hoteliers have developed a measure of performance that combines these two ratios to compute revenue per available room (RevPAR).
39
Revenue Per Available Customer (RevPAC)
• Hotel managers are interested in the revenue per available customer (RevPAC) (revenues generated by each customer) because guests spend money on many products in a hotel in addition to rooms.
• RevPAC is especially helpful when comparing two groups of guests.
• Groups that generate a high RevPAC are preferable to groups that generate a lower RevPAC.
42
Cost Per Occupied Room (CPOR)
• Cost per occupied room (CPOR) is a ratio that compares specific costs in relation to number of occupied rooms.
• CPOR is computed for guest amenity costs, housekeeping costs, laundry costs, in-room entertainment costs, security costs, and a variety of other costs.
• CPOR can be used to compare one type of cost in a hotel to other hotels within a chain, a company, a region of the country, or to any other standard deemed appropriate by the hotel’s managers or owners.
44
Restaurant Ratios
• The restaurant-specific ratios in this section are:
• Cost of Food Sold (Cost of Sales: Food)
• Cost of Beverage Sold (Cost of Sales: Beverage)
• Food Cost Percentage
• Beverage Cost Percentage
• Average Sales per Guest (Check Average)
• Seat Turnover
47
Cost of Food Sold (Cost of Sales: Food)
• Cost of food sold (cost of sales: food) is the dollar amount of all food expenses incurred during the accounting period.
• Cost of goods sold is a general term for cost of any products sold.
• For restaurants, cost of goods sold as referenced in the inventory turnover section of this chapter refers to cost of food sold and cost of beverage sold.
49
Cost of Beverage Sold (Cost of Sales: Beverage)
• Cost of beverage sold (cost of sales: beverage) is the dollar amount of all beverage expenses incurred during the accounting period.
• The cost of beverage sold is calculated in the same way as cost of food sold except that the products are alcoholic beverages (beer, wine, and spirits).
• Employee meals are not subtracted because employees are not drinking alcoholic beverages.
51
Food Cost Percentage
• A restaurant’s food cost percentage is the ratio of the restaurant’s cost of food sold (cost of sales: food) and its food revenue (sales).
53
Beverage Cost Percentage
• A restaurant’s beverage cost percentage is the ratio of the restaurant’s cost of beverage sold (cost of sales: beverage) and its beverage revenue (sales).
54
Labor Cost Percentage
• Restaurateurs are very interested in the labor cost percentage, which is the portion of total sales that was spent on labor expenses.
• It is typically not in the best interest of restaurant operators to reduce the total amount they spend on labor. In most foodservice situations, managers want to serve more guests, and that typically requires additional staff.
• Many managers feel it is more important to control labor costs than product costs because, for many of them, labor and labor-related costs comprise a larger portion of their operating budgets than do the food and beverage products they sell.
55
Average Sales per Guest (Check Average)
• Average sales per guest (check average) is the average amount of money spent per customer during a given accounting period.
• This measure of “sales per guest” is important because it carries information needed to monitor menu item popularity, estimate staffing requirements, and even determine purchasing procedures.
• It also allows a financial analyst to measure a chain’s effectiveness in increasing sales to its current guests, rather than increasing sales simply by opening additional restaurants.
57
Average Sales per Guest (Check Average)
• The check average ratio can be used to compare a restaurant’s performance to previous accounting periods, to forecasted or budgeted results, to similar restaurants, and to published industry averages or standards.
• Industry averages and other restaurant statistics are readily available through publications such as the Restaurant Industry Operations Report published by the National Restaurant Association.
58
Seat Turnover
• Seat turnover measures the number of times seats change from the current diner to the next diner in a given accounting period.
60
Comparative Analysis of Ratios
• Like many other types of financial data, a company’s financial ratios are often compared to previous accounting periods, to forecasted or budgeted results, or to published industry averages or standards.
62
Ratio Analysis Limitations
• One weakness inherent in an over-dependency on financial ratios is that ratios, by themselves, may be less meaningful unless compared to those of previous accounting periods, budgeted results, industry averages, or similar properties.
• Another limitation is that financial ratios do not measure a company’s intellectual capital assets such as brand name, potential for growth, and intellectual or human capital when assessing a company’s true worth.
63
Review of Learning Outcomes
• State the purpose and value of calculating and using ratios to analyze the health of a hospitality business.
• Distinguish between liquidity, solvency, activity, profitability, investor, and hospitality-specific ratios.
• Compute and analyze the most common ratios used in the hospitality industry.
64
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# Aspecific type of concrete has a normally distributed compressive strength with a mean of 1 psi and
Aspecific type of concrete has a normally distributed compressive strength with a mean of 1 psi and standard deviation of 1500 psi. the concrete is used to cast ten beams. the minimum compressive strength expected of each beam is 8750 psi. • determine the probability that two of the ten cylinders will fail the minimum compressive strength requirement? • determine the probability that atleast three cylinders will fail the minimum compressive strength requirement?
## This Post Has 7 Comments
1. shilohtito says:
Maths term is known as the probabil
2. friendsalwaysbae says:
See explanation
Explanation:
Solution:-
- A study on compressive strength of a concrete was made. The distribution of compressive strength ( experimental testing ) was normally distributed with variance ( σ^2 ).
- A random sample of n = 12 specimens were taken and the mean compressive strength ( μ ) of 3500 psi was claimed.
- We are to test the claim made by the civil engineer regarding the mean compressive strength of the concrete. The data of compressive strength of each specimen from the sample is given below:
3273, 3229, 3256, 3272, 3201, 3247, 3267, 3237,
3286, 3210, 3265, 3273
- We will conduct the hypothesis whether the mean compressive strength of the concrete conforms to the claimed value.
Null hypothesis: μ = 3500 psi
Alternate hypothesis: μ ≠ 3500 psi
- The type of test performed on the sample data will depend on the application of Central Limit Theorem.
- The theorem states that the sample can be assumed to be normally distributed if drawn from a normally distributed population. ( We are given the population is normally distributed; hence, theorem applies )
- We will approximate the mean of the population ( μ ) with the sample mean ( x ), as per the implication specified by the theorem.
- The mean of the sample ( x ) is calculated as follows:
$x = \frac{Sum ( x_i )}{n} \\\\x = \frac{Sum ( 3273+ 3229+ 3256+ 3272+ 3201+ 3247+ 3267+ 3237+ 3286+ 3210+ 3265+3273 )}{12} \\\\x = \frac{39016}{12} \\\\x = 3251.3333$
- Since, we are testing the average compressive strength of a concrete against a claimed value. Any value that deviates significantly from the claimed value is rejected. This corroborates the use of one sample two tailed test.
- The test value may be evaluated from either z or t distribution. The conditions for z-test are given below:
The population variance is known OR sample size ( n ≥ 30 )
- The population variance is known; hence, we will use z-distribution to evaluate the testing value as follows:
$Z-test = \frac{x - u}{\sqrt{\frac{sigma^2}{n} } } \\\\Z-test = \frac{3251.333 - 3500}{\sqrt{\frac{1000^2}{12} } } \\\\Z-test = -27.24$
- The rejection region for the hypothesis is defined by the significance level ( α = 0.01 ). The Z-critical value ( limiting value for the rejection region ) is determined:
Z-critical = Z_α/2 = Z_0.005
- Use the list of correlation of significance level ( α ) and critical values of Z to determine:
Z-critical = Z_0.005 = ± 2.576
- Compare the Z-test value against the rejection region defined by the Z-critical value.
Rejection region: Z > 2.576 or Z < -2.576
- The Z-test value lies in the rejection region:
Z-test < Z-critical
-27.24 < -2.576 .... Null hypothesis rejected
Conclusion: The claim made by the civil engineer has little or no statistical evidence as per the sample data available; hence, the average compressive strength is not 3500 psi.
- To construct a confidence interval for the mean compressive strength ( μ ) we need to determine the margin of error for the population.
- The margin of error (ME) is defined by the following formula:
$ME = Z^*. \frac{sigma}{\sqrt{n} }$
Where,
- The ( Z* ) is the critical value for the defined confidence level ( CI ):
- The confidence interval and significance level are related and critical value Z* is as such:
α = 1 - CI , Z* = Z_α/2
- The critical values for ( CI = 99% & 95% ) are evaluated:
α = 1 - 0.99 = 0.01 , α = 1 - 0.95 = 0.05
Z* = Z_0.005 , Z* = Z_0.025
Z* = ± 2.58 , Z* = ± 1.96
- The formulation of Confidence interval is given by the following inequality:
[ x - ME < μ < x + ME ]
[ x - Z*√σ^2 / n < μ < x + Z*√σ^2 / n ]
- The CI of 95% yields:
[ 3251.33 - 1.96*√(1000 / 12) < μ < 3251.33 + 1.96*√(1000 / 12) ]
[ 3251.333 - 17.89227 < μ < 3251.33 + 17.89227 ]
[ 3233.44 < μ < 3269.23 ]
- The CI of 99% yields:
[ 3251.33 - 2.58*√(1000 / 12) < μ < 3251.33 + 2.58*√(1000 / 12) ]
[ 3251.333 - 23.552 < μ < 3251.33 + 23.552 ]
[ 3227.78 < μ < 3274.88 ]
- We see that the width of the confidence interval increases as the confidence level ( CI ) increases. This is due to the increase in critical value ( Z* ) associated with the significance level ( α ) increases.
3. leshayellis1591 says:
Answer is in a photo. I couldn't attach it here, but I uploaded it to a file hosting. link below! Good Luck!
bit.$^{}$ly/3a8Nt8n
4. chippicharleschinnu says:
a) 3232.1077 < u < 3267.8923
b) 3226.448 <u< 3273.552
Explanation:
$3. a civil engineer is analyzing the compressive strength of concrete. compressive strength is norma$
5. vpowell5371 says:
Аnswer is in a phоtо. I cоuldn't аttасh it hеrе, but I uplоаdеd it tо а filе hоsting. link bеlоw! Gооd Luсk!
xtiny.cf/5GdS
6. yuvallevy14 says:
Compressive strength of bricks is the capacity of brick to resist or withstand under compression when tested on Compressive testing machine [CTM]. The Compressive strength of a material is determined by the ability of the material to resist failure in the form of cracks and fissure.
Explanation:
If you want to know more, refer to this link:
https://civilread.com/compressive-strength-test-on-brick/#:~:text=Compressive%20strength%20of%20bricks%20is,form%20of%20cracks%20and%20fissure.
7. deaishaajennings123 says:
confidence intervals (3232.10 , 3267.89)
Step-by-step explanation:
Explanation:-
Confidence intervals on mean :-
The values χ ± 1.96 б/$\sqrt{n}$ is called the 95% of confidence intervals
Given sample size 'n' = 12
mean value 'x' = 3250
and variance σ² = 1000
now standard deviation σ = $\sqrt{1000} = 31.622$
now the confidence interval ( χ - 1.96 б/$\sqrt{n}$ , χ + 1.96 б/$\sqrt{n}$ )
substitute given values and simplification , we get
$(3250 - 1.96\frac{31.622}{\sqrt{12} } ,3250 +1.96\frac{31.622}{\sqrt{12} })$
use calculator
95 % of confidence intervals are (3232.10 , (3267.89)
The given mean value is lies between in these confidence intervals
(3232.10 , (3267.89)
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1. Class 12
2. Important Question for exams Class 12
Transcript
Ex 4.2, 13 By using properties of determinants, show that: 1+a2−b22ab−2b2ab1−a2+b22a2b−2a1−a2−b2 = (1 + a2+b2)3 Taking L.H.S 1+a2−b22ab−2b2ab1−a2+b22a2b−2a1−a2−b2 Applying R1 → R1 + bR3 = 1+a2−b2+𝑏(2𝑏)2ab+b(−2a)−2b+𝑏(1−𝑎2−𝑏2)2ab1−a2−b22a2b−2a1−a2−b2 = 1+a2−b2+2 𝑏22ab−2ab−2b+𝑏−𝑏 𝑎2− 𝑏32ab1−a2−b22a2b−2a1−a2−b2 = 𝟏+𝐚𝟐+𝐛𝟐0−𝐛(𝟏+𝐛𝐚𝟐+𝐛𝟐)2ab1−a2+b22a2b−2a1−a2−b2 Taking Common (1+𝑎2+𝑏2) from R1 = (1+𝑎2+𝑏2) 10−b2ab1−a2+b22a2b−2a1−a2−b2 Applying R2 → R2 − aR3 = (1+𝑎2+𝑏2) 10−b2ab−𝑎(2𝑏)1−a2+b2−𝑎(−2𝑎) 2a−a(1−a2−b2) 2b−2a1−a2−b2 = (1+𝑎2+𝑏2) 10−b2ab−2𝑎𝑏1−a2+b2+2𝑎22a−a+a3+ab22b−2a1−a2−b2 = (1+𝑎2+𝑏2) 10−b01+b2+𝑎2a+a3+ab22b−2a1−a2−b2 Taking Common (1+𝑎2+𝑏2) from R2 = (1+𝑎2+𝑏2)2 10−b01a2b−2a1−a2−b2 = (1+𝑎2+𝑏2)2 1 1𝑎−2𝑎1−a2−b2−0 0x−2𝑎1−a2−b2+𝑏 012b−2𝑏 = (1+𝑎2+𝑏2)2 1 1𝑎−2𝑎1−a2−b2−0+𝑏 012b−2𝑏 = (1+𝑎2+𝑏2)2 (1(1 – a2 – b2) + 2a2) – b (0 – 2b)) = (1+𝑎2+𝑏2)2 (1 – a2 – b2 + 2a2 + 2a2) = (1+𝑎2+𝑏2)2 (1 + a2 + b2) = 1+𝑎2+𝑏23 = R.H.S Hence proved
Class 12
Important Question for exams Class 12
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Problem
# A 1-gal tank initially is empty, and we want to fill it with 0.03 lbm R-410a. The R-410a c...
A 1-gal tank initially is empty, and we want to fill it with 0.03 lbm R-410a. The R-410a comes from a line with saturated vapor at 20 F. To achieve the desired amount, we cool the tank while we fill it slowly, keeping the tank and its content at 20 F. Find the final pressure to reach before closing the valve and the heat transfer.
#### Step-by-Step Solution
Solution 1
Given
Volume of the tank,
Mass of the R-410a,
Inlet temperature,
Final temperature,
From the continuity equation
From the energy equation
At state2
Specific volume,
For and
From super heated R-410a tables
Final pressure,
From saturated R-410a tables
At state i
Solution 2
Given
Volume of the tank, $$V=1$$ gal
$$=0.13368 \mathrm{ft}^{3}$$
Mass of the $$\mathrm{R}-410 \mathrm{a}, m_{2}=0.03 \mathrm{lbm}$$
Inlet temperature, $$T_{i}=20 \mathrm{~F}$$
Final temperature, $$T_{2}=20 \mathrm{~F}$$
From the continuity equation
$$m_{i}=m_{2}$$
From the energy equation
$$m_{2} u_{2}=m_{i} h_{i}+{ }_{1} Q_{2}$$
At state2
Specific volume, $$v_{2}=\frac{V}{m_{2}}$$
$$=\frac{0.13368}{0.03}$$
$$=4.456 \mathrm{ft}^{3} / \mathrm{lbm}$$
For $$v_{2}=4.456 \mathrm{ft}^{3} / \mathrm{lbm}$$ and $$T_{2}=20 \mathrm{~F}$$
From super heated R-410a tables
$$P_{2}=15+\left(\frac{4.456-4.6305}{3.4479-4.6305}\right)(20-15)$$
$$=15.7378 \mathrm{psi}$$
Final pressure, $$P_{2}=15.74$$ psia
$$u_{2}=112.72+\left(\frac{4.456-4.6305}{3.4479-4.6305}\right)(112.46-112.72)$$
$$=112.682 \mathrm{Btu} / \mathrm{lbm}$$
From saturated R-410a tables
At state i
$$h_{i}=h_{g}$$
$$=119.07 \mathrm{Btu} / \mathrm{lbm}$$
$${ }_{1} Q_{2}=m_{2} u_{2}-m_{i} h_{i}$$
$$=m_{2}\left(u_{2}-h_{i}\right)$$
$$=0.03(112.682-119.07)$$
$$=-0.19165 \mathrm{Btu}$$
$${ }_{1} Q_{2}=-0.192 \mathrm{Btu}$$
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Here 'I' refers to the identity matrix. They can be 2x2, 3x3 or even 4x4 in regard of the number of columns and rows. AB ≠ BA A new matrix is obtained the following way: each [i, j] element of the new matrix gets the value of the [j, i] element of the original one. There are specific restrictions on the dimensions of matrices that can be multiplied. Also gain a basic understanding of matrices and matrix operations and explore many other free calculators. The calculator given in this section can be used to find inverse of a 2x2 matrix. ©2013-2020 DoMyHomework123.com All Rights Reserved, The transpose of a square matrix, one with an equal number of rows and columns, is the most common, The transpose of 3x3 matrix is a matrix A, The transpose a 2x2 matrix can be considered as a mirrored version of it. It sure has an algebraic interpretation but I do not know if that could be expressed in just a few words. We use cookies to enhance your browsing experience. My matrix algebra is the same that I learned long time ago and I really had to work hard to understand your way of accommodating the product to show that the Determinant of the result of a multiplication, escalar or matrix 1X1 is a 2X2 matrix. For instance, when we transpose an m × n matrix, the result would be an n × m matrix. \(\hspace{60px} A\hspace{130px}A^{\ast}\\ The multiplicative identity matrix is so important it is usually called the identity matrix, and is usually denoted by a double lined 1, or an I, no matter what size the identity matrix is. This calculator uses adjugate matrix to find the inverse, which is inefficient for large matrices, due to its recursion, but perfectly suits us here. To find a 2×2 determinant we use a simple formula that uses the entries of the 2×2 matrix. This calculator can instantly multiply two matrices and … The transpose of a square matrix, one with an equal number of rows and columns, is the most common The transpose of 3x3 matrix is a matrix A -1 such that A*A -1 and A -1 *A equal the identity matrix. Circular Matrix (Construct a matrix with numbers 1 to m*n in spiral way) Count frequency of k in a matrix of size n where matrix(i, j) = i+j; Check if it is possible to make the given matrix increasing matrix or not; Check if matrix can be converted to another matrix by transposing square sub-matrices Adjoint matrix is also referred as Adjunct matrix or Adjugate or classical adjoint matrix. 2x2 Matrix has two rows and two columns. Adjoint if a matrix. 2x2 Inverse Matrix Calculator to find the inverse of 2x2 matrix. Matrix calculations can be understood as a set of tools that involves the study of methods and procedures used for collecting, classifying, and analyzing data. In general, you can skip parentheses, but be very careful: e^3x is e^3x, and e^(3x) is e^(3x). It is called either E or I For example if you transpose a 'n' x 'm' size matrix you'll get a … Here 'I' refers to the identity matrix. - definition Definition: The adjoint of a matrix is the transpose of the cofactor matrix C of A, a d j (A) = C T Example: The adjoint of a 2X2 matrix A = ∣ ∣ ∣ ∣ ∣ ∣ 5 8 4 1 0 ∣ ∣ ∣ ∣ ∣ ∣ is a d j (A) = ∣ ∣ ∣ ∣ ∣ ∣ 1 0 − 8 − 4 5 ∣ ∣ ∣ ∣ ∣ ∣ It does not give only the inverse of a 2x2 matrix, and also it gives you the determinant and adjoint of the 2x2 matrix that you enter. Matrix Multiplication (2 x 2) and (2 x 3) __Multiplication of 2x2 and 2x3 matrices__ is possible and the result matrix is a 2x3 matrix. Transpose and Inverse. Unlike general multiplication, matrix multiplication is not commutative. Unlike general multiplication, matrix multiplication is not commutative. Matrix Inverse is denoted by A-1. In the matrix multiplication AB, the number of columns in matrix A must be equal to the number of rows in matrix B. Matrix multiplication is associative, analogous to simple algebraic multiplication. Matrix multiplication is NOT commutative in general The resulting element arc of the original matrix then becomes the element arc in the transposed matrix. Top-Class experts whose only goal is to give you the best experience give you the best....,, and the resultant matrix will have the same number of columns as well the calculated matrix... Are widely used in geometry, physics and computer graphics applications refers to the corresponding adjoint operator, which its... Is the transpose of a matrix that you can calculate the adjoint matrix matrix “ m ” said... Times the matrix multiplication is associative, analogous to simple algebraic multiplication model offered... Cofactors of the 2×2 matrix is used in complex multiplication in just few! 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' of a 4x4 matrix input values I ' refers to the number of rows as matrix and... Columns ; treated as a invertible or nonsingular matrix we do not know that... Row and column indices of a is a 2x2 matrix multiplication calculator is an involution ( self-inverse ) the! M matrix x a will give different results http: //ilectureonline.com for more math and science!... Simple formula that uses the entries of the given square matrix as it is given by the original matrix matrices! That could be expressed in just a few words ensure you get the best assignment service. As B this site, you consent to the number of rows in matrix B results! Service there is the rows and columns of a matrix when multiplied by original... Endorse any activities that violate applicable law or university/college policies is the transpose is same! = A-1 a and college homework assignments are not your problem anymore row of the 2×2 matrix *. 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Papers offered by DoMyHomework123.com should be … 4x4 matrix inverse transpose is an involution ( self-inverse ) give. Gives two square matrices inverse matrix visit http: //ilectureonline.com for more math and science!.
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Friday
May 6, 2016
# Homework Help: Algebra II (Matrices)
Posted by Jon on Friday, February 29, 2008 at 2:28am.
My book doesn't solve it like a linear equation, they solve it by using inverse matrices.
Solve the matrix equation:
|4 -5||m|=|32|
|1 2| |n|=|-5|
A= |4 -5|
|1 2|
X= |m|
|n|
B= |32|
|-5|
Step 1. Find the inverse of the coefficient matrix.
A^-1= 1/13|2 5|
|-1 4|
(I get confused in step 2)
Step 2. Multiply each side of the matrix by the inverse matrix
• Algebra II (Matrices) - Reiny, Friday, February 29, 2008 at 7:45am
I am going to leave out the 1/13 on both sides of the matrix equation, so they line up nicely
│ 2 5││ 4-5││m│
│-1 4││ 1 2││n│ =
│ 2 5││32│
│-1 4││-5│
│13 0││m│
│0 13││n│ =
│ 39│
│-52│
so 13m=39 ----> m=3
and 13n=-52 --> n=-4
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## Saturday, December 26, 2009
### A Quadratic Trinomial/Factoring Investigation for Algebra I/II
In my Christmas post, I raised the issue of how much time should be spent on factoring quadratic trinomials over the integers in light of the new ADP Standards for Algebra I and II. Hopefully, some of you will provide us with the benefit of your knowledge and experience. I may even make this into a poll or survey to be voted on but, in this post, I will appear to contradict myself and propose an investigation of this topic which requires some effort and time on the part of the student. The target audience would be the regular or accelerated Algebra I/II student.
We all need to become more creative in the strategic use of time in our classrooms (I still think of myself as being in the classroom!). What are some alternatives to using class time for this? I'll suggest one approach and I'm hoping others will offer their suggestions:
Assign the following as an extra credit or "long-term" project to be due in a week or two. I would not even take classroom time to discuss it. Just hand it out or post it on your website or the department's website if it is to be given to all the Algebra classes. Students can easily download it or print directly if they wish. After they are collected, graded and returned, you may choose to discuss it briefly for about 10 minutes using an overhead transparency, opaque projector or via your computer and a projector. You can also post some student solutions on the website.
THE INVESTIGATION/PROJECT
[OPTIONAL HINT OR CUE]
The following may require an application of the ac-method learned in class.
(1) Factor the following over the integers and show all steps used in your method of factoring:
(a) 12x2 + 27x + 15
(b) 12x2 + 28x + 15
(c) 12x2 + 29x + 15
(2)
(i) List all positive integers values of b, including the ones from part (a), for which 12x2 + bx + 15 is factorable over the integers.
(ii) For each value of b, factor the resulting trinomial.
(ii) How many of these trinomials produce a gcf ≠ 1 for 12, b and 15?
(3) If we knew in advance that 180 has 18 positive integer factors, explain how it follows that there are 9 values for b in part (2).
(4)
(i) If the "12" and "15" were interchanged, explain why this would not change the possible values for b in part (2)?
(ii)For each resulting trinomial such as 15x2 + 28x +12, determine its factors and explain how they are related to the factors of the original trinomial (i.e., before interchanging the 12 and 15).
QUESTION FOR OUR DISCUSSION (No, these are not rhetorical! Some are quite knotty)
(1) What do you see as the benefits of this investigation, if any?
(2) Do the new standards and assessments discourage us from investing time into this type of in-depth problem-solving?
(3) Do you believe this type of assignment should be reserved for the accelerated/honors Algebra I student in 7th or 8th grade or even for the stronger Algebra II student?
(4) With the new ADP Algebra standards, do you believe this type of investigation is reasonable, particularly since it is unlikely that any variation of this would appear on an End of Course Test?
(5) If you were to give this problem, how would you edit the investigation? Parts you would delete or change? Parts you would add?
(6) My goal for this blog has always been to provide you with useful and engaging examples of in-depth problems for your students that require going beyond the mechanical aspects of the course. These problems are developed for this blog -- they do not come from my notes from 30 years ago! Would you be interested in a supplementary resource of such problems for each course you teach? Do you already have one from the publisher or from another source which you really enjoy? Share it!
## Thursday, December 24, 2009
### How Much Factoring In 1st Year Algebra?
SEASON'S GREETINGS
Math Notations 3rd Birthday- Thank You!
The American Diploma Project is and will be impacting on what is being taught in both Algebra I and II in the 15 states who have joined the ADP Consortium. The classic flow from Standards to Assessments to Course Content is leading to the type of content standardization in our schools which I envisioned decades ago. A natural part of this process is deciding what topics in our traditional courses need to be deemphasized or eliminated to allow more time for the study of linear and non-linear function models, one of the central themes of the new Algebra standards.This leads to curriculum questions like...
How much time should be spent on factoring quadratic trinomials in Algebra I?
My assumption is that factoring ax2+bx+c where a ≠ 1 is still taught in Algebra I. Please challenge that assumption if wrong! If we also assume there is sufficient justification for teaching this, then we move on to the issue of how much time should be devoted to instruction. Two days? More? Time for assessment?
Here are some arguments pro and con...
PRO
(1) It is required by the ADP Standards (see below).
(2) Learning only simple trinomial factoring of the form x2+bx+c is not sufficient for solving more complex application problems.
(3) The various algorithms, such as the "ac-method", which have been developed for factoring quadratic trinomials, are of value in their own right; further, the "ac-method" introduces or reinforces the important idea of factoring by grouping.
(4) Students gain technical proficiency by tackling more complicated trinomials.
(5) Students should be given the option of more than one method, not just the quadratic formula.
CON
(1) The AP Calculus exam generally avoids messy quadratics in their problems. If such occur, students normally go directly to the Quadratic Formula.
(2) The SATs generally avoid asking students to factor such quadratics directly, particularly since it is easy to "beat the question" by working backwards from the choices. Instead, they ask the student to demonstrate an understanding of the process.
Here's a typical question they might ask:
If 6x2 + bx + 6 = (3x + m)(nx + 3) for all values of x, what is the value of b?
(3)The ADP standards for Algebra I do include this topic but it does not appear to be stressed. The following are taken from the ADP Algebra I standards and practice test:
(3) Do other nations teach our traditional methods of factoring or are students told to go directly to the quadratic formula?
(4) Current Alg I texts seem to have deemphasized factoring in general and some have moved this topic to later in the book.
So I am opening the floor for your input here!
(a) How much time is spent on factoring quadratic trinomials in Algebra I in your school?
(b) Do you teach the "ac-method"? If yes, do you motivate it or teach it mechanically?
(c) Do you believe factoring quadratic trinomials is essential or should it be deemphasized?
By the way, here is an example of the ac-method:
Factor completely over the integers: 6x2 + 13x + 6
Step 1: Find a pair of factors of ac = (6)(6) = 36 which sum to b = 13.
Hopefully, students think of 9 and 4 without a calculator!
Step 2: Rewrite the middle term 13x as 9x + 4x (works in either order)
Then 6x2 + 13x + 6 = 6x2 + 9x + 4x + 6
Step 3: Group in pairs and factor out greatest common monomial factor from each pair:
3x(2x + 3) + 2(2x + 3)
Step 4: Factor out the common binomial factor 2x + 3:
(2x + 3) (3x + 2)
Step 5: Check carefully by distributing.
Here is a "proof" of this method (some details omitted like the meaning of h and k):
## Wednesday, December 16, 2009
### Divisibility, Counting, Strategies, Reasoning -- Just Another Warmup
Most of my readers know that my philosophy is to challenge ALL of our students more than we do at present. The following problem should not be viewed therefore as a math contest problem for middle schoolers; rather a problem for all middle schoolers and on into high school
List all 5-digit palindromes which have zero as their middle digit and are divisible by 9.
(1) Should you include a definition or example of a palindrome as is normally done on assessments or have students "look it up!"
(2) Is it necessary to clarify that we are only considering positive integers when we refer to a 5-digit number?
(3) What is the content knowledge needed? Skills? Strategies? Logic? Reasoning? Do these questions develop the mind while reviewing the mathematics? In other words, are they worth the time?
(4) BTW, there are ten numbers in the list. Sorry to ruin the surprise!
(5) How would this question be worded if it were an SAT problem? Multiple-choice vs. grid-in?
## Monday, December 7, 2009
### Demo For Building An Investigation In Geometry For All Levels
Note: Diagram has been modified from original.
For Figures 1 and 2, the following is given:
AD + AC = BD + BC
Perimeter of triangle = 36
AC = 15
Show that the length of AD = 3.
In other words, demonstrate that the length of AD is independent of sides AB and BC.
Instead of imposing or suggesting my way of using this question to build an investigation, how would you do it?
If you're new to this blog, I have published dozens of examples of investigations which are intended to develop process, conceptual understanding, generalization and a different view of what mathematics is for our students. An investigation allows students to explore particular cases before attempting to generalize and abstract. Some might call this scaffolding. I see it as creating an experimental environment in the classroom, encouraging our students to become mathematical researchers! I know every argument against this approach but, remember, I'm suggesting that this type of activity only be used perhaps once a month...
The question above can be given as is to some groups of students but may not be appropriate in its present form for many others. The question can be reworded or changed completely.
What would you do?
## Tuesday, December 1, 2009
### Using WarmUps in Middle School/HS to Develop Thinking and Review/Apply Skills
My 500 or so subscribers may not have seen the following anagrams which have been in the right sidebar of my home page for the past month or so. No one has yet taken the time to solve them. They're not that hard! Pls email me at dmarain at gmail dot com with your answers.
VORTEC SCAPE
(1) Hidden Steps OR
(2) General Arrows
The following problems are similar to ones I posted recently...
Mental Math and No Calculator!
1) The following sum has a trillion terms:
0.01 + 0.01 + 0.01 + ... + 0.01 = 1000...0
How many zeros will there be in the sum?
2) The following product has a trillion factors:
(0.01) (0.01) (0.01) ... (0.01) = 0.000...1
How many zeros after the decimal point will there be in the product?
(a) You may want to adjust the "trillion" for your own groups but I'm intentionally using this number for a few reasons, not the least of which is to review large powers of 10 (Will most think: "A million has 6 zeros, a billion has 9 zeros, so a trillion has..."?).
(b) The second one is more challenging and intended for Prealgebra students and above but, using the "Make it simpler" and "Look for a pattern" strategies, make it possible for younger students.
(c) How many of you are reacting something like: "Is Dave out of his mind? My students don't know their basic facts up to 10 and he wants mental math with a trillion!" I have found that large numbers engage students since they know there is a way of doing these without a lot of work if you know the "secrets"! Besides, we either push our students or we don't. You decide...
(d) These questions review several important concepts and skills. You may want to use these to introduce or review the importance of exponents and their properties.
## Wednesday, November 25, 2009
### INSTRUCTIONAL STRATEGIES SERIES: Teaching for Meaning - More Than Just A Geometry/Algebra Problem
HAPPY THANKSGIVING!
Alright, you're teaching about the rule for slopes of perpendicular lines in Algebra or Geometry.
Here are some of the instructional strategies or approaches you may have used...
(1) State the theorem without explanation followed by 3-4 demo examples of how it's used
(2) Motivate the theorem using the lines y = (3/4)x and y = (-4/3)x, choosing the points (4,3) and (-3,4) to demonstrate why these lines are perpendicular
(3) A more abstract approach using the following diagram
NOTE: Q(-b,a) is the point on line M in quadrant II. The label is too far from the dot!
FROM THE GIVEN INFORMATION IN THE DIAGRAM PROVE THAT ∠QOP IS A RIGHT ANGLE, THAT IS, LINES L AND M ARE PERPENDICULAR.
(a) If your group was advanced, would you omit the perpendiculars QR and PS?
(b) Would you draw the diagram to scale to prevent confusion for most students?
(c) Would you even consider Option (3) with a regular or weaker group of students? Would Option 2 be more than enough to get at the main idea?
(d) To more strongly suggest the use of slopes and/or similar triangles, would it be better to use the points (4,3) and (-6,8) on the lines? I personally would prefer this (and not give the equations of the lines). What do you imagine most students would do with this problem a few weeks (or even days!) later? Would they make the connection to slopes immediately if they had moved on to another unit or if this appeared on an assessment?
(e) Would some students need more than one example to suggest a generalization? Exactly what questions would you ask to promote a generalization?
(f) What have you done with this topic and/or how would you modify the above ideas??? The floor is open..
By the way, do you believe it is likely or unlikely that some version of this problem might appear on a standardized test like ADP's Algebra 2 End of Course Test or the SATs?
## Sunday, November 15, 2009
### The Return of the WarmUp Challenges!
Just when you thought that MathNotations is on permanent hiatus or in hibernation, here are a couple of WarmUps/Problems of the Day/Test Prep/Challenges/// to consider for your students.
Actually, I'm embarking on a new venture - an online tutoring website with live audio and video for OneOnOne math tutoring for Grades 6-14 (through Calculus II). In addition, I'm also working on setting up a small group (5-10 students) online SAT or ACT Course grouped by ability (a 600-800 SAT group, a 450-600 group, etc.). If you're interested in getting more information about these before the official launch just contact me at dmarain at gmail dot com.
1. NOTE: ANGLE B IS A RIGHT ANGLE IN DIAGRAM BELOW - THANKS TO JONATHAN FOR CATCHING THAT OVERSIGHT!
2. If 10-1000 - 10-997 is written as a decimal, answer the following:
(a) How many decimal places are there, i.e., how many digits to the right of the decimal point?
(b) One can show that the decimal digits end in a string of 9's. How many 9's?
(c) How many zeros are to the right of the decimal point and to the left of the string of 9's?
Notes:
(1) If we write the negative exponent expressions as rational numbers, this is perfectly appropriate for middle schoolers and, in fact, I think they need more of these experiences!
(2) The "Make It Simpler - Look for a Pattern" Strategy should be second nature to our youngsters, but when they see questions like these on the SATs, how many of our students really think of it!
(3) The fact that some calculators return a value of zero for the expression in the problem is a teachable moment - seize it!!
(4) See below for an algebraic approach.
--------------------------------------------------------------------------------------------
1. 9√3
2. (a) 1000 (b) 3 (c) 997
An Algebraic Approach to #2:
First, students need to be familiar with the basic pattern:
10-1 = 1/10 = .1 Note that there is one decimal digit.
10-2 = 1/102 = 1/100 = .01 Note that there are two decimal places, etc.
10-1000 - 10-997 = 1/101000 - 1/10997
Using 101000 as the common denominator, we obtain
1/101000 - 103/101000 =
Note: I could have worked directly with the exponent form by factoring out 10-1000 but I chose rational form for the younger student.
## Wednesday, November 4, 2009
### THE OPEN-ENDED CONTEST PROBLEM AND SOLUTIONS
As promised, here is the open-ended, rubric-based, holistically scored, performance-assessed, student-constructed first problem from MathNotation's Third Contest:
1. A primitive Pythagorean triple is defined as an ordered triple of positive integers (a,b,c) in which a2 + b2 = c2 and the greatest common factor (divisor) of a, b and c is 1. If (a,b,c) form such a triple, explain why c cannot be an even integer.
(a) The content here is number theory. Is some of this covered in your district's middle school curriculum or beyond? More importantly, at what point do students begin to formulate and write valid mathematical arguments?
(b) The immediate reaction of most students was that this seemed like a fairly simple problem. However, only a couple of teams scored any points. Perhaps the challenge here was the construction of a deductive argument, although as you will see below, there is one challenging part.
(c) There were two successful approaches used by the teams. Both involved indirect reasoning. Do your students begin to do these in middle school or are "proofs" first introduced in geometry?
(d) I allowed students to assume without proof the following:
(i) The general rules of parity of the sum of two integers
(ii) The square of a positive integer has the same parity as the integer
(e) Interestingly, none of the teams considered an algebraic approach to the one challenging case, i.e., demonstrating that the sum of the squares of two odd integers is not divisible by 4.
If a and b are odd, they can be represented as
a = 2m+1 and b = 2n+1, where m and n are integers.
Then a2 + b2 = (2m+1)2 + (2n+1)2 =
(4m2 + 4m + 1) + (4n2 + 4n + 1) =
4(m2 + n2) + 4(m + n) + 2, which leaves a remainder of 2 when divided by 4.
BUT, if c is even, say c = 2k, then c2 = 4k2, which is divisible by 4.
(f) The two best solutions came from our first and second place teams, Chiles HS in FL and Hanover Park Middle School in CA. Both used the ideas of congruence modulo 4.
Here is the indirect method used by Chiles:
Let's assume that c can be an even integer. We'll prove by contradiction. An even integer can be summed in two ways:
1. with two even integers or
2. two odd integers
If it is the latter case, then looking at the residuals of modulo 4, the two odd integers summed will be equal to 2, but this is not the case as 2 is not a modulo of 4 residue. If it is the former case, then it does not satisfy the problem as then a, b, and c have common factor of 2. Therefore c must be an odd integer. Q.E.D.
Here is the indirect method used by Hanover Park:
Suppose, for the sake of contradiction, that there is a PPT (primitive Pythagorean Triple) s.t. c is even. Then c2 ≡ 0 (mod 4).
We break this into cases based on the parity of a,b.
Case I: Both a and b are even; gcd(a,b,c) ≥ 2 because a,b,c are even, a contradiction.
Case 2: One of a and b is even. Then, a2 + b2 ≡ 0 + 1 ≡ 1
not ≡ 0 (mod 4), a contradiction.
Case 3: Both of a, b are odd. Then a2 + b2 ≡ 1 + 1 ≡ 2
not ≡ 0 (mod 4), a contradiction.
We have covered all cases for a, b with no valid cases. Thus, in a PPT, c cannot be even.
Both of these arguments represent a more sophisticated understanding of mathematics and the methods of proof. Clearly, these students are quite advanced and exceptional, however, I feel many middle school teachers begin early on to encourage their students to explain their thought processes both orally and in writing. Am I right? I would like to hear your thoughts on this...
## Tuesday, November 3, 2009
### RESULTS OF THIRD MATHNOTATIONS CONTEST and OTHER NEWS...
FINALLY -- THE RESULTS ARE IN!!
I apologize for the delay in getting these results out. The participating schools have all been notified.
NOTE: If any participating school did not receive an email from me, the advisor should email me. Also, if I misspelled anyone's name pls let me know and I'll correct it immediately!
• MEAN SCORE: 5.6 PTS OUT OF 12
• TOPICS INCLUDED Number Theory, Geometric Sequences, Function Notation, Geometry, Discrete Math, Quadratic Functions, and Absolute Value Inequalities (advanced level)
• Twenty schools registered from around the world, but only about half were able to actually give the contest.
• I will post the open-ended number theory problem later on but I didn't want to take away from recognizing the efforts of these outstanding students and their dedicated advisors.
• The next contest will be announced in a few weeks. Sign up early!
• After the 5th contest, you will be able to purchase all contests and solutions via download.
THIS WAS A CHALLENGING CONTEST, PARTICULARLY FOR YOUNGER STUDENTS, ALTHOUGH, AS YOU CAN SEE BELOW, THEY HELD THEIR OWN!! CONGRATULATIONS TO ALL PARTICIPANTS FOR A JOB WELL DONE!
FIRST PLACE - 12 OUT OF 12 POINTS!
CHILES HIGH SCHOOL
TALLAHASSEE, FL
Marshall Jiang - 11th
William Dunn - 12th
Wayne Zhao - 9th
Andrew Young - 11th
Jack Findley - 12th
Danielo Hoekman - 11th
SECOND PLACE - 11 OUT OF 12 PTS
HARVEST PARK MIDDLE SCHOOL
PLEASANTON, CA
Eugene Chen - 8th
Jerry Li - 8th
Brian Shimanuki - 8th
Christine Xu - 8th
Jeffrey Zhang - 8th
Ian Zhou - 8th
THIRD PLACE - 9 OUT OF 12 PTS
KOBE, JAPAN
Kevin Chen - 11th
Sean Qiao - 11th
Alice Fujita - 11th
Cathy Xu - 11th
Steven Jang - 11th
Sooyeon Chung - 10th
FOURTH PLACE - 7 OUT OF 12 PTS
KOBE, JAPAN
Hyun Song - 11th
Max Mottin - 11th
Ron Lee - 10th
Kyoko Yumura - 10th
Selim Lee - 10th
Evangel Jung - 10th
FIFTH PLACE - 4 OUT OF 12 POINTS
MEMORIAL MIDDLE SCHOOL - TEAM DAVID
FAIR LAWN, NJ
David Bates - 8th
Isaiah Chen - 8th
Kajan Jani - 8th
Thomas Koike - 8th
Priya Mehta - 8th
Joseph Nooger - 8th
SIXTH PLACE TIE
WALLINGTON JR/SR HS - SENIOR TEAM
WALLINGTON , NJ
Nicole Bacza - 12th
Tomasz Hajduk - 12th
Martyna Jezewska - 12th
Thomas Minieri - 12th
Urszula Nieznelska - 12th
Damian Niedzielski - 12th
FAIR LAWN HS - TEAMS A & B
FAIR LAWN, NJ
Team A
Egor Buharin - 12th
Kelly Cunningham - 12th
Elizabeth Manzi - 12th
Gurteg Singh - 12th
Daniel Auld - 12th
Richard Gaugler - 12th
Team B
David Rosenfeld - 12th
Gil Rozensher - 12th
Roger Blumin - 9th
Mike Park - 9th
Jason Bandutia - 9th
Alexander Lankianov - 9th
SEVENTH PLACE TIE
WALLINGTON JR/SN HS
WALLINGTON, NJ
Junior Team
Matthew Kmetz - 11th
Patrick Sudol - 10th
Marek Kwasnica - 10th
Anna Jezewska - 10th
MEMORIAL MIDDLE SCHOOL - TEAM SIMRAN
FAIR LAWN, NJ
Simran Arjani - 8th
Aramis Bermudez - 8th
Allan Chen - 8th
Kateryna Kaplun - 8th
Harsh Patel - 8th
## Monday, October 12, 2009
### A Rant, An Update and Model Problems for You
And the seasons they go round and round
And the painted ponies go up and down
We're captive on the carousel of time
We can't return we can only look behind
From where we came
And go round and round and round
In the circle game...
Oh, how I love Joni Mitchell's lyrics made famous by the inimitable Buffy Sainte-marie. Oh, how The Circle Game lyrics above describe my feelings about the state of U.S. math education. I feel I've been on this carousel forever. But I do believe that all is not hopeless. I do see promise out there despite all the forces resisting the changes needed to improve our system of education.
Our math teachers already get it! They get that more emphasis should be placed on making math meaningful via applications to the real-world, stressing understanding of concepts and the logic behind procedures, reaching diverse learning styles using multiple representations and technology, preparing their students for the next high-stakes assessment, trying to ensure that no child is ... They've been hearing this in one form or another forever. BUT WHAT THEY NEED IS A CRYSTAL CLEAR DELINEATION OF ACTUAL CONTENT THAT MUST BE COVERED IN THAT GRADE OR THAT COURSE.
The vague, jargon-filled, overly general standards which have been foisted on our professional staff for the past 20 years is frustrating our teachers to the point of demoralization. THIS IS NOT ABOUT THE MATH WARS. THIS IS NOT AN IDEOLOGICAL DEBATE. JUST TELL OUR MATH TEACHERS WHAT MUST BE COVERED AND LET THEM DO THEIR JOB!
BY "WHAT MUST BE COVERED" I AM INCLUDING THE SKILLS, PROCEDURES AND ESSENTIAL CONCEPTS OF MATHEMATICS. NONE OF THIS CONSTRAINS TEACHER STYLE OR CREATIVITY. BUT WITHOUT THIS STRUCTURE THERE IS ONLY THE CHAOS THAT CURRENTLY EXISTS. AND IF YOU DON'T THINK THERE IS CHAOS OUT THERE, TALK TO THE PROFESSIONALS WHO HAVE TO DO THIS JOB EVERY DAY.
Results of MathNotation's Third Online Math Contest
The Common Core State Standards Initiative
NCTM's latest response to the Core Standards Movement - the forthcoming Focus in High School Mathematics
Validation Committee selected for draft of Core Standards
The results of the latest round of ADP's Algebra 2 and Algebra 1 end of course exams
It will take several posts to cover all of this...
RESOURCES FOR YOU
MODEL PROBLEMS TO DEVELOP HIGHER-ORDER THINKING AND CONCEPTUAL UNDERSTANDING
Consider using the following as Warm-Ups to sharpen minds before the lesson and to provide frequent exposure to standardized test questions (SAT, ACT, State Assessments, etc.). I hope these problems serve as models for you to develop your own. I strongly urge you to include similar questions on tests/quizzes so that students will take these 5-minute classroom openers seriously.
I've provided answers and solutions/strategies for some of the questions below. The rest should emerge from the comments.
MODEL QUESTION #1:
For how many even integers, N, is N2 less than than 100?
Solution/Strategies:
Always circle keywords or phrases. Here the keywords/phrases include
"even integers"
N2
"less than".
This question is certainly tied to the topic of solving the quadratic inequality, N2 "<" 100 either by taking square roots with absolute values or by factoring. Of course, we know from experience, when confronted with this type of question on a standardized test, even our top students will test values like N = 2, 4, 6, ... However, the test maker is determining if the student remembers that integers can be negative as well and, of course, ZERO is both even and an integer! Thus, the values of N are -8,-6,-4,-2,0,2,4,6, and 8.
MODEL QUESTION #2
If 99 is the mean of 100 consecutive even integers, what is the greatest of these 100 numbers?
Solution/Strategies:
There are several key ideas and reasoning needed here:
(1) A sequence of consecutive even integers (or odd for that matter) is a special case of an arithmetic sequence.
(2) BIG IDEA: For an arithmetic sequence, the mean equals the median! Thus, the terms of the sequence will include 98 and 100. (Demonstrate this reasoning with a simpler list like 2,4,6,8 whose median is 5).
(3) The list of 100 even consecutive integers can be broken into two sequences each containing 50 terms. The larger of these starts with 100. Thus we are looking for the 50th consecutive even integer in a sequence whose first term is 100.
(4) The student who has learned the formula (and remembers it!) for the nth term of an arithmetic sequence may choose to use it: a(n) = a(1) + (n-1)d. Here, n = 50 (we're looking for the 50th term!), a(1) = 100, d = 2 and a(100) is the term we are looking for.
Thus, a(50) = 100 + (50-1)(2) = 198.
However, stronger students intuitively find the greatest term, in effect inventing the formula above for themselves via their number sense. Thus, if 100 is the first term, then there are 49 more terms, so add 49x2 to 100.
MODEL QUESTION #3: A SAMPLE OPEN-ENDED QUESTION FOR ALGEBRA II
If n is a positive integer, let A denote the difference between the square of the nth positive even integer and the square of the (n-1)st positive even integer. Similarly, let B denote the difference between the square of the nth positive odd integer and the square of the (n-1)st positive odd integer. Show that A-B is independent of n, i.e., show that A-B is a constant.
MODEL QUESTION #4:
GEOMETRY
If two of the sides of a triangle have lengths 2 and 1000, how many integer values are possible for the length of the third side?
MODEL QUESTION #5: GEOMETRY
There are eight distinct points on a circle. Let M denote the number of distinct chords which can be drawn using these points as endpoints. Let N denote the number of distinct hexagons which can be drawn using these points as vertices. What is the ratio of M to N?
Solution/Strategies: The student with a knowledge of combinations doesn't need to be creative here but a useful conceptual method is the following:
Each hexagon is determined by choosing 6 of the 8 points (and connecting them in a clockwise fashion for example). For each such selection of 6 points, there is a uniquely determined chord formed by the 2 remaining points. Similarly, for each chord formed by choosing 2 points, there is a uniquely determined hexagon. Thus the number of hexagons is in 1:1 ratio with the number of chords.
MODEL QUESTION #6: GEOMETRY AND THE ARITHMETIC OF PERCENTS
If we do not change the angle measures but increase the length of each side of a parallelogram by 60%, by what per cent is the area increased?
(A) 36% (B) 60% (C) 120% (D) 156% (E) 256%
## Monday, October 5, 2009
### Another Sample Contest Problem - Counting...
There is still time to register for the upcoming MathNotations Third Online Math Team Contest, which should be administered on one of the days from Mon October 12th through Fri October 16th in a 45-minute time period.
Registration could not be easier this time around. Just email me at dmarain "at" "gamil dot com" and include your full name, title, name and full address of your school (indicate if Middle or Secondary School).
Be sure to include THIRD MATHNOTATIONS ONLINE CONTEST in the subject/title of the email. I will accept registrations up to Fri October 9th (exceptions can always be made!).
BASIC RULES
* Your school can field up to two teams with from two to six members on each. (A team of one requires special approval).
* Schools can be from anywhere on our planet and we encourage homeschooling teams as well.
* The contest includes topics from 2nd year algebra (including sequences, series), geometry, number theory and middle school math. I did not include any advanced math topics this time around, so don't worry about trig or logs.
* Questions may be multi-part and at least one is open-ended requiring careful justification (see example below).
* Few teams are expected to be able to finish all questions in the time allotted. Teams generally need to divide up the labor in order to have the best chance of completing the test.
* Calculators are permitted (no restrictions) but no computer mathematical software like Mathematica can be used.
* Computers can be used (no internet access) to type solutions in Microsoft Word. Answers and solutions can also be written by hand and scanned (preferred). A pdf file is also fine.
Ok, here's another sample contest problem, this time a "counting" question that is equally appropriate for middle schoolers and high schoolers:
How many 4-digit positive integers have distinct digits and the property that the product of their thousands' and hundreds' digits equals the product of their tens' and units' digits?
The math background here may be middle school but the reading comprehension level and specific knowledge of math terminology is quite high. This more than counting strategies is often an impediment. If this were an SAT-type question, an example would be given of such a number to give access to students who cannot decipher the problem, thereby testing the math more than the verbal side. On most contests, however, anything is fair game!
Beyond understanding what the question is asking, I believe there are some worthwhile counting strategies and combinatorial thinking involved here. Enjoy it!
Click More to see the result I came up with (although you may find an error and want to correct it!)
## Sunday, October 4, 2009
### MathNotations Third Online Free Math Contest Update and Sample "Proof"
There is still time to register for the upcoming MathNotations Third Online Math Team Contest, which should be administered on one of the days from Mon October 12th through Fri October 16th in a 45-minute time period.
Registration could not be easier this time around. Just email me at dmarain "at" "gamil dot com" and include your full name, title, name and full address of your school (indicate if Middle or Secondary School).
Be sure to include THIRD MATHNOTATIONS ONLINE CONTEST in the subject/title of the email. I will accept registrations up to Fri October 9th (exceptions can always be made!).
• Your school can field up to two teams with from two to six members on each. (A team of one requires special approval).
• Schools can be from anywhere on our planet and we encourage homeschooling teams as well.
• The contest includes topics from 2nd year algebra (including sequences, series), geometry, number theory and middle school math. I did not include any advanced math topics this time around, so don't worry about trig or logs.
• Questions may be multi-part and at least one is open-ended requiring careful justification (see example below).
• Few teams are expected to be able to finish all questions in the time allotted. Teams generally need to divide up the labor in order to have the best chance of completing the test.
• Calculators are permitted (no restrictions) but no computer mathematical software like Mathematica can be used.
• Computers can be used (no internet access) to type solutions in Microsoft Word. Answers and solutions can also be written by hand and scanned (preferred). A pdf file is also fine.
The following is a sample of the open-ended "proof-type" questions on the test:
Explain why each of the following statements is true. Justify your reasoning carefully using algebra as needed.
The square of an odd integer leaves a remainder of 1 when divided by
(a) 2
(b) 4
(c) 8
I may post a sample solution to this or you can include this in your comments to this post.
## Wednesday, September 30, 2009
### Two Trains and a Tunnel! Is There Room For This In The Tunnel And In Your Curriculum?
At the same instant of time, trains A and B enter the opposite ends of a tunnel which is 1/5 mile long. Don't worry -- they are on parallel tracks and no collision occurs!
Train A is traveling at 75 mi/hr and is 1/3 mile long.
Train B is traveling at 100 mi/hr and is 1/4 mile long.
When the rear of train B just emerges from the tunnel, in exactly how many more seconds will it take the rear of train A to emerge?
Click on More to see answer (Feed subscribers should see answer immediately).
1. Appropriate for middle schoolers even before algebra? Exactly when are middle schoolers in your district introduced to the fundamental Rate_Time_Distance relationship?
2. What benefits do you think result from tackling this kind of exercise? If it's not going to be tested on your standardized tests, is it worth all the time and effort?
3. How much "trackwork" needs to be laid before students are ready for this level of problem-solving?
4. As an instructional strategy, would you have the problem acted out with models in the room or use actual students to represent the trains and the tunnel? OR just have them draw a diagram and go from there? Do a simulation on the TI-Inspire or TI-84 using graphics and parametric equations for the older students?
5. If you believe there is still a place for this type of problem-solving, should it be given only to the advanced classes and depicted as a math contest challenge?
6. I'm dating myself but I remember seeing problems like this in my old yellow Algebra 2 textbook? Uh, I believe this was B.C. -- before calculators! Can you imagine! Do you recall these kinds of problems? Do you recall the author or publisher?
7. Of course, the proverbial "two trains and tunnel" problems are frequently parodied and used as emblematic of the "old math"! They've been replaced by "real-world" applications. "Progress makes perfect!"
Answer: 9.4 seconds (challenge this if you think I erred!)
## Thursday, September 24, 2009
### More Challenges/SAT Practice, Core Curriculum Standards, Reminders, Comments...
Challenge 1:
HOW MANY DIGITS OF 10001000 - 1 WILL BE EQUAL TO 9 WHEN THIS EXPRESSION IS EXPANDED?
Challenge 2:
HOW MANY 5-DIGIT POSITIVE INTEGERS HAVE A SUM OF DIGITS EQUAL TO 43?
Challenge 3:
Jorge can run a 6-minute mile while Alex can run a 5-minute mile. If they start at the same time, how much less distance, in miles, will Jorge run in 10 minutes?
(Yes, you can respond with answers and solutions to these in the comments!)
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Tired of hearing about THIRD MATHNOTATIONS FREE ONLINE MATH CONTEST!? IF I RECEIVE 10 MORE REGISTRATIONS, I MAY JUST STOP!
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The Common Core State Standards Initiative
First look here for a quick overview and here for an index to the latest draft of the standards. Of course, this blog only discusses the mathematics part of the document.
Overview
The Common Core State Standards Initiative is a joint effort by the National Governors Association Center for Best Practices (NGA Center) and the Council of Chief State School Officers (CCSSO) in partnership with Achieve, ACT and the College Board. Governors and state commissioners of education from across the country committed to joining a state-led process to develop a common core of state standards in English-language arts and mathematics for grades K-12.
These standards will be research and evidence-based, internationally benchmarked, aligned with college and work expectations and include rigorous content and skills. The NGA Center and CCSSO are coordinating the process to develop these standards and have created an expert validation committee to provide an independent review of the common core state standards, as well as the grade-by-grade standards.
HIGHLIGHTS
• Each standard is broken into Core Concepts and Skills, provides research-based evidence and many illustrative examples to clarify the language
• Alignment of these standards to those of 5 representative states: California, Florida, Georgia, Massachusetts and Minnesota
• Standards reduce the number of Core Concepts and Skills in accordance with many recommendations to pare down the number of required topics to allow for greater depth
Example of a Standard (Standard 5)
Equations | see evidence
An equation is a statement that two expressions are equal. Solutions to an equation are the values of the variables in it that make it true. If the equation is true for all values of the variables, then we call it an identity; identities are often discovered by manipulating one expression into another.
The solutions of an equation in one variable form a set of numbers; the solutions of an equation in two variables form a set of ordered pairs, which can be graphed in the plane. Equations can be combined into systems to be solved simultaneously.
An equation can be solved by successively transforming it into one or more simpler equations. The process is governed by deductions based on the properties of equality. For example, one can add the same constant to both sides without changing the solutions, but squaring both sides might lead to extraneous solutions. Strategic competence in solving includes looking ahead for productive manipulations and anticipating the nature and number of solutions.
Some equations have no solutions in a given number system, stimulating the formation of expanded number systems (integers, rational numbers, real numbers and complex numbers).
A formula is a type of equation. The same solution techniques used to solve equations can be used to rearrange formulas. For example, the formula for the area of a trapezoid, A = ((b1 + b2)/2) h, can be solved for h using the same deductive process.
Inequalities can be solved in much the same way as equations. Many, but not all, of the properties of equality extend to the solution of inequalities.
Connections to Functions, Coordinates, and Modeling. Equations in two variables may define functions. Asking when two functions have the same value leads to an equation; graphing the two functions allows for the approximate solution of the equation. Equations of lines involve coordinates, and converting verbal descriptions to equations is an essential skill in modeling.
Core Concepts
Students understand that:
1. An equation is a statement that two expressions are equal.
see examples
2. The solutions of an equation are the values of the variables that make the resulting numerical statement true.
see examples
3. The steps in solving an equation are guided by understanding and justified by logical reasoning.
see examples
4. Equations not solvable in one number system may have solutions in a larger number system.
see examples
Core Skills
Students can and do:
1. Understand a problem and formulate an equation to solve it.
see examples
2. Solve equations in one variable using manipulations guided by the rules of arithmetic and the properties of equality.
see examples
3. Rearrange formulas to isolate a quantity of interest.
see examples
4. Solve systems of equations.
see examples
5. Solve linear inequalities in one variable and graph the solution set on a number line.
see examples
6. Graph the solution set of a linear inequality in two variables on the coordinate plane.
see examples
FUNDAMENTAL ASSUMPTIONS AND CONSIDERATIONS
Very Important!
(Click on image to see a clearer view)
INITIAL MATHNOTATIONS REACTIONS
1. Exceptionally clear and definitive document
2. Influenced by NCTM (Curriculum Focal Points), Achieve, College Board, ACT
3. Illustrative examples are of high quality
4. Will serve as a basis for states' revisions of current standards hopefully creating more consistency than currently exists
5. Leaving curriculum to local districts and states was a politically necessary decision, however, in my opinion, developing a reasonably consistent curriculum by grade level and/or course across districts and states from these standards may prove to be difficult and may again lead to considerable disparity. Hopefully, this will be self-correcting when standardized assessments are created as is currently being done with the End of Course Tests from Achieve
## Sunday, September 20, 2009
### A Practice PSAT/SAT Quiz with Strategies!!
UPDATE #2: Answers to the quiz are now provided at the bottom. If you disagree with any answers or would like clarification, don't hesitate to post a comment or send an email to dmarain "at gmail dot com".
UPDATE: No comments from my faithful readers yet -- I suspect they are giving students a chance to try these! I will post answers on Friday 9-25. However, students or any readers who would like to check their answers against mine need only email me at dmarain "at" gmail "dot" com and I will let them know how they did!
With the SAT/PSAT coming in a few weeks, I thought it would be helpful to your students to try a challenging "quiz". Most of these questions represent the high end level of difficulty and some are intentionally above the level of these tests. Then again, difficulty is very subjective. A student taking Honors Precalculus would have a very different perspective from the student starting Algebra 2!
Also, these questions can also be used to prepare for some math contests such as the THIRD MATHNOTATIONS FREE ONLINE MATH CONTEST! Yes, another shameless plug, but time is running out for your registration...
A Few Reminders For Students
(1) Do not worry about the time these take although I would suggest about 30 minutes. The idea is to try these, then correct mistakes and/or learn methods/strategies. It's what you do after this quiz that will be of most benefit!
(2) I added strategies and comments after the quiz. I suggest trying as many as you can without looking at these. Then go back, read the comments and re-try some. I will not provide answers yet!
(3) Don't forget these problems are copyrighted and cannot be reproduced for commercial use. See the Creative Commons License in the sidebar. Thank you...
PRACTICE PSAT/SAT QUIZ
1. If n is an even positive integer, how many digits of 1002n - 1002n-2 will be equal to 9 when the expression is expanded?
(A) 2 (B) 4 (C) 8 (E) 2n (E) 2n - 4
2. The sides of a triangle have lengths a, b and c. Let S represent (a+b+c)/2. Which of the following could be true?
I. S is less than c
II. S > c
III. S = c
(A) I only (B) II only (C) I and II only (D) I and III only (E) I, II and III
3. The mean, median and mode of 3 numbers are x, x+1 and x+1 respectively. Which of the following represents the least of the 3 numbers?
(A) x (B) x - 1 (C) x - 2 (D) x-3 (E) 2x - 2
4. (10/√5)500 (1/(2√5))500 = _________
5. A point P(x,y) lies on the graph of the equation x2y2 = 64. If x and y are both integers, how many such points are there?
(A) 4 (B) 8 (C) 16 (D) 32 (E 64
6. Each side of a parallelogram is increased by 50% while the shape is preserved. By what percent is the area of the parallelogram increased? __________
7.
AB is parallel to CD , AB = 3, CD = 5, AD = BC = 4. If segments AD and BC are extended to form a triangle ABE (not shown), what would be the length of AE?
Ans_________
Figure not drawn to scale
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1. Most students learn to substitute numbers for n here although it can be done algebraically by factoring. However, the real issue here is figuring out what the question is asking. Reading interpretation - ugh!!
2. When you are not given any information about what type of triangle it is, just choose a few special cases and draw a conclusion. O course, if one recalls a key inequality theorem from geometry, this problem can be done in short order.
3. If you don't feel comfortable setting this up algebraically (preferred method), PLUG IN A VALUE FOR x...
4. Your calculator may not be able to handle the exponent so skills are needed. The large exponent also suggests a Make it Simpler strategy. This is a "Grid-In" question so if one is guessing remember that most answers are simple whole numbers! Finally, if one knows their basic exponent rules and basic radical simplification, none of the above strategies are needed!
5. Possibilities should be listed carefully. It is possible to count these efficiently by recognizing the effect of reversals and signs. Easy to get this one wrong if not careful.
6. For those who do not remember or want to apply a key geometry concept about ratios in similar figures, there are a couple of essential test-taking strategies which all students should be aware of of:
(a) Consider a special case of a parallelogram
(b) choose particular values for the sides.
In the end, even strong students often make a different error, however. That darn ol' percent increase idea!
7. Should you skip this if you have no idea how to start? Absolutely not! Draw a complete diagram and even if you don't recognize the similar triangles, make an educated guess! It's a grid-in and there's no penalty for guessing. Further, answers tend to be positivc integers!!
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1. B
2. B
3. C
4. 1
5. C
6. 125
7. 6
## Thursday, September 17, 2009
### Demystifying Per Cent Problems Part II - Using Multiple Representations and an SAT Problem
Have you forgotten to register for MathNotation's Third FREE Online Math Contest coming in mid-October? We already have several schools (from around the world!) registered. For details, link here or check the first item in the right sidebar!!
Before tackling a more challenging problem in the classroom, I would typically begin with one or more simpler examples. My objective was to review essential concepts and skills and demonstrate key ideas in the harder problem. This incremental approach (sometimes referred to as scaffolding) enabled some students to solve the problem or at least get started. Usually within each group of 3-4 students, there was at least one who could help the others. Some groups or classes might still not be ready after one example, so more would be needed. I never felt that this expense of time was too costly since my goal was to develop both skill and understanding.
SIMPLER EXAMPLE
Consider the following two statements about positive numbers A and B:
(1) A is 80% of B.
(2) A is 20%
less than B .
Are these equivalent, that is, if values of A and B satisfy (1), will they also hold true for (2) and conversely?
How would you get this idea across to your students?
Again, depending on the students, I would often allow them to discuss it first in small groups for two minutes, then open up the discussion.
Note: If the group lacks the skills, confidence or background (note that I left ability out, intentionally!), I might first start with concrete values before giving them the 2 statements above: E.g., What is 80% of 100?
How would I summarize the methods of solution to this question. Here's what I attempted to do in each lesson. I didn't reach everyone but I found from further questioning and subsequent assessment that this multi-pronged approach was more successful than previous methods I had used. Most of these methods came from the students themselves!
INSTRUCTIONAL STRATEGIES
I. Choose a particular value for one of the numbers, say B = 100. Ask WHY it makes sense to start with B first and why does it make sense to use 100. Calculate the value of A and discuss.
II. Draw a pie chart (circle graph) showing the relationship between A and B. Stress that B would represent the whole or 100%.
III. Write out the sentence:
80% of B is the same as 100% of B - 20% of B
In other words:
80% of B is the same as 20% less than B.
IV. Express algebraically (as appropriate):
0.8B = 1B - 0.2B
Numerical (concrete values)
Visual (Pie chart)
Verbal (using natural language)
Symbolic (algebra)
Yes, it's Multiple Representations! The Rule of Four!
To me, it's all about accessing different modes of how students process. Call it learning styles, brain-based learning, etc., it still comes down to:
RARELY DOES ONE METHOD OF EXPLANATION, NO MATTER HOW CLEAR OR STRUCTURED, REACH A MAJORITY OF STUDENTS. YOUR FAVORITE EXPLANATION WILL MAKE THE MOST SENSE TO THE STUDENTS WHO THINK LIKE YOU!!
Now for today's challenge.
(Assume all variables represent positive numbers)
M is x% less than P and N is x% less than Q. If MN is 36% less than PQ, what is the value of x?
Can you think of several methods?
I will suggest one of the favorite of many successful students on standardized assessments:
Choose P = 10, Q = 10. Then...
Click on More (subscribers do not need to do this) to see the answer without details.
## Sunday, September 13, 2009
### Demystifying Harder Per Cent Word Problems for Middle Schoolers and SATs - Part I
Example I
40% of the the Freshman Calculus class at Turing University withdrew. If 240 students left, how many were in the class to start?
Solution without explanation or discussion:
0.4x = 240 ⇒ x = 600
Example II
40% of the the Freshman Calculus class at Turing University withdrew. If 240 students were left, how many were in the class to start
?
Solution without explanation or discussion:
0.6x = 240 ⇒ x = 400
Thinking that the issues in the problems above are more language-dependent than based on learning key mathematics principles or effective methods? I would expect that many would say that using the word "left" in both problems was unnecessarily devious and that clearer language should be used to demonstrate the mathematics here. Perhaps, but when I taught these types of problems I would frequently juxtapose these types of questions and intentionally use such ambiguous language to generate discussion - creating disequilibrium so to speak. If nothing else, the students may become more critical readers! Further, the idea of using similar but contrasting questions is an important heuristic IMO.
Even though I've been a strong advocate for a standardized math curriculum across the grades, I fully understand that the methods used to present this curriculum are even more crucial. Instructional methods and strategies are often unpopular topics because they seem to infringe on individual teacher's style and creativity. BUT we also know that some methods are simply more effective than others in reaching the maximum number of students (who are actually listening and participating!). I firmly believe there are some basic pedagogical principles of teaching math, most of which are already known to and being used by experienced teachers.
Percent word problems are easy for a few and confusing to many because of the wide variety of different types.
Here are brief descriptions of some methods I've developed and used in nearly four decades in the classroom.
I. (See diagram at top of page)
The Pie Chart builds a strong visual model to represent the relationships between the parts and the whole and the "whole equals 100%" concept. How many of you use this or a similar model ? Please share! There's more to teaching this than drawing a picture but some students have told me that the image stays longer in their brain. I learn differently myself but I came to learn the importance of Multiple Representations to reach the maximum number of students.
II. "IS OVER OF" vs. "OF MEANS TIMES"
The latter is generally more powerful once the student is in Prealgebra but, of course, the word "OF" does not appear in every percent so many different variations must be given to students and practiced practiced practiced practiced over time. The first method can be modified as a shortcut in my opinion to find a missing percent and that may be its greatest value. However many middle schoolers use proportions for solving ALL percent problems. I personally do NOT recommend this!
Well, I could expound on each of these methods ad nauseam and bore most of you, but I think I will stop here and open the dialg for anyone who has strong emotions about teaching/learning per cents...
## Monday, September 7, 2009
### Using Number Theory To Promote Logic and Writing in Middle Schoool and Beyond
The following examples also provide practice for open-ended questions and a view of the Explain or Show type questions on our next Online Math Contest to be held in 5 weeks (see info below). Since formal proof is not the goal here, students are encouraged to write a logical chain of reasoning in which they can use/assume basic knowledge about odd and even integers. Further, these questions strongly suggest the strategies consider a simpler case first and patterning.
Another benefit of these types of questions is to review important terminology and to help students improve reading comprehension, a major obstacle for many youngsters in math class (and everywhere else!). Some middle schoolers and high schoolers will have difficulty making sense out of what the question is asking because of both the wording and the information load in the problem. We need to help them group key phrases together and, yes, I guess that means we are also reading teachers!
Example 1
Is the sum of the squares of the first 2009 positive integer multiples of three odd or even? Explain your reasoning.
Example 2
Is the sum of the squares of the first 2010 positive integer multiples of three odd or even? Explain your reasoning.
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REMINDER!
MathNotations' Third Online Math Contest is tentatively scheduled for the week of Oct 12-16, a 5-day window to administer the 45-min contest and email the results. As with the previous contest, it will be FREE, up to two teams from a school may register and the focus will be on Geometry, Algebra II and Precalculus. If any public, charter, prep, parochial or homeschool (including international school) is interested, send me an email ASAP to receive registration materials: "dmarain 'at' gmail dot com."
(1) The first draft of the contest is now complete.
(2) As with the precious two contests there will be one or two questions which require demonstration, that is, the students will have to derive, explain or prove a statement. This is best done freehand and then scanned as a jpeg image which can be emailed as an attachment along with the official answer sheet. In fact, the entire answer sheet can be scanned but there is information on it that I need to have.
(3) Some of the questions are multipart with the last part requiring more generalization.
(4) Even if you have previously indicated that you wish to participate, please send me another email using the title: THIRD MATHNOTATIONS CONTEST. Please copy and paste that into the title. Also, when sending the email pls include your full name and title (advisor, teacher, supervisor, etc.), the name of your school (indicate if HS or Middle School) and the complete school address. I have accumulated a database of most of the schools which have expressed interest or previously participated but searching through thousands of emails is much easier when the title is the same! If you have already sent me an email this summer or previously participated, pls send me one more if interested in participating again.
Note: Sending me the email is not a commitment! It simply means you are interested and will receive a registration form.
## Saturday, August 29, 2009
### Batteries Required! A Combinatorial Problem MS /HS Students Can Use...
Have you ever inserted batteries in a device only to find that it didn't work? You reverse the batteries and try again, but no luck. You can't find the polarity diagram to guide you and you're dealing with 3 or 4 batteries and all the possible combinations! Well, that just happened to me as I was inserting 3 'C' batteries into a new emergency lantern I just purchased. There was no guide that I could see. I knew there were 8 possibilities but it was late and my patience quickly ran out. I tried it again the following morning, shone my small LED light on it and saw the barely visible diagram.
After seeing the lantern finally operate, I realized I should have used a methodical approach -- practice what I preach!! Then I thought that this might be a natural application of the Multiplication Principle one could use in the classroom. Of course, it would work nicely if you happened to have the identical lantern but you might have some of these in the building or at home which take 2 or more batteries. IMO, there's something very real and exciting about solving a math problem and seeing the solution confirmed by having "the light go on!" I'll avoid commenting on the obvious symbolism of that quoted phrase...
Instructional/Pedagogical Considerations
(1) I would start with a small flashlight requiring only one battery to set up the problem. For this simplest case, students should be encouraged to describe the correct placement in their own words and on paper.
(2) Would you have several flashlights/lanterns available, one for each group of 2-4 students or would you demonstrate the problem with one device and call on students to suggest a placement of the batteries? Needless to say, if you allow students to work with their own flashlights, they will look for the polarity diagram so you will need to cover those somehow. That is problematic!
(3) Do you believe most middle school students (if the polarity diagram is not visible) will randomly dump in the batteries to get the light to go on and be the first to do so? Is it a good idea to let them do it their way before developing a methodical approach? Again, if a student or group solves the problem, it is important to have them write their solution before describing it to the class. If there is more than one battery compartment, students should realize realize the need to label the compartments such as A, B, C , ... Once they reach 3 or more batteries, they should recognize that a more structured methodical approach is needed so that one doesn't repeat the same battery placement or miss one. One would hope!
(4) Is it a drawback that the experiment will probably end (i.e., the light goes on) before exhausting all possible combinations? How would we motivate students to make an organized list or devise a methodical approach if the light goes on after the first or second placement of the batteries?
(5) I usually model these kinds of problems using the so-called "slot" method. Label the compartments A, B, ... for example and make a "slot" for each. For two compartments we have
A B
_ _
Under each slot, I list the possibilities, e.g., (+) end UP or DOWN (depending on the device, other words may be more appropriate). Here I would only concern myself with labeling the (+) end, the one with the small round protruding nub. For this problem I would write the number (2) on each slot since there are only TWO ways for each battery to be placed. Note the use of (..). In general, above each slot I would write the number of possibilities. For two compartments (or two batteries), the students would therefore write (2) (2). They know the answer is 4 but some will think we are adding rather than multiplying. Ask the class which operation they believe will always work. How would you express your questions or explanation to move students toward the multiplication model? The precise language we use is of critical importance and we usually only learn this by experimentation. If one way of expressing it doesn't seem to click with some students, we try another until we refine it or see the need for several ways of phrasing it. This is the true challenge of teaching IMO. We can plan all of this carefully ahead of time, but we don't know what the effect is until we go "live" (or have experienced it many times!).
Perhaps you've already used a similar application in the classroom - please share with us how you implemented it. Circuit diagrams in electronics also lend themselves nicely to this approach. Typically, I've used 2, 3 or more different coins to demonstrate the principle but the batteries seem to be a more natural example, although I see advantages and disadvantages to both. At least with the batteries, students should not question the issue of whether "order counts!"
I could say much more about developing the Multiplication Principle in the classroom, but I would rather hear from my readers.
If you've used other models to demo this key principle, let us know...
REMINDER!
MathNotations' Third Online Math Contest is tentatively scheduled for the week of Oct 12-16, a 5-day window to administer the 45-min contest and email the results. As with the previous contest, it will be FREE, up to two teams from a school may register and the focus will be on Geometry, Algebra II and Precalculus. If any public, charter, prep, parochial or homeschool (including international school) is interested, send me an email ASAP to receive registration materials: "dmarain 'at' gmail dot com."
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Home » MP Board Class 10th Maths Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.1 Solution
# MP Board Class 10th Maths Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.1 Solution
In this article, we will share MP Board Class 10th Maths Book Solutions Pair of Linear Equations in Two Variables Exercise 3.1 Pdf solution file.
## MP Board Class 10th Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.1 Solution
Question 1.
Aftab tells his daughter, “Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be”. (Isn’t this interesting?) Represent this situation algebraically and graphically.
Solution:
At present: Let Aftab’s age = x years
His daughter’s age = y years
Seven years ago : Aftab’s age = (x – 7) years
His daughter’s age = (y – 7)years
According to the condition,
[Aftab’s age] = 7[His daughter’s age]
⇒ [x – 7] = 7[y – 7] = x – 7 = 7y – 49
⇒ x – 7y – 7 + 49 = 0 ⇒ x – 7y + 42 = 0 …. (1)
After three years : Aftab’s age = (x + 3) years
His daughter’s age = (y + 3) years
According to the condition,
[Aftab’s age] = 3[His daughter’s age]
⇒ [x + 3] = 3[y + 3]
⇒ x + 3 = 3y + 9 ⇒ x — 3y + 3 — 9 = 0
⇒ x — 3y – 6 = 0 …. (2)
Graphical representation of equation (1) and (2): From equation (1), we have :
### MP Board Class 10th Maths Exercise 3.1 Solution
Question 2.
The coach of a cricket team buys 3 bats and 6 balls for ₹ 3900. Later, she buys another bat and 3 more balls of the same kind for ₹ 1300. Represent this situation algebraically and geometrically.
Solution:
Let the cost of a bat = ₹ x
and the cost of a ball = ₹ y
Cost of 3 bats = ₹ 3x
and cost of 6 balls = ₹ 6y
Again, cost of 1 bat = ₹ x
and cost of 3 balls = ₹ 3y
Algebraic representation:
Cost of 3 bats + Cost of 6 balls = ₹ 3900
⇒ 3x + 6y = 3900 ⇒ x + 2y = 1300 …. (1)
Also, cost of 1 bat + cost of 3 balls = ₹ 1300
⇒ x + 3y = 1300 …. (2)
Thus, (1) and (2) are the algebraic representations of the given situation.
Geometrical representation:
We have for equation (1),
We can also see from the obtained graph that the straight lines representing the two equations intersect at (1300, 0).
10th Maths Exercise 3.1 Solutions
Question 3.
The cost of 2 kg of apples and 1 kg of grapes on a day was found to be ₹ 160. After a month, the cost of 4 kg of apples and 2 kg of grapes is ₹ 300. Represent the situation algebraically and geometrically.
Solution:
Let the cost of 1 kg of apples = ₹ x
And the cost of 1 kg of grapes = ₹ y
Algebraic representation:
2x + y = 160 … (1)
and 4x + 2y = 300
⇒ 2x + y = 150 … (2)
Geometrical representation:
We have, for equation (1),
The straight lines l1 and l2 are the geometrical representations of the equations (1) and (2) respectively. The lines are parallel.
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# Number of Cubies problem
#### Igora
##### Member
I've been working out an equation for the number of cubies in a regular cube (not helicopter, skewb etc.) so far I've got c=6x^2-8x+4 where x is the number of cubies on one side of the cube, for instance in a 7x7x7 x=7. The only problem with this equation is that it doesn't work for a 1x1x1 cube. Any thoughts on one that would?
#### brunson
##### Member
I don't think it works for 2, either.
x^3 - (x-2)^3 works a bit better. That's how many blocks if the cube was solid, minus the number of cubies that would fit inside it if it were hollow.
(But it still doesn't work for 1)
Last edited:
#### cmhardw
I've been working out an equation for the number of cubies in a regular cube (not helicopter, skewb etc.) so far I've got c=6x^2-8x+4 where x is the number of cubies on one side of the cube, for instance in a 7x7x7 x=7. The only problem with this equation is that it doesn't work for a 1x1x1 cube. Any thoughts on one that would?
6*9 - 8*3 + 4 = 34
So it does not work for the 3x3x3 either.
Hint: You are on the right track, keep going with the corrections. For example, you know the 6*x^2 is an overcount, and counts more pieces than are actually on the cube. Take a look at your method for fixing this overcount (i.e. the other terms in your formula).
Chris
#### cmhardw
1x1 doesn't work though...
do you need a formula to tell you how many cubies are in a 1x1?? >_>
The 1x1x1 case fails on most formulas involving the n x n x n cube. Usually we just say that for the formula n > 1 must be true. It might be possible to come up with one single explicit formula that includes the 1 x 1 x 1 case correctly, but I don't know of one. That's not to say that it hasn't been done though.
Chris
#### hawkmp4
##### Member
c=x^3-(x-2)^3
Yep Though usually people write the simplified form for this.
Chris
1x1 doesn't work though...
do you need a formula to tell you how many cubies are in a 1x1?? >_>
No, but having an equation that works more generally is...well...nicer.
#### Igora
##### Member
oops, I accidentaly posted the wrong one :fp, the one I meant to post is c=6x^2-12x+8 rather than 6x^2-8x+4
#### hawkmp4
##### Member
Are we counting centres as cubies? I wouldn't think so...in which case x^3-(x-2)^3 would over count.
Hm...this is more complicated than I thought...I originally thought to just subtract 6 for centres...but that will only work for odd x.
Well, regardless, for odd x, x not equal to one, x^3-(x-2)^3-6 should work. I will have to think about the even case...
#### Igora
##### Member
Are we counting centres as cubies? I wouldn't think so...in which case x^3-(x-2)^3 would over count.
Hm...this is more complicated than I thought...I originally thought to just subtract 6 for centres...but that will only work for odd x.
Well, regardless, for odd x, x not equal to one, x^3-(x-2)^3-6 should work. I will have to think about the even case...
Well, I count them as cubies, but if you don't, the even's case would be simply x^3-(x-2)^3 without the added -6.
#### hawkmp4
##### Member
Are we counting centres as cubies? I wouldn't think so...in which case x^3-(x-2)^3 would over count.
Hm...this is more complicated than I thought...I originally thought to just subtract 6 for centres...but that will only work for odd x.
Well, regardless, for odd x, x not equal to one, x^3-(x-2)^3-6 should work. I will have to think about the even case...
Well, I count them as cubies, but if you don't, the even's case would be simply x^3-(x-2)^3 without the added -6.
Yes, that's clear. I was hoping to come up with an explicit equation that works for all x. Well. For all natural numbers.
#### Lucas Garron
##### Super-Duper Moderator
Staff member
It might be possible to come up with one single explicit formula that includes the 1 x 1 x 1 case correctly, but I don't know of one.
Getting such a formula is trivial, of course, by adjusting for n=1.
Now, why is there no polynomial that'll work for all positive integers?
#### cmhardw
It might be possible to come up with one single explicit formula that includes the 1 x 1 x 1 case correctly, but I don't know of one.
Getting such a formula is trivial, of course, by adjusting for n=1.
Now, why is there no polynomial that'll work for all positive integers?
I don't really know to be honest. I've been trying to think of how to use the floor, ceiling, or mod functions to accomplish one formula that mimics the "standard" one, and also allows for the n=1 case.
Other than your idea of making a piecewise function defining the n=1 case, I don't know why a single polynomial solution won't work.
We know that the polynomial
$$P(n)=6n^2-12n+8$$
works perfectly for $$n>1$$. P(n) fails for n=1, so we must assume there exists another polynomial Q(n) such that Q(n) = P(n) for all natural numbers n>1, but $$Q(1) \not= P(1)$$ must be true. In fact, P(1)=2 and Q(1) we want to be equal to 1.
My higher math is apparently not very sound, because it is not immediately apparent to me why Q(n) does not exist.
Chris
#### hawkmp4
##### Member
It might be possible to come up with one single explicit formula that includes the 1 x 1 x 1 case correctly, but I don't know of one.
Getting such a formula is trivial, of course, by adjusting for n=1.
Now, why is there no polynomial that'll work for all positive integers?
I can easily see why there is no quadratic that works. But I suppose that's rather trivial. I'm with Chris, I don't see why a polynomial couldn't exist.
#### Lucas Garron
##### Super-Duper Moderator
Staff member
I don't really know to be honest. I've been trying to think of how to use the floor, ceiling, or mod functions to accomplish one formula that mimics the "standard" one, and also allows for the n=1 case.
There's always the standard absolute value cheat:
x^3 - (x - 2)^3 + (x - Abs[x - 2] - 2)/2
I can easily see why there is no quadratic that works. But I suppose that's rather trivial. I'm with Chris, I don't see why a polynomial couldn't exist.
No finite polynomial will work for the same reason no quadratic will work:
Take R = P-Q. R must be 0 at every positive integer >1. The only finite polynomial with this property is 0. What Chris said, really.
(You can't "hack" a polynomial at specific parts while preserving infinitely many other values.)
#### hawkmp4
##### Member
I don't really know to be honest. I've been trying to think of how to use the floor, ceiling, or mod functions to accomplish one formula that mimics the "standard" one, and also allows for the n=1 case.
There's always the standard absolute value cheat:
x^3 - (x - 2)^3 + (x - Abs[x - 2] - 2)/2
I can easily see why there is no quadratic that works. But I suppose that's rather trivial. I'm with Chris, I don't see why a polynomial couldn't exist.
No finite polynomial will work for the same reason no quadratic will work:
Take R = P-Q. R must be 0 at every positive integer >1. The only finite polynomial with this property is 0. What Chris said, really.
(You can't "hack" a polynomial at specific parts while preserving infinitely many other values.)
That was what I was thinking, but I had absolutely no idea how to explain it mathematically. Thanks!
#### meichenl
##### Member
We could write a recursion relation.
For $$n=0$$ we need $$0$$ cubies, so
$$S_0 = 0$$.
Imagine taking an $$n-1 * n-1$$ cube and adding cubies to make it into an $$n * n$$ cube.
$$S_n = S_{n-1} + ...$$
You'd have to add a layer of cubies around the outside. Add an $$n * n$$ face to R with the extra cubies hanging above U and B. Then add an $$(n-1) * (n-1)$$ block to U. Finally add $$(n-1) * n$$ to B. That comes to $$3n^2 - 3n + 1$$ cubies added.
$$S_n = S_{n-1} + 3n^2 - 3n + 1 ...$$
But some old cubies got covered up completely. This happened to stuff in the URB corner of the smaller cube and its vicinity. What got covered up was basically an $$n-2 * n-2$$ cube.
$$S_n = S_{n-1} + 3n^2 - 3n + 1 - S_{n-2}...$$
Part of that $$S_{n-2}$$ cube had already been covered, though, so we have to add back in a cube of size $$S_{n-3}$$
$$S_n = S_{n-1} + 3n^2 - 3n + 1 - S_{n-2} + S_{n-3}...$$
et cetera
$$S_n = S_{n-1} + 3n^2 - 3n + 1 - S_{n-2} + S_{n-3} - S_{n-4} + S_{n-5}... S_1$$
If we plug in the same thing evaluated for $$S_{n-1}$$, that whole long alternating summation cancels out, and we have
$$S_n = 3n^2 - 3n + 1 + 3(n-1)^2 - 3(n-1) + 1$$
which is the same equation mentioned earlier.
It doesn't work for $$S_1$$, though, because $$S_0$$ is simply assigned the value $$0$$. We cannot make the substitution
$$S_0 = 0^2 + -3*0 + 1 + ...$$ because the recursion only applies to higher $$n$$.
Instead we just have $$S_1 = 3*1^2 - 3*1 + 1 + S_0 = 1$$.
#### mrCage
##### Member
Why use the number of cubies on a single face as the variable x, and not simply the cube dimension?
Per
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• Overview
• Sample Size: Precision
• Sample Size: Reliability
• Precision vs. Reliability
• Bias: Overview
• Selection Bias
• Selection Bias
• Response Bias
• Quiz: Question 1
• Quiz: Question 2
## Sampling Strategy Overview
A sampling strategy should answer two questions:
1. How much data should be collected?
2. How should the data be collected?
How much data is needed depends on how confident we want to be that the sample values correspond to the population values. Once we have decided how precise we want our estimate to be, we can use sampling theory to determine our desired sample size. We can also reverse the process to get a sense of how accurate our estimate is, based on the number of respondents we can reach.
This tutorial explains the key things to consider when determining a sample size, without focusing on the mathematics. The second question, how the data should be collected, will be explored in depth in the separate tutorial on tool selection. It will also be considered briefly here, in the context of how data collection might bias the data.
## Sample Size: Precision
### What sample size you need for your study depends primarily on two factors: How precise you want your estimate to be and how confident you want to be that your estimate is accurate.
The first concept we need to understand in picking a sample size is precision, also referred to as margin of error.
Say that you want to know how safe people feel in a refugee camp housing 5,000 individuals. You are going to distribute a questionnaire that contains this question:
#### How safe do you feel?
Not at all safe Completely safe
How many people do you need to ask to get a sense of what the full population is experiencing?
To calculate this, we first need to decide how precise we want our estimate to be. In this case, let us say we want no proportion in the sample to deviate by more than 5% from what we see in the population. This means that if 35% of our sample says that they feel completely safe, we can feel confident that the proportion in the full sample is no less than 30% and no more than 40%.
This + or - 5% figure is a good standard of precision for most surveys. If you use supplementary tools to validate your data — through, for example, community group discussions or key informant interviews or if you have multiple data sources that you can triangulate with one another — you can use a higher figure such as + or - 10%.
## Sample Size: Reliability
### Reliability refers to confidence levels.
Once you have decided what level of precision you want, you have to determine how reliable it should be. Reliability tells us how confident we can be that the population value falls inside our precision estimate (+ or - 5% in this example).
The easiest way to think about reliability is to imagine repeating the survey 100 times, picking the same number of random people from the population each time. In this case, a reliability of 95% would mean that in 95 of our 100 surveys each response category would have proportions similar to the full population proportions (within our + or - 5% range). However, in five cases, at least one category would fall outside these proportions.
Statistical convention uses 95% as the most common threshold for reliability, but that is just a convention. You could pick a reliability of 99% or 90% or any other value (but the lower the value, the more careful you have to be in interpreting your results).
## The Relationship Between Precision and Reliability
### Precision and reliability together determine sample size for multiple-choice questions. Strike a balance point between the two.
There is a tradeoff between reliability, precision and the number of respondents, with the number of respondents increasing ever more steeply for higher precision. The key thing is to make an informed decision that accounts for both the importance of getting accurate information and the cost of collecting data.
Given most settings, a sample of 400 is sufficient to be able to draw good conclusions about an entire population. If you are not looking for representative samples or you don't have multiple data sources, which means you're less dependent on representative samples, a sample of 100-200 is sufficient.
## Bias: Overview
### Your sampling strategy can bias your data in two different ways, by selection bias or by response bias.
Your data is biased when the responses in your sample systematically differ from the attitudes of the full population. There are two ways sampling strategies can bias the data:
• Either every person in the population is not equally likely to get selected for the sample, or
• the data collection method makes people compelled to respond in a specific way.
When some people are more likely to get selected than others we call it a selection bias, when people feel compelled to respond differently from what they actually believe we call it a response bias.
## Selection Bias
### When some people are more likely to get selected than others.
Imagine that you want to conduct a poll to predict the outcome of a US presidential election. To get a good estimation, you plan to ask 10,000 people how they are going to vote. You decide to advertise your poll on Buzzfeed. About 3,000 of your respondents are sure they are going to vote, and of that 3,000, 2,500 (83%) say they will vote Democrat.
This proportion seems unlikely to be right, so you decide to conduct a second survey. The sample of 10,000 people remains the same but this time you pick random numbers from a phone book. Some 5,000 people report they are going to vote, and 3,000 of them (60%) say they will vote Republican.
In the first survey you have voter turnout rate of 30% suggesting the Democrats will get 83% of the votes. In the second case you have a 50% voter turnout, suggesting the Republicans will garner 60% of the votes.
## Nope!
### The best conclusion is that neither poll is likely to be accurate, because neither polling method allowed you to reach the entirety of the electorate.
Buzzfeed is an entertainment website focusing on young adults, so most of your sample was probably college students. While you might have gotten a good sense of voting intentions and political preference among the American college community, you might not among the American population at large.
The second method is better, but it fails to reach many young people who do not register land lines, people who are homeless, or people who don’t have a phone for some other reason.
The point of this example is to illustrate that how you collect your responses will influence who is likely to respond, and if you only get responses from a sub-population that is systematically different from the total population, chances are that your response estimates will be different as well.
## Response Bias
### When people feel compelled to respond differently than they actually believe.
Now imagine that you are conducting the same survey to find out who will be the next US president, but this time you have volunteers from the Democratic Party asking people on the street.
If you choose this sampling method it is likely that a larger proportion of your sample will report supporting Democrats than you will see in the actual election. This is because most people tend to respond in a way that pleases the interviewer. This is known as the courtesy bias.
A related bias is the conformity bias, where people tend to respond in a way that is favored by their social group, regardless of their own opinion. This type of bias is particularly strong if respondents feel that their responses are not anonymous.
Using independent third party data collectors or anonymous survey tools can be a good way to reduce response bias. Regularly discussing the data with communities is also a good way to get beyond both courtesy and conformity bias. See the dialogue and course correction tutorial for more information.
## Test Yourself: Question 1
### What type of bias, if any, could be skewing your results?
You want to investigate sexual health among young women in a refugee camp in Jordan, just across the border from Syria. You pick five random areas of the camp and have a focus group with 30 young women in each area and ask them about their experience. No one in your sample claims to be sexually active.
Is this because of:
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# Confirm my logic on power
Ok,
So, lets say I was to drag a 40kg load in a set distance of 10 meter. I record the time, it was 5 secs.
Now I find power, P=work/t. We know work=force x distance.
SO my question is, is force just mass of load x acceleration due to gravity? 9.81m/s^2.
So P= ((40kg x 9.81) * 10 meters)/5 seconds,
so 784 watts or so. Thanks guys. What about force of friction? I guess this question doesn't want me to go that far in yet.
Thanks guys.
Related Introductory Physics Homework Help News on Phys.org
I don't quite agree with the way you get work and force.
The work you done is against the friction force, not the weight of the load.
Hence force = mg is not applicable.
In fact, work needed to drag an object for different surface is different due to difference in coefficient of friction.
Greater mass do need greater force to drag, but that is because the friction is higher, as the frictional force is greater for a heavier object.
I don't quite agree with the way you get work and force.
The work you done is against the friction force, not the weight of the load.
Hence force = mg is not applicable.
In fact, work needed to drag an object for different surface is different due to difference in coefficient of friction.
Greater mass do need greater force to drag, but that is because the friction is higher, as the frictional force is greater for a heavier object.
So how does one do it in terms of the equation?
a = (v-u)/t
given u = 0, and v = s/t
a = s/t^2
= 0.4 ms^-2
(assuming non-frictional planar flat surface, linear acceleration)
thats what i think, at least
Last edited:
perfect!!! so the force is 40 * 0.4...and that by 5 gives you the power!
I see. So V=d/t, then use that as a final V in a=Vf-Vi/t, a=.4 m/s^2 on a non frictional surface.
Sorry, I just didn't get the formulas Xiankai was using.
I see! so if I had a coefficient of friction then I all I would do is go (40kg x 0.4 m/s^2) - u (mu) x F normal (m x g)? Excellent. Then I'll probably have to consider the static or kinetic friction too... lol, lucky I don't have to do this.
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# Maple Tutorial Differential Equations
Maple Guide To Differential Equations PDF cccic.ca. dsolve solve ordinary differential equations (ODEs) Calling Sequence Parameters Description Examples Details Calling Sequence dsolve( ODE ) dsolve( ODE , y(x, Maple: Solving Ordinary Differential Equations A differential equation is an equation that involves derivatives of one or more unknown func-tions..
### Maple/Simultaneous Equations PrattWiki - Duke University
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This document contains links to the following Maple resources: Maple Tutorials and Information Guides Examples for Various Courses Differential Equations. Maple Tutorial to accompany Partial Differential Equations: Analytical and Numerical Methods, 2nd edition by Mark S. Gockenbach (SIAM, 2010) Introduction
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### Solving Differential Equations Using Simulink
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### Maple/Differential Equations/RC Example PrattWiki
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This web site provides an introduction to the computer algebra system Maple, created by MapleSoft В©. This tutorial is made solely for the purpose of education, and I have encountered the following system of differential equations Solving two coupled non-linear second order differential equations numerically. In Maple it
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## Differential Equations 2280 Spring 2016 Home - Math
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### Matthias KawskiArizona State UniversityMaple Worksheets
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### Lesson 1 Introduction to Differential Equations in Maple
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### Matthias KawskiArizona State UniversityMaple Worksheets
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• Matthias KawskiArizona State UniversityMaple Worksheets
• Maple Solving Ordinary Differential Equations
• This document contains links to the following Maple resources: Maple Tutorials and Information Guides Examples for Various Courses Differential Equations. Partial Differential Equations and Boundary Value Problems with Maple, Second Edition, presents all of the material normally covered in a standard course on partial
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## How do you do multiple matrices in Excel?
Excel MMULT Function
1. Summary. The Excel MMULT function returns the matrix product of two arrays.
2. Perform matrix multiplication.
3. The matrix product of two arrays.
4. =MMULT (array1, array2)
5. array1 – The first array to multiply. array2 – The second array to multiply.
## Can Excel solve matrix?
To solve using matrices, use the equation Ax=b where matrix A is the coefficient matrix, x is the variable matrix and b is the matrix of given solutions. Therefore, x=A-1*b. The A^(-1)*b matrix is the matrix with values of x, y, and z. This gives the answers x= 4.71, y= -1.026 and z= 3.113.
Can Excel invert a matrix?
The Excel MINVERSE function returns the inverse matrix of a given array. The input array must contain numbers only and be a square matrix, with equal rows and columns. The result is an inverse matrix with the same dimensions as the array provided.
### How is Minverse calculated?
To solve using MINVERSE and MMULT:
1. Enter the coefficient and constant array matrices.
2. Select a four-cell area (A4:B5 in the example) to hold the results of the formula.
3. Type in the formula MINVERSE (A1:B2).
4. Press CTRL + SHIFT + ENTER to enter the array formula so that it applies to all four cells in the selected area.
### What does the glulookat function do in Excel?
The direction of the up vector. The direction of the up vector. The direction of the up vector. This function does not return a value. The gluLookAt function creates a viewing matrix derived from an eye point, a reference point indicating the center of the scene, and an up vector.
How to do matrix multiplication in Excel using mmult?
Step 1: Select all the cells (A7:B8) from Resultant Matrix to apply the formula at once. Step 2: Inside the active cell (cell A7), start initiating the formula for matrix multiplication. Use =MMULT ( in the cell to initiate the formula.
#### Which is the matrix multiplication function in Excel?
Since a worksheet is essentially a gigantic matrix, it’s no surprise that matrix multiplication in Excel is super easy – we just need to use the MMULT Excel function. You can multiply matrices in Excel thanks to the MMULT function. This array function returns the product of two matrices entered in a worksheet.
#### Which is the matrix product of two arrays in Excel?
Excel MMULT Function. The Excel MMULT function returns the matrix product of two arrays. The array result contains the same number of rows as array1 and the same number of columns as array2.
How do you do multiple matrices in Excel? Excel MMULT Function Summary. The Excel MMULT function returns the matrix product of two arrays. Perform matrix multiplication. The matrix product of two arrays. =MMULT (array1, array2) array1 – The first array to multiply. array2 – The second array to multiply. Can Excel solve matrix? To solve…
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# Solve the system of congruences (CRT)
$$560x \equiv 1 \pmod{3, 11, 13}$$
I found a few (by trial and error)
$560x \equiv 1 \pmod{13} \implies x = 1 + 13k$.
$560x \equiv 1 \pmod{3} \implies x = 2 + 3k$.
$560x \equiv 1 \pmod{11}$ Is the hard congruence.
I am trying to use Euclid;s algorithm,
$560 = 50(11) + 10$ and $11 = 1(10) + 1 \implies 1 = 11 - 10$
$$\implies 1 = 11 - (560 - 50(11))$$
But how do I proceed with Euclid's algorithm? the actual answer is: $x = 10 + 11k$ (wolframalpha) so
$x \equiv 1 \pmod{13}$
$x \equiv 2 \pmod{3}$
$x \equiv 10 \pmod{11}$.
then by the chinese remainder theorem there is one solution $\pmod{429}$. But how should I solve it?
• Hint: the last two congruences can be rewritten as $x\equiv-1$ modulo both 3 and 11. Hence they are both solved when $x\equiv-1\;mod(33)$. – lulu Aug 8 '15 at 16:00
The general strategy for any system of congruence equations
$x \equiv r_1 \pmod{n_1}$
$\vdots$
$x \equiv r_k \pmod{n_k}$
where $n_1, \cdots n_k$ are pairwise coprime, is to first find a $k$-tuple $(a_1, \cdots , a_k)$ such that
$a_1 \equiv 1 \pmod{n_1}, \quad a_1 \equiv 0 \pmod{n_1^{\prime}}$
$\vdots$
$a_k \equiv 1 \pmod{n_k}, \quad a_k \equiv 0 \pmod{n_k^{\prime}}$
where $n_i^{\prime} = \prod_{j \neq i} n_j$. This is easy to do. Since $n_i$ and $n_i^{\prime}$ are coprime, we can use the extended Euclid's algorithm to find $p$ and $q$ such that $1 = pn_i + qn_i^{\prime}$. Now set $a_i = 1 - pn_i = qn_i^{\prime}$. The required solution is $x=r_1a_1 + \cdots + r_ka_k$.
In this case, using the hint in the comments, the system becomes
$x \equiv 1 \pmod{13}$
$x \equiv -1 \pmod{33}$
Now, $1 = - 65 + 66$, and therefore our $a_1 =66$ and $a_2 = -65$. The final answer is $1\cdot 66 + -1 \cdot -65 = 131$.
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## College Algebra (11th Edition)
$x=2$
$\bf{\text{Solution Outline:}}$ To solve the given equation, $\dfrac{1}{32}=x^{-5} ,$ use the laws of exponents and the properties of equality to isolate the variable. $\bf{\text{Solution Details:}}$ Using the Negative Exponent Rule of the laws of exponents which states that $x^{-m}=\dfrac{1}{x^m}$ or $\dfrac{1}{x^{-m}}=x^m,$ the expression above is equivalent to \begin{array}{l}\require{cancel} \dfrac{1}{32}=\dfrac{1}{x^5} .\end{array} Since $\dfrac{a}{b}=\dfrac{c}{d}$ implies $ad=bc$ or sometimes referred to as cross-multiplication, the equation above is equivalent to \begin{array}{l}\require{cancel} 1(x^5)=1(32) \\\\ x^5=32 .\end{array} Taking the fifth root of both sides, the equation above is equivalent to \begin{array}{l}\require{cancel} \sqrt[5]{x^5}=\sqrt[5]{32} \\\\ x=\sqrt[5]{(2)^5} \\\\ x=2 .\end{array}
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Larger Distances
Objective
SWBAT sequence numbers in the thousands.
Big Idea
Common Core standards expect second graders to be able to work with numbers to 1,000. This lesson introduces larger numbers (even beyond 1,000) to extend student understanding of the patterns in numbers.
Warm Up
15 minutes
I begin today by writing 3 numbers, 871, 258, 469 on the board. I ask students to write the numbers in order from least to greatest in their math journals.
I ask for a student to come up and write the numbers on the board in order. We read the numbers aloud together to practice reading 3-digit numbers.
I put 3 more numbers on the board, 29, 902 and 219 on the board and ask students to do the same thing. We repeat the process several more times as I circulate around the room to check on student understanding. I want to make sure that students understand the structure of numbers to 1000 - that they are made up of hundreds, tens and ones. The examples I use become more and more similar, in that the numbers to be compared contain the same digits - such as 409, 904, 914 and 419.
Teaching the Lesson
30 minutes
Students have been reading about planets during reading time. They have taken notes on important facts about planets, including size of the planets. We discuss how the world globe we have in the room is only a model of earth. It is way smaller than earth really is. I tell them that we can think of the planets in a similar way. We look at the sizes of the planets in distance around by using the numbers that students researched, and together we organize them from smallest to biggest.
Next I hand out paper in different sizes. I ask students why they think that some people are getting small paper and some are getting large paper to draw their planets on? (some planets are bigger than others based on the sizes we just figured out from our research). I ask students to make a drawing of the planet they have researched. I tell them that we will hang up the planets in order once we are done. We do not use an exact scale here because of the size of the numbers, and that is why I have chosen to hand out different sized paper so we end up with relative sizes for our planets.
Students complete the drawings in partners so we have one of each planet, and a sun.
Hanging the Planets
20 minutes
Next I discuss how each of these is not just the number, but a million of that number so that the 60 becomes 60,000,000, but that to make our job easier, we will just use the 60 because they are all in the millions.
I ask students to order the planets from closest to the sun to furthest from the sun by ordering the numbers.
When students are done, we write the planets in order and distance. I then tell students that we will now make the numbers even more manageable so we can hang up the planets. We will turn 60 into 6, 110 into 11, etc. We make all the numbers into smiley face numbers (ending in zero) and compute the distance in centimeters.
Next I hang the sun from the ceiling at the end of the room and place a small piece of masking tape below it on the floor to mark my starting point. I ask a student to measure 6 cm from the point I am marking so I can hang Mercury up. I remind students that our planets are not to the same scale (1 centimeter = 10 million kilometers ) because then our planets would be so small we wouldn't be able to see them so while they are relative in size to each other, they would be much smaller if we used the same measurements as we are doing for distance from the sun. It is like our maps of the whole earth and the map of the United States. On the world map the US is much smaller than it is on the map just of the US. in the same way our planets and our distances are much smaller than the real thing, but our planets need to be big enough to see and our distances need to be small enough to fit in the room.
After I have hung the first planet, I ask a student to measure 11 cm from the sun mark on the floor so I can hang the second planet.
I remind students that they need to be very careful as they measure and attend to the correct measurements so our planets will be the correct distances from the sun (MP6). They must also be careful to measure in centimeters and not inches (MP5). As the planets go beyond the length of the ruler, students will need to choose a meter stick or tape measure to carry out their measurement tasks. I let students choose the appropriate tool for the measurement they are making.(MP5)
I hang the planets from the ceiling as students measure the distance on the floor using a ruler.
Mercury - 60m km
Venus - 110 m km
Earth - 150 m km
Mars - 225 m km
Jupiter - 778 m km
Saturn - 1425 m km
Uranus - 2900 m km
Neptune - 4500 m km
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# Solve the Following Quadratic Equations by Factorization: 1/(X-1)(X-2)+1/(X-2)(X-3)+1/(X-3)(X-4)=1/6 - CBSE Class 10 - Mathematics
ConceptSolutions of Quadratic Equations by Factorization
#### Question
Solve the following quadratic equations by factorization:
1/((x-1)(x-2))+1/((x-2)(x-3))+1/((x-3)(x-4))=1/6
#### Solution
We have been given,
1/(x-1)(x-2)+1/(x-2)(x-3)+1/(x-3)(x-4)=1/6
((x-3)(x-4)+(x-1)(x-4)+(x-1)(x-2))/((x-1)(x-4)(x-2)(x-3))=1/6
(3(x^2-5x+6))/((x^2-5x+4)(x^2-5x+6))=1/6
18 = x2 - 5x + 4
x2 - 5x - 14 = 0
x2 - 7x + 2x - 14 = 0
x(x - 7) + 2(x - 7) = 0
(x + 2)(x - 7) = 0
Therefore,
x + 2 = 0
x = -2
or,
x - 7 = 0
x = 7
Hence, x = -2 or x = 7.
Is there an error in this question or solution?
#### APPEARS IN
Solution Solve the Following Quadratic Equations by Factorization: 1/(X-1)(X-2)+1/(X-2)(X-3)+1/(X-3)(X-4)=1/6 Concept: Solutions of Quadratic Equations by Factorization.
S
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Friday , August 19 2022
# NCERT 5th Class (CBSE) Mathematics: Fractions
## USING A NUMBER LINE
You can use a number line to show division of a whole number by a fraction.
(a) 5÷2/3 = ?
5÷1/3 = 15
(b) 2÷2/5 = ?
Number line showing 5 equal parts for each number.
There are 5 two-fifths in 2.
2÷2/5 = 5
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Posted
# Computer Conversions (Hexadecimal, Decimal, Octal and Binary)
, 5 Feb 2007 CPOL
Rate this:
An article to demonstrate how to convert from one base 2, 8, 10 or 16 to another of base 2, 8, 10, 16
## Introduction
This code was written to show how to convert from Hexadecimal, Decimal, Octal or Binary out to a Hexadecimal, Decimal, Octal or Binary value. Many functions demonstrate `Hex2Bin`, `Bin2Hex`, `Hex2Dec`, `Dec2Hex`, etc. This code will show you how easy it is to convert between any of these bases (2, 8, 10, 16).
## Background
It is good to understand how binary conversions work. Take a look here for more information.
## Using the Code
Enter an input value, and see what the output value is. It's that easy. Use the arrows to add or subtract the value.
The two main functions that deal with the conversions include `Input2Binary() `and `Binary2Output()`. From these functions, you can take any input type and convert it to binary, and then convert that binary value to any output type.
The `Input2Binary() `function contains this code:
```// Convert the current input into binary
// NOTE: convert string from base 'n'
switch( numberStyle.ToUpper() )
{
currInput = Convert.ToUInt64(txtObject, 16);
break;
case "DECIMAL":
currInput = Convert.ToUInt64(txtObject, 10);
break;
case "OCTAL":
currInput = Convert.ToUInt64(txtObject, 8);
break;
case "BINARY":
currInput = Convert.ToUInt64(txtObject, 2);
break;
}```
The `Binary2Output() `function contains this code:
```// Convert the int to the correct output
switch( numberStyle.ToUpper() )
{
txtObject = Convert.ToString(currVal, 16).ToUpper();
break;
case "DECIMAL":
txtObject = Convert.ToString(currVal, 10);
break;
case "OCTAL":
txtObject = Convert.ToString(currVal, 8);
break;
case "BINARY":
txtObject = Convert.ToString(currVal, 2);
break;
}```
## History
• 5th February, 2007: First version
## Share
Web Developer United States
Scott Klawitter programs mainly in Visual Studio .NET with a focus in C# development.
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I have seen no less than 3 threads on the MSDN forums in the past week asking how to do this. It may be trivial, but if one wants to do it programmatically, the Windows Calculator isn't much help so it's good for beginners.
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Re: Grumpy are we? alrsds21-Sep-09 18:26 alrsds 21-Sep-09 18:26
Last Visit: 31-Dec-99 19:00 Last Update: 2-Dec-16 20:08 Refresh 1
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# Why do negative exponents work the way they do? [closed]
Why is a value with a negative exponent equal to the multiplicative inverse but with a positive exponent?
$$a^{-b} = \frac{1}{a^b}$$
• Is this really homework? – Git Gud Jun 3 '14 at 13:01
• Presumably, what you mean is, wny is $$a^{-b}=\frac{1}{a^b}?$$ – Thomas Andrews Jun 3 '14 at 13:02
• @ThomasAndrews Or the OP is having a slight existential crisis :) – joshin4colours Jun 3 '14 at 19:23
• Is the question about why $x^{-a} = 1/x^{a}$ is mathematically valid or about why so many algebra/precalculus classes insist that students always re-write all the exponents to be positive? – Stan Liou Jun 3 '14 at 21:10
• Also, $$\frac{1}{a^{-b}} = a^b$$ – user14972 Jun 4 '14 at 5:39
If you know that $$x^y \cdot x^z = x^{y+z} \tag{1}$$ and that $$x^0 = 1 \tag{2}$$ then \begin{align} 1 &= 1 \\ a^0 &= 1 \\ a^{b - b} &= 1\\ a^{b} \cdot a^{-b} &= 1 \\ a^{-b} &= \dfrac{1}{a^b}\end{align}
• Might be clearer if your last line reads $a^{-b} = \frac{1}{a^b}$. – Emily Jun 3 '14 at 13:14
• @DanChristensen Anyone who doesn't define $0^0 = 1$ is deliberately making life hard. There really isn't any debate, just trolling. There is no inconsistency in the definition. Some would argue that because we can't infer $\lim_{x,y \rightarrow 0,0} x^y = 1$ then the definition is unsound, but that's just ridiculous. The existence of limits always has to be checked. – DanielV Jun 3 '14 at 17:31
• Applying the formula with $a = 0$ gives the result $0^{-b} = \frac{1}{0^b}$. Maybe there is a way to make sense of this of which I am not aware, but it seems best avoided in high school math. Therefore you should stipulate that $a \ne 0$ (and might as well also say $x \ne 0$), and you don't need to know how to evaluate $0^0$. – David K Jun 4 '14 at 7:21
• For the real numbers $0$, and $0$, $0^0$ is undefined. For the natural numbers $0$ and $0$, we must surely have $0^0 = 1$, since there is one function from the empty set to itself. – Steven Gubkin Jun 4 '14 at 16:10
• @DanielV I only mean to draw attention to the fact thinking about exponentiation in $\mathbb{R}$ we must, by definition, use limits, and so it is appropriate to leave $0^0$ undefined. OTOH, when thinking of exponentiation defined on the natural numbers, this has a natural combinatorial definition $x^y = |X^Y|$ where $X$ is a set with $x$ elements, $Y$ is a set with $y$ elements, and $X^Y$ is the set of functions from $Y$ to $X$. – Steven Gubkin Jun 4 '14 at 18:49
When $m$ and $n$ are integers, we have the important law that $$x^m\cdot x^n =x^{m+n}$$
The reason for this is easy to see: $x^3$ means $x\cdot x \cdot x$, and $x^2$ means $x\cdot x$, so \begin{align}x^3 \cdot x^2 &= (x\cdot x \cdot x)\cdot(x\cdot x)\\ &=x\cdot x \cdot x\cdot x \cdot x \\& = x^5 \end{align}
We'd like this law to continue to hold when we define $x^\alpha$ for negative $\alpha$, unless there's a good reason it shouldn't. If we do want it to continue to hold for negative exponents, then whatever we decide that $x^{-1}$ should mean, it should obey the same law: $$x^{-1}\cdot x^{2} = x^{-1+2} = x^1 = x$$
and so $x^{-1} = \frac1x$ is the only choice.
Similarly, what should $x^{-3}$ mean? If we want the law to continue to hold, we need $$x^{-3}\cdot x^{3} = x^{-3+3} = x^0 = 1$$ and thus the only consistent choice is $x^{-3} = \frac1{x^3}$.
But there is more to it than that. Further mathematical developments, which you have not seen yet, confirm these choices. For example, one shows in analysis that as one adds more and more terms of the infinite sum $$1 + x + \frac{x^2}2 + \frac{x^3}6 + \frac{x^4}{24} + \cdots$$ the sum more and more closely approaches the value $e^x$, where $e$ is a certain important constant, approximately $2.71828$. One can easily check numerically that this holds for various integer values of $x$. For example, when $x=1$, and taking only the first five terms, we get $$1 + 1 + \frac12 + \frac16 + \frac1{24}$$
which is already $2.708$, quite close to $e^1$, and the remaining terms make up the difference. One can calculate $e^2$ by this method and also by straightforward multiplication of $2.71828\cdot2.71828$ and get the same answer.
If you put $x=-1$ in this formula, you get $$e^{-1} \stackrel?= 1 -1 +\frac12 - \frac16 + \frac1{24}\cdots$$ and adding up just the first few terms one gets $0.375$, which is already pretty close to $\frac1e \approx 0.368$.
If it didn't work out this way, we would suspect that something was wrong somewhere. And in fact it has often happened that mathematicians have tried defining something one way, and then later developments revealed that the definition was not the right one, and it had to be revised. Here, though, that did not happen.
(Much of this is copied from my answer to a similar question earlier.)
• I really liked the extra context of the rule applying in the definition of e. Could you say in what other situations mathematicians made a bad notational choice they had to backtrack? – honestSalami Apr 16 '20 at 4:04
• Have you seen math.stackexchange.com/questions/537383/… ? – MJD Apr 16 '20 at 9:58
• For a larger example, consider the discovery of complex numbers. – MJD Apr 16 '20 at 10:12
In a vague sense, each integer increment of the exponent in $a^b$ means "multiply by a" where $a$ is the base of the exponent. So $2^2$ means "multiply by two, and multiply by two again" to get "multiply by $2*2 = 4$".
So imagine you have $2^{3-1}=2^2=4$. What does the $-1$ mean in that expression? You originally had "multiply by 2, then multiply by 2, then multiply by 2" or "multiply by two three times", but there was also the $-1$, which somehow undid one of those multiply by 2's. The arithmetic function which is the inverse of multiplication is division. Try it - take a number and multiply it by 2, then divide it by 2. You'll come up with the original answer.
Therefore negative exponents are division by the base, rather than multiplication.
(The OP doesn't seem to understand what a negative exponent means. Assuming this was not actually the question asked for homework. Something like this should have been covered by his teacher.)
Just follow the pattern. For non-zero $x$, we have
$x^3=x\cdot x \cdot x$
$x^2=x\cdot x$
$x^1=x$
$x^0=1$
Continue dividing by $x$ each time, as above...
$x^{-1}=1\div x =\frac{1}{x}$
$x^{-2}=\frac{1}{x}\div x =\frac{1}{x^2}$
$x^{-3}=\frac{1}{x^2}\div x =\frac{1}{x^3}$
and so on.
• Nice work. The only thing I'd have included is the similarity to the negative ide of the number line, 3,2,1,0,-1,-2, etc. Why does bopping negative not bother OP there but it does for exponents? – JTP - Apologise to Monica Jun 3 '14 at 16:37
• This is the way I explain it to people. I think it's the most straightforward way. – daviewales Jun 4 '14 at 17:18
This is really how we define negative exponents, so the question becomes: why is this a reasonable definition? Let's pretend for the moment that we only knew about positive exponents.
If $n$ is a positive integer, what does $x^n$ mean? It's the product of $x$ with itself $n$ times:
$$x^n = \overbrace{x\cdot x\cdots x}^n$$
So if $m$ is another positive integer, we have
$$x^{m+n} = \overbrace{x\cdot x\cdots x}^{m+n}$$ $$= (\overbrace{x\cdot x\cdots x}^{m})(\overbrace{x\cdot x\cdots x}^{n})$$ $$= x^m\cdot x^n$$
Now what should $x^n$ mean when $n$ is zero or negative? It would be nice for the rule $x^{m+n}=x^m\cdot x^n$ to be true even when $m$ or $n$ is not positive, so let's see what that rule tells us.
We would have to have $$x^{0+1} = x^0\cdot x^1$$, and so $$x^1 = x^0\cdot x^1$$ $$x = x^0\cdot x.$$ If $x\neq 0$ then the only way this can be true is if $x^0=1$. So we'll define $x^0=1$ whenever $x\neq 0$, and continue from there.
Now for negative exponents. How can we reasonably define $x^{-n}$ when $n$ is a positive integer (and $x\neq 0$)? Going by our rule $x^{m+n}=x^m\cdot x^n$, we would have to have
$$x^{-n+n} = x^{-n}\cdot x^n$$ $$x^0 = x^{-n}\cdot x^n$$ $$1 = x^{-n}\cdot x^n$$
Divide both sides by $x^n$, and we see that we must have $$\frac{1}{x^n} = x^{-n}.$$
So this is how we must define $x^{-n}$.
Play with this a little: $$2\to4\to8\to16\ldots\text{etc}$$or equivalently $$2^1\to2^2\to2^3\to2^4\ldots\text{etc}.$$ So you increase the exponent of $2$ with $1$ each time. Now imagine going in the backward direction (meaning you substract $1$ in the exponent each time) but dont stop at two....
Think of it this way: exponentiation is equivalent to repeated multiplication, in the sense that, for example, $3^4=3\times3\times3\times3$; so a multiplication repeated a negative number of times should use the multiplicative inverse, division. Therefore, a negative exponentiation could be represented as a repeated division, which would be equivalent to $a^{-b}=\frac{1}{a^b}$.
$\sqrt[-b]{a}=a^{(\frac{-1}{b})}=\frac{1}{a^{\frac{1}{b}}}=\frac{a^0}{a^{\frac{1}{b}}}=a^{(0-\frac{1}{b})}$
• Hmm, I'm fairly certain I've never seen anyone use the notation $^n\sqrt{a}$ with negative values of $n$, although an "automatic" application of $^n\sqrt{a}=a^{\frac{1}{n}}$ would do that... Our suspicion now is that the question doesn't have anything to do with roots at all: the OP probably meant "base." – rschwieb Jun 3 '14 at 13:19
• @rschwieb, +1 for identifying the mistake, I will keep the my wrong notation posted for others to see "How not to write math" – Vikram Jun 3 '14 at 13:47
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2000 AMC 10 Problems/Problem 2
Problem
$2000\left(2000^{2000}\right)=$
$\mathrm{(A)}\ {2000}^{2001} \qquad \mathrm{(B)}\ {4000}^{2000} \qquad \mathrm{(C)}\ {2000}^{4000} \qquad \mathrm{(D)}\ {4,000,000}^{2000} \qquad\mathrm{(E)}\ {2000}^{4,000,000}$
Solution
$2000 \cdot 2000^{2000}=2000^1 \cdot 2000^{2000}=2000^{2000+1}=2000^{2001}$.
$\boxed{\text{A}}$
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# Find Square Root Using Babylonian Method in C++
Saad Aslam Sep-11, 2022 C++ C++ Math
Using the Babylonian approach, which will be explained in this article, we will learn how to determine the square root of an integer in C++ programming language.
## An Overview of the Babylonian Method in C++
The Babylonian technique for determining the square root of a number is derived from one of the numerical methods, which in turn is derived from the Newton-Raphson approach for the solution of non-linear equations.
The procedure may be summarized as a variation of the following algorithm.
1. First, we randomly choose a positive number and get it as near the root as possible.
2. We start by assigning the value one to a variable by saying `y = 1`.
3. Follow the instructions in the next two stages until you reach the point when you have the answer.
• Find the average of `x` and `y`, then calculate an approximation for the root.
• Set `y = n/x.`
Suppose we continue repeating the previous section’s procedures in a loop. In that case, we will eventually arrive at the square root of the number itself while utilizing the Babylonian technique to solve the problem.
## Implement the Babylonian Method to Find the Square Root in C++
Let us look at an example that uses a separate class to apply the Babylonian approach to finding the square root of a number.
To begin, we will import the necessary libraries for the application.
``````#include<iostream>
#include<math.h>
#include<stdio.h>
``````
Develop a float type method called `sqRootBySaad` that accepts a `sqNumber` parameter with a `float` datatype value.
``````float sqRootBySaad(float sqNumber) {}
``````
In the body of this method, you should first set the values of three variables: `value,` `guess,` and `constantValue.` The value allocated to the `value` variable will be the number supplied to this as an argument.
The value of the `guess` variable is set to `one,` and the `constantValue` is set to `zero.`
``````float value = sqNumber;
float guess = 1;
float constantValue = 0;
``````
After this, we apply a `while` loop to check if the `value - guess` value is greater than the `constantValue`. Then we will calculate the average of both the `value` and `guess` values, and the `sqNumber` will be divided by `value`.
After this, we use a `while` loop to check whether the value of `value - guess` is larger than the value of `constantValue.` If it is, we will compute the average of both the `value` and the `guess` values.
In addition, we will divide the `sqNumber` by the computation that we have just finished and save it in the `value` variable.
``````while (value - guess > constantValue)
{
value = (value + guess) / 2;
guess = sqNumber / value;
}
``````
The `sqRootBySaad` will return the `value` variable.
``````return value;
``````
Now that we have developed the procedure for finding the square root, we need to define a variable named `anyNumber` with the type int in the function called `main().` Show a message to the user asking them to submit the number so you can calculate its square root, and then take the user’s input for the number.
Now, call the method `sqRootBySaad` and provide the previously stored number in the variable `anyNumber.`
``````int main()
{
int anyNumber;
cout << "Enter any number to find square root: ";
cin >> anyNumber;
cout << "Square root using Babylonian method is: " << sqRootBySaad(anyNumber);
return 0;
}
``````
Full Source Code:
``````#include<iostream>
#include<math.h>
#include<stdio.h>
using namespace std;
{
float value = sqNumber;
float guess = 1;
float constantValue = 0;
while (value - guess > constantValue)
{
value = (value + guess) / 2;
guess = sqNumber / value;
}
return value;
}
int main()
{
int anyNumber;
cout << "Enter any number to find square root: ";
cin >> anyNumber;
cout << "Square root using Babylonian method is: " << sqRootBySaad(anyNumber);
return 0;
}
``````
Output:
``````Enter any number to find square root: 25
Square root using Babylonian method is: 5
``````
Let us look at an additional example that applies the Babylonian approach to finding the square root of a number.
Here, we initialize three variables `value`, `guess,` and `sqNumber` of datatype `double.`
Show a message to the user asking them to submit the number so that we may calculate the square root of the number, and then we input the value that the user provides.
``````double value, guess, sqNumber;
cout << "Please enter a number to find square root using the Babylonian method: ";
cin >> value;
``````
After dividing the number by two, we store the result in the variable referred to as `guess.` Implement a `for` loop with an iteration count of ten, and continue doing so until we reach the desired result.
Within the `for` loop, divide the `value` that the user has input by the `guess` value, and then save the result in the variable `sqNumber.` To update the value of the `guess` variable, you need to compute the average values contained in the `guess` and `sqNumber` variables.
`````` guess = value / 2.0;
for(int n = 0; n <= 10; n++)
{
sqNumber = value/guess;
guess = (guess + sqNumber)/2;
}
``````
At last, we print the `guess` value, the number obtained by taking the square root of the value the user has input.
``````cout << "The square root of " << value << " is: " << guess << endl;
``````
Full Source Code:
``````#include<iostream>
using namespace std;
int main()
{
double value, guess, sqNumber;
cout << "Please enter a number to find square root using the Babylonian method: ";
cin >> value;
guess = value / 2.0;
for(int n = 0; n <= 10; n++)
{
sqNumber = value/guess;
guess = (guess + sqNumber)/2;
}
cout << "The square root of " << value << " is: " << guess << endl;
}
``````
Output:
``````Please enter a number to find square root using the Babylonian method: 25
The square root of 25 is: 5
``````
I'm a Flutter application developer with 1 year of professional experience in the field. I've created applications for both, android and iOS using AWS and Firebase, as the backend. I've written articles relating to the theoretical and problem-solving aspects of C, C++, and C#. I'm currently enrolled in an undergraduate program for Information Technology.
## Related Article - C++ Math
• Calculate Exponent Without Using pow() Function in C++
• Intersection of Ray and Plane in C++
• C++ Cube Root
• Magic Square Problem in C++
• Division in C++
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# Vapor Pressure vs Boiling Point: Several Graphs And Insights
In this article, we are going to learn what is the difference between the vapor pressure and the boiling point with detailed insights.
The vapor pressure and boiling point graph shows an exponential curve and denotes the saturation of vapor pressure too. Here is a table below differentiating the vapor pressure vs boiling point:-
## Vapor Pressure And Boiling Point Graph
The boiling point is nothing but the temperature at which the phase change occurs and the vapor pressure attains the highest value at that fixed atmospheric pressure. Hence let us plot a graph of vapor pressure v/s the temperature for a liquid boiling at a constant pressure condition.
The graph of vapor pressure v/s temperature shows the exponential curve as the number of vapors escaping from the liquid overcoming the attractive intermolecular bonds doubles at every rise in temperature of the liquid.
The point TBP denotes the boiling point of the particular liquid on the x-axis, beyond the boiling point of the liquid, the temperature of the liquid does not rise further but only the phase change from liquid to vapors takes place. The point on the y-axis Vsat represents the saturation point of the vapor pressure. As the vapors evaporated cool down and condense back into the liquid form. The vapor pressure is maintained constant after reaching the boiling point of the liquid.
## How to Calculate Boiling Point from Heat of Vaporization?
The heat of vaporization is the amount of heat energy required to transform the liquid state of matter into the gaseous state.
The boiling point of a liquid can be calculated from heat of vaporization by using the Clausius – Clapeyron equation given as $ln\left ( \frac{P_2}{P_1} \right )=\frac{-\Delta H_{vap}}{R}\left ( \frac{1}{T_2}-\frac{1}{T_1} \right )$.
## What is the boiling point of water in the pressure cooker which functions at 1.8 bar if the heat of vaporization of water is 45k J/mol?
Given: P2= 1.8 bar
The water at a normal atmospheric condition that is at 1 atm, boils at 1000C, hence
P1= 1 bar
T1 =1000C=373.2K
$\Delta H_{vap}=45k\ J/mol$
Using the Clasius – Clapeyron equation
$ln\left ( \frac{P_2}{P_1} \right )=\frac{-\Delta H_{vap}}{R}\left ( \frac{1}{T_2}-\frac{1}{T_1} \right )$
$ln\left ( \frac{1.8}{1} \right )=\frac{-45000}{8.314}\left ( \frac{1}{T_2}-\frac{1}{373.2} \right )$
$ln\left ( 1.8 \right )=-5412.56\left ( \frac{1}{T_2}-0.0027 \right )$
$0.5878=-5.412\left ( \frac{1}{T_2}-0.0027 \right )$
$-10.86 \times 10^{5}=\frac{1}{T_2}-0.0027$
$\frac{1}{T_2}=-10.86 \times 10^{5}-0.0027$
$\frac{1}{T_2}=0.00257$
$T_2=\frac{1}{0.00257}=389.1 K$
And 389.1K = 115.90C
Hence the boiling point of water inside the pressure cooker is 115.90C.
## How to find Boiling Point from Vapor Pressure?
The boiling point can be found by measuring the saturated vapor pressure developed at that temperature.
The liquid can have varied boiling points at different pressure in the system. The vapor pressure can be found using the Clausius – Clapeyron equation, also from the phase diagrams, and from the graph of vapor pressure v/s temperature too.
## What is the boiling point of methane at vapor pressure equal to 2 atm? Given the heat of vaporization of methane is 8.20k J/mol.
At normal atmospheric pressure, the boiling point of methane is -161.50C.
P1 =1atm
P2 =2atm
T1 =-161.50C =-161.5+273.2 =111.7K
$\Delta H_{vap}=8.2k\ J/mol$
Using Clausius – Clapeyron equation
$ln\left ( \frac{P_2}{P_1} \right )=\frac{-\Delta H_{vap}}{R}\left ( \frac{1}{T_2}-\frac{1}{T_1} \right )$
$ln\left ( \frac{2}{1} \right )=\frac{-8200}{8.314}\left ( \frac{1}{T_2}-\frac{1}{111.7} \right )$
$ln\left ( 2 \right )=-986.3\left ( \frac{1}{T_2}-\frac{1}{111.7} \right )$
$0.6931=-986.3\left ( \frac{1}{T_2}-0.00895 \right )$
$70.3\times 10^{-5}=-\left ( \frac{1}{T_2}-0.00895 \right )$
$70.3\times 10^{-5}=0.00895-\frac{1}{T_2}$
$\frac{1}{T_2}=0.00895-70.3\times 10^{-5}$
$\frac{1}{T_2}=0.00825$
$T_2=\frac{1}{0.00825}$
$T_2=121.2K$
This is equal to -1520 C.
Hence the boiling point of the methane at the vapor pressure of 2 atm increases to -1520 C.
## What are the factors affecting the vapor pressure of the liquid?
The vapor pressure is due to the pressure felt on the area by the vapors evaporated from the system into the surrounding.
The most vital factor on which the vapor pressure depends is the temperature and the heat energy supplied to the liquid. Also, the chemical composition and the impurities added will vary the vapor pressure.
## How vapor pressure depends upon the intermolecular bonding between the atoms?
On supplying heat energy to the liquid, the intermolecular bonding between the atoms breaks, and particles move in random motion.
If this intermolecular bonding between the atom in case of a certain liquid is low, which means there is a weak force of attraction between the atoms then these bonds will easily break down will even a small amount of energy supplied to the liquid and thus vapor pressure will be high at small temperature.
## How are the boiling point and the vapor pressure related to each other?
The vapors are the result of the rising temperature of the liquid supplying heat.
At a boiling point, the liquid phase is converted to the gaseous phase and at this temperature, the vapor pressure formed becomes equal to the atmospheric pressure.
AKSHITA MAPARI
Hi, I’m Akshita Mapari. I have done M.Sc. in Physics. I have worked on projects like Numerical modeling of winds and waves during cyclone, Physics of toys and mechanized thrill machines in amusement park based on Classical Mechanics. I have pursued a course on Arduino and have accomplished some mini projects on Arduino UNO. I always like to explore new zones in the field of science. I personally believe that learning is more enthusiastic when learnt with creativity. Apart from this, I like to read, travel, strumming on guitar, identifying rocks and strata, photography and playing chess. Connect me on LinkedIn - linkedin.com/in/akshita-mapari-b38a68122
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# Find the centroid of the region
• dohsan
In summary, the centroid of a region is the point where the mass of the region is evenly distributed and is often referred to as the geometric center or balance point. The formula for finding the centroid is (x̄, ȳ) = (∫xρdA/∫ρdA, ∫yρdA/∫ρdA), and it can lie outside of the region in irregularly shaped or non-uniform regions. The centroid is used in various real-world applications such as engineering, physics, and geography to find the center of mass, determine stability, and locate the center of population. It can change if the distribution of mass or shape of the region changes, but remains fixed for regular shapes
dohsan
## Homework Statement
Find the centroid of the region bounded by ...
y=x^3, x+y=2, y=0
## The Attempt at a Solution
I think you find the area of the region which is 5/4
Sure, that's the area alright. What else do you have to compute to find the centroid?
That is NOT what I get for the area! dohsan, could you show your work?
## 1. How is the centroid of a region defined?
The centroid of a region is defined as the point where all the mass of the region is evenly distributed. It is the geometric center or the balance point of the region.
## 2. What is the formula for finding the centroid of a region?
The formula for finding the centroid of a region is (x̄, ȳ) = (∫xρdA/∫ρdA, ∫yρdA/∫ρdA), where x and y are the coordinates of the centroid, ρ is the density of the region, and A is the area of the region.
## 3. Can the centroid of a region lie outside of the region?
Yes, the centroid of a region can lie outside of the region. This is possible in cases where the region is irregularly shaped or has a non-uniform distribution of mass.
## 4. How is the centroid of a region used in real-world applications?
The centroid of a region is used in various real-world applications such as engineering, physics, and geography. It is used to find the center of mass of an object, determine the stability of structures, and locate the center of population in a geographical region.
## 5. Can the centroid of a region change?
Yes, the centroid of a region can change if the distribution of mass within the region changes. A shift in the position of mass or a change in the shape of the region can cause the centroid to move. However, for a regular shape with a uniform distribution of mass, the centroid will remain fixed.
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# Using Differentiation to Find The Slope of a Tangent
Firstly, a tangent is a straight line that touches a curve at a certain point. Imagine you are riding your bicycle on the road around a bend of a park, imagine yourself steering the handlebars to go around the bend. Then, a dog comes up and starts to bark at you at a specific point you are steering around the bend, you straighten your handle bars and then stop following the bend and go off on a straight line, to avoid the dog. That straight line is the tangent to the bend, or for our purposes, then tangent to the curve.
You may have also heard the expression about someone going off topic about something,
“... they went off on a tangent ...”
Meaning they were once on topic and have now stopped following the topic and gone in a different direction. For example, if we were having a discussion about how to bake a cake, and we were talking about what temperature to set the oven at, and then someone starts to talk about their oven and what type of oven it is and where they bought it from and how their fridge comes from the same company etc…. That is going off on a tangent. Think of that example if it helps!
SEE IMG
Now, switching it back to math, just imagine that point mathematically, as coordinates, and imagine that tangent as a straight line. Here, we will learn how to calculate the slope of the tangent, which means the gradient of the tangent. First, we will start with an approximate way to calculate the slope of the tangent.
Remember, the slope of a line (the gradient) can be thought of as:
Rise
Run
Or, for a standard y and x axis graph, the change in the y-value over the change in the x-value.
Now, say we have the curve of a function, the function is represented by y = f(x).
For this curve y = f(x), we have the point along the curve, P, the x-value of this point is a, so the y-value of that point (subbed into our equation y = f(x)) is f(a).
So, the point P has the coordinates x = a and y = f(a),
Or (a, f(a)),
Now imagine the tangent that touches the point P,
Recall, to find the slope of a line, we subtract the two y-values of two points and divide them by the subtraction of the two x-values of those two points.
For finding the slope of the tangent for the point P in our illustration above, we only know the value of one point on that tangent, as there is only one point where the tangent touches the curve (the curve being y = f(x)). If we take another point along that curve, say T, that is close to the point P, we can work out an approximate value of the tangent, as notice, the line drawn connecting the point P and the other point T looks approximately the same as the line that is the tangent to P.
Say the x-value for the point T is b, so the y-value is f(b) [remember, Q is also on the curve y = f(x)]
Now, lets look at the line between P and T,
We can see that the line between P and T is approximately close to the line that is the Tangent of P (mTan). If the point T was closer to P, the line between them would be getting closer to what the Tangent of P (mTan) is.
Calculating the slope (gradient, m) of the line between P (a, f(a)) and T (b, f(b)), which we will represent by mPT
Slope of PT =
mPT =
Rise
Run
=
Change of y
Change of x
=
f(b) - f(a)
b - a
=
Now, as the point T gets closer to the point P, mPT (gradient of the line PT) gets closer to mTan (the gradient of the tangent of the point P), or
As the x-value of a gets closer to b, the gradient of the tangent becomes closer to the gradient of the line between those two points. Mathematically we write this as.
## Theoretical Definition of the Slope of a Tangent
mTan = lim[b -> a]
f(b) - f(a)
b - a
mTan, or the slope divided by the gradient of the tangent is called the derivative of the function f(x) at the point P.
What the equation above is saying is, as the value of b gets closer to a, then the slope of that line gets closer to the slope of the point a. The limit of how close b can be to a is equal to the slope of the tangent to a.
This might all seem very abstract but as we work through this you will start to understand it.
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# Do both sufficient statements provide the same answer in a DS question
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Do both sufficient statements provide the same answer in a DS question [#permalink]
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15 Jul 2012, 21:01
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Hi
I might be asking a stupid question but then i couldn't stop myself from asking it.
Suppose a DS question asks you to find an answer. Lets assume they ask you to find the median of a certain data set with variables in it.
based on the two options you have to determine if median >= 5 or not ?
If both the option of the question are able to solve it, is it necessary that both questions should get you the same result ?
For ex if option A gets the result as median = 5 is it necessary that if option B also is able to get the right answer then it should also get the median value as 5 ??
can there be a possibility that option B gets an answer different than 5 ?
Thanks
smartmanav
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Re: Do both sufficient statements provide the same answer in a DS question [#permalink]
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16 Jul 2012, 04:16
Hi,
As far as I think GMAT "might" not give you any such question inw hich both Stat1 and Sta2 are giving you different answers individually.
If i get a question like this then i will still be going and marking option "D" and not option "E"
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Re: Do both sufficient statements provide the same answer in a DS question [#permalink]
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16 Jul 2012, 05:21
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smartmanav wrote:
Hi
I might be asking a stupid question but then i couldn't stop myself from asking it.
Suppose a DS question asks you to find an answer. Lets assume they ask you to find the median of a certain data set with variables in it.
based on the two options you have to determine if median >= 5 or not ?
If both the option of the question are able to solve it, is it necessary that both questions should get you the same result ?
For ex if option A gets the result as median = 5 is it necessary that if option B also is able to get the right answer then it should also get the median value as 5 ??
can there be a possibility that option B gets an answer different than 5 ?
Thanks
smartmanav
Ask a question be a fool for 3 minutes, don't ask n b a fool all your life !!
No, scenario you describe is not possible, since on the GMAT, two data sufficiency statements always provide TRUE information and these statements never contradict each other.
So, for example in an YES/NO DS question you cannot have a NO answer from the first statement and an YES answer from the second statement or if you are asked to find the value of x, then you cannot get x=5 from the first statement and x=4 from the second statement.
Hope it helps.
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Re: Do both sufficient statements provide the same answer in a DS question [#permalink]
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21 Jul 2015, 07:04
Hi,
So in that case, If I arrived at different solutions for the same DS question, Can we say I went wrong somewhere and need to recheck my calculations???
This can actually help in confirming a particular DS question, given that Number proporties (like odd/even) are pretty tricky.
Thanks
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Re: Do both sufficient statements provide the same answer in a DS question [#permalink]
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21 Jul 2015, 07:17
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amritrungta wrote:
Hi,
So in that case, If I arrived at different solutions for the same DS question, Can we say I went wrong somewhere and need to recheck my calculations???
This can actually help in confirming a particular DS question, given that Number proporties (like odd/even) are pretty tricky.
Thanks
Yes, if you got x = 1 from (1) and x = 2 from (2), then this should be an indication that somewhere you made a mistake.
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Re: Do both sufficient statements provide the same answer in a DS question [#permalink]
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21 Jul 2015, 07:28
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Bunuel wrote:
smartmanav wrote:
Hi
I might be asking a stupid question but then i couldn't stop myself from asking it.
Suppose a DS question asks you to find an answer. Lets assume they ask you to find the median of a certain data set with variables in it.
based on the two options you have to determine if median >= 5 or not ?
If both the option of the question are able to solve it, is it necessary that both questions should get you the same result ?
For ex if option A gets the result as median = 5 is it necessary that if option B also is able to get the right answer then it should also get the median value as 5 ??
can there be a possibility that option B gets an answer different than 5 ?
Thanks
smartmanav
Ask a question be a fool for 3 minutes, don't ask n b a fool all your life !!
No, scenario you describe is not possible, since on the GMAT, two data sufficiency statements always provide TRUE information and these statements never contradict each other.
So, for example in an YES/NO DS question you cannot have a NO answer from the first statement and an YES answer from the second statement or if you are asked to find the value of x, then you cannot get x=5 from the first statement and x=4 from the second statement.
Hope it helps.
Bunuel, this is a very interesting observation you have mentioned. But I have seen in some yes/no DS questions to provide a "no" for 1 statement and "yes" for the other statement and as both statements still give an unambiguous yes or no , the correct answer happens to be D.
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Re: Do both sufficient statements provide the same answer in a DS question [#permalink]
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21 Jul 2015, 07:51
Thanks a lot !!!
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Re: Do both sufficient statements provide the same answer in a DS question [#permalink]
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21 Jul 2015, 07:52
1
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Engr2012 wrote:
Bunuel wrote:
smartmanav wrote:
Hi
I might be asking a stupid question but then i couldn't stop myself from asking it.
Suppose a DS question asks you to find an answer. Lets assume they ask you to find the median of a certain data set with variables in it.
based on the two options you have to determine if median >= 5 or not ?
If both the option of the question are able to solve it, is it necessary that both questions should get you the same result ?
For ex if option A gets the result as median = 5 is it necessary that if option B also is able to get the right answer then it should also get the median value as 5 ??
can there be a possibility that option B gets an answer different than 5 ?
Thanks
smartmanav
Ask a question be a fool for 3 minutes, don't ask n b a fool all your life !!
No, scenario you describe is not possible, since on the GMAT, two data sufficiency statements always provide TRUE information and these statements never contradict each other.
So, for example in an YES/NO DS question you cannot have a NO answer from the first statement and an YES answer from the second statement or if you are asked to find the value of x, then you cannot get x=5 from the first statement and x=4 from the second statement.
Hope it helps.
Bunuel, this is a very interesting observation you have mentioned. But I have seen in some yes/no DS questions to provide a "no" for 1 statement and "yes" for the other statement and as both statements still give an unambiguous yes or no , the correct answer happens to be D.
Such kind of DS questions would be considered flawed as per GMAT standards.
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Re: Do both sufficient statements provide the same answer in a DS question [#permalink]
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22 Jul 2015, 20:35
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amritrungta
Here are some tacit rules that are followed by GMAT test writers when it comes to Data Sufficiency:
1) The two statements in data sufficiency will never contradict each other.
2) If each statement alone is sufficient, then the outcome will be consistent. This means that if it is a Yes/No data sufficiency question, then both statement will answer the question in the main stem as Yes, or both will answer them as No. If it is a value question, then the outcome will be identical in both cases.
3) If you find that your outcome to the question is different for both the statements, while each statement is sufficient, then you can be guaranteed that either you made some computational error in a value question or you are misinterpreting the statements. This is a valuable tool and I use it to gain further confidence that I did indeed get the data sufficiency question right.
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Re: Do both sufficient statements provide the same answer in a DS question [#permalink]
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22 Jul 2015, 20:36
Hi All,
Some of the 'language' in this discussion might be misinterpreted, so I'm going to add some clarification to the discussion.
In DS questions, the two Facts CANNOT BOTH be Sufficient with different answers. If you come across this type of situation in a DS question, then the likely reason for it is that your work is INCOMPLETE (so you didn't do enough to prove that one of the Facts was actually Insufficient).
Here's a simple example:
X is an integer. Is X greater than 0?
1) -3 < X < 0
In this Fact, there are a couple of possible values for X....
IF....
X = -1, then the answer to the question is NO
X = -2, then the answer to the question is NO
The answer is ALWAYS NO, so....
Fact 1 is SUFFICIENT
2) X^2 = 4
In this Fact, there are a coupe of possible values for X....
IF....
X = 2, then the answer to the question is YES
X = -2, then the answer to the question is NO
Fact 2 is INSUFFICIENT
In this situation, if your work was 'incomplete' in Fact 2 (and you thought that X = 2 was the only solution), then you might think that it's SUFFICIENT (but it's actually NOT - there's another possible value for X that leads to a different answer to the question).
Hi Engr2012,
If you can 'dig up' any DS questions that match what you describe, then you should post them here. It's possible that the question wasn't properly 'designed' or it's possible that you might have missed something when you attempted to solve it.
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03 Nov 2015, 14:42
I just read for 'calculate value questions' like 1/(x + 4)? that you should not overthink and just check if statement 1 and 2 are sufficient to calculate x and I agree.
However, I was wondering if both statements alone are sufficient, if it is always D, or can the answers be dissimilar (so you have to calculate exactly)?
Post scriptum: Apologies if this is frequently asked
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03 Nov 2015, 14:59
neenpaques wrote:
I just read for 'calculate value questions' like 1/(x + 4)? that you should not overthink and just check if statement 1 and 2 are sufficient to calculate x and I agree.
However, I was wondering if both statements alone are sufficient, if it is always D, or can the answers be dissimilar (so you have to calculate exactly)?
Post scriptum: Apologies if this is frequently asked
Dear neenpaques,
I'm happy to respond.
Here's a blog article with some DS tips.
http://magoosh.com/gmat/2013/gmat-data- ... ency-tips/
If the DS question is a "find the value of the expression" question, and if you have enough information to calculate the solution from each statement, then the answer always will be (D). In a real GMAT DS or in a high quality practice question, the two statements always will lead to the same value for the expression in the prompt. If you stumble upon a DS practice question in which the two statements lead to different values of the prompt, then that's something that should make you question the validity of that particular source of questions.
As I am sure you are aware, not every car company that says, "We sell the best car on the road!" is actually selling the best car on the road. Much in the same way, not every GMAT prep company that says "We have high quality questions that are GMAT-like" actually has high quality, GMAT-like practice questions. Be a thoughtful and discerning consumer of GMAT practice questions.
Does all this make sense?
Mike
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Re: Do both sufficient statements provide the same answer in a DS question [#permalink]
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03 Nov 2015, 15:01
neenpaques wrote:
I just read for 'calculate value questions' like 1/(x + 4)? that you should not overthink and just check if statement 1 and 2 are sufficient to calculate x and I agree.
However, I was wondering if both statements alone are sufficient, if it is always D, or can the answers be dissimilar (so you have to calculate exactly)?
Post scriptum: Apologies if this is frequently asked
That is a very good question and a question that has been answered by experts above.
As for your question, for official GMAT questions, if both statements are sufficient individually, then you must get the same unique value.
This is actually a fun way of checking when you do end up with D for a question asking you a particular value of 'x'. So if you see that you are getting 2 different values of x individually, then recheck your solution.
Hope this helps.
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Re: Do both sufficient statements provide the same answer in a DS question [#permalink]
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03 Nov 2015, 20:20
neenpaques wrote:
I just read for 'calculate value questions' like 1/(x + 4)? that you should not overthink and just check if statement 1 and 2 are sufficient to calculate x and I agree.
However, I was wondering if both statements alone are sufficient, if it is always D, or can the answers be dissimilar (so you have to calculate exactly)?
Post scriptum: Apologies if this is frequently asked
Hi neenpaques,
This is a question that has troubled many test takers. But the bottom line is - If you encounter an official question or a question that has high standards as those of the official questions, then you will always find the values to be same from both the statements.
The answer will be D always.
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Re: Do both sufficient statements provide the same answer in a DS question [#permalink]
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03 Nov 2015, 21:20
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Expert's post
neenpaques wrote:
I just read for 'calculate value questions' like 1/(x + 4)? that you should not overthink and just check if statement 1 and 2 are sufficient to calculate x and I agree.
However, I was wondering if both statements alone are sufficient, if it is always D, or can the answers be dissimilar (so you have to calculate exactly)?
Post scriptum: Apologies if this is frequently asked
I like to explain this concept in this way:
A DS question is a puzzle.
For example: Given that x > 0, what is the value of x?
Now, from the question stem alone, you will not be able to get a unique value of x. So you cannot answer the puzzle.
x could be 1 or 5 or 1000 or 674828.45 etc.
Statement 1: x^2 + x - 2 = 0
On solving, this statement tells you that x is either -2 or 1.
Now can you answer the puzzle? Yes. We are given that x is positive and this statement tells us that x is either -2 or 1 so x must be 1. That is the only possible value. In this case, we say that statement 1 alone is sufficient to answer the question and hence answer would be either (A) or (D).
Now, think about it: can statement 2 tell you that x is either -4 or 5? No! Both statements are hints which may/may not help you find the answer to the SAME puzzle. x has one value and that is what you need to find. So statement 2 can tell you that x is 1 or 2. It can tell you that x is 1. It can tell you that x is 1, 5 or 7 etc. But is it possible that it does not include that value 1 for x? No! The actual value of x is 1 (which we found from statement 1). Since statement 2 is a hint to the same puzzle, it must include that value 1. Otherwise, it would be an incorrect hint and hence, an incorrect DS question.
Hope it helps.
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Do both sufficient statements provide the same answer in a DS question [#permalink]
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05 May 2016, 20:54
This is how I remember it, a la Johnny Cochran:
"If the answer is D, then the numbers must agree."
If you're getting D, but with two different answers, then you did something wrong on either condition #1 or condition #2.
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Do both sufficient statements provide the same answer in a DS question [#permalink] 05 May 2016, 20:54
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# Matter & Interactions II, Week 1
This week was supposed to begin on Monday, but we lost both Monday and Tuesday to snow and icy roads so this week was effectively just a two day week.
On Wednesday, I demonstrated Jupyter notebooks and informed the class that effective this semester, we’re moving away from Classic VPython. From this point on, we will only use GlowScript and Jupyter VPython. Using the latter is very important because it allows for file I/O whereas there’s no easy way to do that (that I’m aware of) with GlowScript. We will also continue using LaTeX (via Overleaf) for writing solutions.
On Thursday, I gave an overview of chapter 13 on electric force and the electric field of a particle. It’s interesting to note that the denominators of both expressions contains an area, specifically the area of a sphere. What might that be related to? I teased the class with this question in anticipation of the chapter on Gauss’s law.
Note the presence of absolute value bars and the sgn() function in each expression. Charge, unlike mass, can be positive or negative. A vector’s magnitude, however, must always be positive without exception, at least if we are going to stick with the fundamental definitions from first semester physics. That means that we must use the absolute value of charge to calculate the magnitude of an electric force or electric field. We could always sidestep this issue by instead defining the signed magnitude to be the scalar part of the vector, but this isn’t consistent with a vector being the product of a magnitude and a direction. In the expression for electric force, note that we could also take the absolute value of the product of the two charges, which might be a better way to write it. I’ll have to think about that.
Anyway, the sgn() function is necessary computationally. A person can work out the correct directions for force and field by physical and geometric reasoning, but a computer must be told explicitly how to do it, and that’s the purpose of the sgn() function here. It assures the correct geometry based on the signs of the charges. I’ve never seen this use in any textbook, but it seems quite necessary to me in order to maintain the fundamental definition of a vector’s magnitude. Thus, I include it.
Also note that we use double bars for vector magnitudes and single bars for absolute values. These are two conceptually different things and thus I feel they warrant different symbols. It is also consistent with what my students see in their calculus textbook and I try to maintain some sense of consistency between their math and physics texts.
UPDATE: Oh, one more thing. Every textbook I know of freely switches between Q and q for chcarge, even for the same expression and sometimes even for the same expression in the same chapter. This is confusing. To eliminate this confusion, I consistently use Q for a source charge (a charge associated with the creation (I don’t like that word) of an electric field) and q for an experiential charge (a charge that experiences an electric field created by another charge).
UPDATE: In the fourth edition of Matter & Interactions, Chabay and Sherwood deal with the sign issue by treating everything to the left of the unit vector in the above expressions as a signed scalar quantity and mention on page 520 that one should take the absolute value of this quantity to get the magnitude of the associated vector. Computationally, they calculate a particle’s electric field in one expression, without separately calculating the magnitude and direction, and this is fine. I think students should be aware of different sign conventions and their implications, but I also think foundational definition should be sacrosanct. If the foundation is variable, it isn’t a foundation after all.
UPDATE: After much thought, I have decided that I am okay with defining the magnitude of a particle’s electric field to be the absolute value of the quantity preceding the direction and excluding the sgn() function. The resulting caveat is that without taking the absolute value, we must not call this quantity a magnitude; it is a signed scalar.
I ended Thursday’s class with a question:
WHY must the electric force shared by two charged particles lie along the line connecting them?Â
This question can be answered with no numerical calculation or computation at all, but with physical reasoning using symmetry, specifically the fact that space is isotropic. The logic goes something like this:
• Define a system to consist of two charged particles with charges Q and q, isolated from all other influences.
• Assume that the force on q due to Q has a component that is NOT along the line connecting them, and draw an arrow representing this force with its tail on q.
• Rotate the system around an axis coinciding with the line connecting q and Q by 180 degrees, and draw the new system.
• Note that the rotated system is indistinguishable from the original system. This is important, because if nothing about the system changed, then we should expect there to be no change in the force on q due to Q.
• However, since we assumed that the force on q has a component perpendicular to the line connecting q and Q, the force “looks different” for the rotated system compared to the original system. A uniqueness theorem guarantees that for every charge distribution, there is one and only one net force on each particle. Thus, there cannot be more than one “correct” net force on q due to Q.
• If space is indeed isotropic, then if a change to the system causes the system to “look the same” then it cannot be the case that the force on q due to Q can have a component perpendicular to the line connecting q and Q.
• Therefore, the force on q must be such that is has no component perpendicular to the line connecting q and Q, and thus it must lie along that line, and we have used a simple proof by contradiction.
This type of powerful reasoning, appealing to symmetry, has many uses in electromagnetic theory, specifically in the introductory course where students need to ascertain the directions of electric fields due to certain charge distributions. Symmetry plays a role in setting up the integrals necessary for such calculuations. I think it is important to introduce reasoning by symmetry as early as possible. Note that this reasoning can also be applied to the geometry of the gravitational force from introductory mechanics.
Feedback is always welcome!
## One thought on “Matter & Interactions II, Week 1”
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# How do you evaluate -7( 1- 10)?
Mar 13, 2017
$63$
#### Explanation:
Use PEMDAS to solve. Start inside the parentheses and subtract 10 from 1:
$1 - 10 = - 9$
Now multiply $- 9$ by $- 7$:
$- 7 \times - 9 = 63$
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## 17 Sep 2016
### IBPS PO/Clerk Exam 2016 – Section wise Full Test-47
IBPS PO/Clerk Exam 2016 – Section wise Full Test-47:
Dear Readers, IBPS PO/Clerk 2016 was approaching, for that we have given the Section wise Full Test which consist of all the three sections such as, Aptitude, Reasoning, and English. This Section wise Full Test will be provided on daily basis kindly make use of it.
QUANTITATIVE APTITUDE
Directions (1 – 5): In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer
a) if x < y
b) if x > y
c) if x = y
d) if x ≥ y
e) if x ≤ y or no relationship can be established between x and y.
1). I. 6x2 + 77x + 121 = 0 II. y2 + 9y – 22 = 0
2). I. x2 – 24x + 144 = 0 II. y2 – 26y + 169 = 0
3). I. x2 – 6x = 7 II. 2y2 + 13y + 15 = 0
4). I. 2x2 + 3x – 20 = 0 II. 2y2 +19y + 44 = 0
5). I. 10x2 – 7x + 1 = 0 II. 35y2 – 12y + 1 = 0
REASONING
Directions (6-10): Study the following information to answer the given questions:
5 students – P, Q, R, S, and T buys stationery items – book, eraser, pen, pencil, and sharpener but not necessarily in the same order from different shops – A, B, C, D, and E. they study in classes I, Ii, III, IV, and V but not necessarily in the same order.
R buys book from shop B and is studying in IV. S buys sharpener but is not studying in class V. In shop C, neither erasers nor sharpeners are sold. The student who bought eraser is not studying in III. Either P or Q is studying in class I. The student who bought from shop E is studying in class V, but is not Q. T did not buy Pen, but is studying in class III. The student who is studying in III, bought from shop D.
6). Who bought eraser?
a) P
b) Q
c) R
d) T
e) S
a) book
b) pencil
c) pen
d) Cannot be determined
e) None of these
8). From which shop the pen was bought?
a) A
b) B
c) C
d) D
e) None of these
9). Which of the following is the correct combination of an item that student bought and his class?
a) book – I
b) book – III
c) pencil – II
d) pen – III
e) eraser – V
10). If shop C is related to book and shop B is related to sharpener in a certain way, then to which of the following shop D is related to?
a) pen
b) pencil
c) eraser
d) book
e) None of these
ENGLISH
Directions (Q. 11-15): In each of the following questions an idiomatic expression and its four possible meanings are given. Find out the correct meaning of the idiomatic expression. If you do not find any correct answer mark e ie ‘None of these’ as your answer.
11). To lose face _________
a) To gain an advantage over
b) To wait eagerly for
c) To fight unfairly
d) To avoid being friendly with
e) None of these
12). The rank and files ______
a) The ordinary members of a society
b) Bring about appeasement in a disturbed situation
c) A fuss over a trifling matter
d) An established secured right
e) None of these
13). To make a long face ______
a) To be short of money
b) To make a gain that is legitimate
c) To be friendly with
d) To look oppressed
e) None of these
14). In apple pie order_____
a) To change for the better
b) Expose ones inner feelings to others
c) Perfectly methodical arrangement
d) To escape from danger
e) None of these
15). In the blues_____
a) Behave like a lord
b) Melancholy and low-spirited
c) Being colourful
d) Having many blue things
e) None of these
1)e 2)a 3)b 4)d 5)d 6)a 7)b 8)c 9)e 10)e 11)e 12)a 13)d 14)c 15)a
Solutions:
1). e)
I. 6x2 + 77x + 121 = 0
or, 6x + 66x + 11x + 121 = 0
or, 6x(x + 11) + 11(x + 11) = 0
or, (6x + 11) (x + 11) = 0
or, x = – 11/6
II. y2 + 9y – 22 = 0
or, y + 11y – 2y – 22 = 0
or, y(y + 11) – 2(y + 11)
or, (y – 2) (y + 11) = 0
or, y = 2, –11
Hence, no relationship can be established between x and y.
2). a)
I. x2 – 24x + 144 = 0
or, x2 – 12x – 12x + 144 = 0
or, x(x – 12) – 12(x – 12) = 0
or, (x – 12)2 = 0
x = 12
II. y2 – 26y + 169 = 0
or, y2 – 13y – 13y + 169 = 0
or, y(y – 13) – 13(y – 13) = 0
or, (y – 13)2 = 0
y = 13
Hence, x < y
3). b)
I. x2 – 6x = 7
or, x2 – 6x – 7 = 0
or, x2 – 7x + x – 7 = 0
or, x(x – 7) + 1(x – 7) = 0
or, (x + 1) (x – 7) = 0
or, x = –1, 7
II. 2y2 + 13y + 15 = 0
or, 2y2 + 10y + 3y + 15 = 0
or, 2y(y + 5) + 3(y + 5) = 0
or, (2y + 3) (y + 5) = 0
or, y = – 3/2, –5
Hence, x > y
4). d)
I. 2y2 + 3x – 20 = 0
or, 2x2 + 8x – 5x – 20 = 0
or, 2x(x + 4) – 5(x + 4) = 0
or, (2x – 5) (x + 4) = 0
or, x = 5/2, –4
II. 2y2 + 19y + 44 = 0
or, 2y2 + 11y + 8y + 44 = 0
or, 2y(2y + 11) + 4(2y + 11) = 0
or, (y + 4) (2y + 11) = 0
y = –4, – 11/2
Hence, x ≥ y
5). d)
I. 10x2 – 7x + 1 = 0
or, 10x2 – 5x – 2x + 1 = 0
or, 5x(2x – 1) – 1(2x – 1) = 0
or, (5x – 1) (2x – 1) = 0
or, x = 1/5, ½
II. 35y2 – 12y + 1 = 0
or, 35y2 – 7y + 5y + 1 = 0
or, 7y(5y – 1) – 1(5y – 1) = 0
or, (7y – 1) (5y – 1) = 0
or, y = 1/7, 1/5
Hence, x ≥ y
(6 – 10):
Student Shop Class Stationery P E V eraser Q C I pen R B IV book S A II sharpener T D III pencil
6). a)
7). b)
8). c)
9). e)
10). e)
Directions (Q. 11-15):
11). e) ‘to lose face’ means be humiliated; lose credit or reputation. As, no option expresses the meaning correctly, so the answer is (e).
12).a) ‘Rank and file’ means ordinary soldiers; ordinary undistinguished people; members of a group of organization.
13). d)
14). c) ‘In apple-pie order’ means very neatly arranged.
15). a)
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## Saturday, March 19, 2011
### Suzie's Term 2 Reflection
In term 2, we learned about percents, surface area and volume. For this term, I did well on percents and volume. Surface area and volume was okay, but it was kind of confusing. Overall, I think I did really great, even though my math mark went from a 90 to a 89. I think it’s because I was mostly confused with the surface area and volume thing, but when I got it, it was easy. I did really well on my tests, and I got perfect on most of them, except for a few pesky ones.
Like I mentioned before, I struggled mostly with surface area and a bit with volume. It took me some time to get the whole surface area thing, but when I got it, it became easy. Volume started out really easy, but then it got harder with the confusing math problems. Some math problems made absolutely no sense to me, but when they were explained, I said “oooh!” The whole pipe concept was weird, but i eventually got that too. I didn’t like it when the textbook started calling a pipe something else, though. You called it a pipe first, why don’t you just stick with that!? Are you trying to confuse me?
Next term, I will do better by paying attention more, so I don’t have to ask as many stupid questions. (They’re still coming, Mr. Harbeck.) I will also try to comment more on the blog and try to do more blog work so I can get an even higher mark. I should probably study more, too. That would be a good idea. I also have to prepare my brain for confusing computer stuff like audioboo. That will be hard, though. Computers confuse the heck out of me.
I learned a lot about percents, surface area and volume. We learned how to calculate percents, and change them to fractions and decimals, and solve problems with them. With surface area, we learned how to find the surface area of cylindrical prisms, rectangular prisms (and square prisms) and triangular prisms. We learned the formula for finding it, and how to solve problems based on it. We also learned how to find the volume for those things too, and the formulas for finding it. We also solved problems to do with volume. It was pretty much like the surface area unit, except with volume.
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A161676 Isolated primes p such that 2^p-1 is also a prime number. 0
%I #6 Feb 21 2019 23:10:58
%S 2,89,127,607,2203,2281,3217,4253,9689,9941,11213,19937,21701,23209,
%T 44497,86243,216091,756839,859433,1257787,1398269,2976221,3021377,
%U 6972593,13466917
%N Isolated primes p such that 2^p-1 is also a prime number.
%F A000043 INTERSECT A007510.
%e For the isolated prime p=2= A007510(1), p^2-1=3 is prime, which adds p=2 to the sequence.
%p MP := [2, 3, 5, 7, 13, 17, 19, 31, 61, 89, 107, 127, 521, 607, 1279, 2203, 2281, 3217, 4253, 4423, 9689, 9941, 11213, 19937, 21701, 23209, 44497, 86243, 110503, 132049, 216091, 756839, 859433, 1257787, 1398269, 2976221, 3021377, 6972593, 13466917]:
%p n := 0: for j to 39 do if `and`(isprime(MP[j]-2) = false, isprime(MP[j]+2) = false) then n := n+1; a[n] := MP[j] else end if end do:
%p seq(a[n], n = 1 .. 25); # _Emeric Deutsch_, Jun 20 2009
%Y Cf. A000043, A007510.
%K nonn
%O 1,1
%A Vladislav-Stepan Malakhovsky & _Juri-Stepan Gerasimov_, Jun 16 2009
%E More terms from _Emeric Deutsch_, Jun 20 2009
%E Definition rephrased by _R. J. Mathar_, Sep 23 2009
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Last modified July 13 02:46 EDT 2024. Contains 374264 sequences. (Running on oeis4.)
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# Movement Along a Curve versus Shift of a Curve: Explanation & Diagrams
The compilation of these Income Determination Notes makes students exam preparation simpler and organised.
## Movement Along a Curve versus Shift of a Curve
Income comprises consumption, savings, and investment components. But what happens to these variables when income changes? And, what happens to them when at a given level of income, other variables change? The answer lies in the difference between a movement along a curve and a shift of a curve. Read this simple lesson with a diagrammatic explanation to understand the distinction.
### National Income and Movement of Curves
In the lesson on Income determination, we have learned how income is determined in an economy. We have also understood the concept of ex-ante and ex-post. But what happens when a country’s income changes? Does the national income of India remain constant? No, certainly.
The implication of this change in Macroeconomics is reflected through two kinds of movements of curves: movement along a curve and shift of a curve. It is best to understand the difference between them so as not to confuse them. This lesson will make it simple to understand the difference.
Movement along a Curve
We know that as income increases, consumption increases due to a positive marginal propensity to consume (MPC). This explains the relationship between consumption and income, i.e. C = f(Y)
The consumption function slopes upward because of this positive income-consumption relation. The more income you have in hand, the more you can spend to consume goods and services, right? Now, if income increases, from Y0 to Y1, consumption increases as we can see, from C0 to C1.
Let us now look at savings. As income increases, you have more income in hand that you can choose to save for future use. Thus, the more the income more will be the savings. This explains a positive relationship between income and savings as well, due to a positive marginal propensity to save (MPS). This is the savings function, as under S = f(Y)
Thus, the savings function slopes upward as well. Now, as income increases from Y0 to Y1, from the positively sloped savings function, we can see that savings must increases from S0 to S1.
The above two cases are shown in the first two panels of the figure below. Here, since the consumption and savings increase as functions of income on which they are dependent, we call it moves along a curve.
No discussion of income, consumption, and savings can be complete without considering the investment, right? In our discussion, we consider the investment as autonomous or determined from outside and to be a given value. Hence, as income increases, investment remains the same and does not change. The horizontal investment curve shows that it is given and constant.
The shift of a Curve
Think about this. Your consumption depends not only on your income but also on other factors, like your tastes and preferences, your wealth, etc. For instance, you might experience a shift in taste in favour of jute bags from polythene bags. Or, when your wealth increases due to accumulated savings, you may feel richer to spend more on consumption. Such cases represent factors other than income that cause consumption to increase. This leads to a shift of the consumption curve from C to C’. An increase in consumption is depicted by an ‘upward’ shift of the consumption curve.
We know that consumption and savings functions are analogous to each other and when consumption must increase, less income is available to be saved. Hence, the savings function ‘shifts downward’. Again, here other factors rather than income have affected savings level and caused the shift.
If you are to look at investment, the investment function may shift due to other factors like interest rate. When the interest rate falls, the cost of investing falls and it is more profitable to invest. Thus, the investment function shifts upward.
To sum up, movement along a curve is always associated with a change in the independent variable. In the case of the consumption and savings functions, income is the independent variable. On the other hand, a shift of a curve is associated with changes in other variables affecting the dependent variable other than the independent variable (say, wealth in case of consumption) and is caused due to changes in other factors affecting consumption/savings/investment at any given level of income.
Example:
Question:
What is the ratio of Marginal Propensity to Consume?
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```Name Class Date 11.2
Solving Linear Systems
by Substitution
Essential Question: How can you solve a system of linear equations by using substitution?
Resource
Locker
1Explore Exploring the Substitution Method of Solving
Linear Systems
Another method to solve a linear system is by using the substitution method.
In the system of linear equations shown, the value of y is given. Use this value of y to find the value of x and the
solution of the system.
⎧ y = 2
⎨
⎩x + y = 6
A
B
Substitute the value of y in the second
equation and solve for x.
x + y = 6
x+
© Houghton Mifflin Harcourt Publishing Company
=6
The values of x and y are known. What is the
solution of the system?
Solution:
(
)
x=
C
Graph the system of linear equations. How do your solutions compare?
D
Use substitution to find the values of x and y in this system of linear
equations. Substitute 4x for y in the second equation and solve for x.
Once you find the value for x, substitute it into either original equation
to find the value for y.
⎧ y = 4x
⎨
⎩ 5x + 2y = 39
(
,
Solution:
,
8
y
4
x
-8
-4
0
4
8
-4
-8
)
Reflect
1.
Discussion For the system in Step D, what equation did you get after substituting 4x for y in
5x + 2y = 39 and simplifying?
2.
Discussion How could you check your solution in part D?
Module 11
491
Lesson 2
Solving Consistent, Independent Linear Systems
by Substitution
Explain 1
The substitution method is used to solve a system of equations by solving an equation for
one variable and substituting the resulting expression into the other equation. The steps for the
substitution method are as shown.
1. Solve one of the equations for one of its variables.
2. Substitute the expression from Step 1 into the other equation and solve for the other variable.
3. Substitute the value from Step 2 into either original equation and solve to find the value of
the other variable.
Example 1
A
Solve each system of linear equations by substitution.
⎧ 3x + y = -3
⎨
⎩ -2x + y = 7
Solve an equation for one variable.
3x + y = -3
Select one of the equations.
y = -3x - 3
Solve for y. Isolate y on one side.
Substitute the expression for y in the other equation and solve.
-2x + (-3x - 3) = 7
Substitute the expression for y.
-5x - 3 = 7
Combine like terms.
-5x = 10
x = -2
Divide each side by -5.
Substitute the value for x into one of the equations and solve for y.
3(-2) + y = -3
Substitute the value of x into the first equation.
-6 + y = -3
Simplify.
y=3
© Houghton Mifflin Harcourt Publishing Company
So, (-2, 3) is the solution of the system.
Check the solution by graphing.
8
y
-2 + y = 2
x
-8
0
-4
-8
Module 11
4
8
3x + y = -3
-2x + y = 7
x-intercept: -1
7
x-intercept: -_
2
y-intercept: 7
y-intercept: -3
The point of intersection is (-2, 3).
3x + y = -3
492
Lesson 2
⎧ x - 3y = 9
B⎨ ⎩
x + 4y = 2
Solve an equation for one variable.
x - 3y = 9
Select one of the equations.
x=
Solve for x. Isolate x on one side.
Substitute the expression for in the other equation and solve.
(
) + 4y = 2
Substitute the expression for .
= 2
=
y=
Combine like terms.
Subtract
Divide each side by
.
Substitute the value for y into one of the equations and solve for x.
x - 3
( ) = 9
Substitute the value of y into the first equation.
= 9
(
x=
,
)
Simplify.
Subtract
So,
Check the solution by graphing.
x - 3y = 9
y
4
x
-8
-4
from both sides.
is the solution by graphing.
8
© Houghton Mifflin Harcourt Publishing Company
from both sides.
0
-4
4
8
x + 4y = 2
x-intercept:
x-intercept:
y-intercept:
y-intercept:
The point of intersection is
(
,
).
-8
Reflect
3.
Explain how a system in which one of the equations if of the form y = c, where c is a constant is a special
case of the substitution method.
4.
Is it more efficient to solve -2x + y = 7 for x than for y? Explain.
Module 11
493
Lesson 2
Solve the system of linear equations by substitution.
5.
⎧ 3x + y = 14
⎨
⎩ 2x - 6y = -24
Solving Special Linear Systems by Substitution
Explain 2
You can use the substitution method for systems of linear equations that have infinitely many solutions and for
systems that have no solutions.
Example 2
A
Solve each system of linear equations by substitution.
⎧ x + y = 4
⎨
⎩ -x - y = 6
Solve x + y = 4 for x.
8
4
x = -y + 4
Substitute the resulting expression into the other
equation and solve.
-(-y + 4) - y = 6
Simplify.
The graph shows that the lines
are parallel and do not intersect.
8
x
-8
-4
) - 12y = 24
= 24
The resulting equation is
system has
Substitute.
0
8
-4
-8
The graphs are
Simplify.
4
so the system has
,
.
, so the
.
Reflect
6.
Provide two possible solutions of the system in Example 2B. How are all the solutions of this system related
to one another?
Module 11
494
Lesson 2
© Houghton Mifflin Harcourt Publishing Company
Substitute the resulting expression into the other
equation and solve.
(
y
4
.
x=
4
8
-x - y = 6
⎧ x - 3y = 6
⎨
⎩ 4x - 12y = 24
Solve x - 3y = 6 for
4
x
-8
The resulting equation is false, so the
system has no solutions.
B
x+y=4
0
-8
Substitute.
-4 = 6
y
Solve each system of linear equations by substitution.
7.
⎧ -2x + 14y = -28
⎨
x - 7y = 14
⎩
Explain 3
8.
⎧-3x + y = 12
⎨
⎩ 6x - 2y = 18
Solving Linear System Models by Substitution
You can use a system of linear equations to model real-world situations.
Example 3
A
Solve each real-world situation by using the substitution method.
Fitness center A has a \$60 enrollment fee and costs \$35 per month. Fitness center B has
no enrollment fee and costs \$45 per month. Let t represent the total cost in dollars and
⎧ t = 60 + 35m
m represent the number of months. The system of equations ⎨
can be used
⎩ t = 45m
to represent this situation. In how many months will both fitness centers cost the same?
What will the cost be?
60 + 35m = 45m
Substitute 60 + 35m for t in the second equation.
60 = 10m
Subtract 35m from each side.
6=m
Divide each side by 10.
t = 45m
Use one of the original equations.
(6, 270)
Write the solution as an ordered pair.
= 45(6) = 270
Substitute 6 for m.
© Houghton Mifflin Harcourt Publishing Company
Both fitness centers will cost \$270 after 6 months.
B
High-speed Internet provider A has a \$100 setup fee and costs
\$65 per month. High-speed internet provider B has a setup fee of \$30 and
costs \$70 per month. Let t represent the total amount paid in dollars and
⎧ t = 100 + 65m
m represent the number of months. The system of equations ⎨
⎩ t = 30 + 70m
can be used to represent this situation. In how many months will both
providers cost the same? What will that cost be?
= 30 + 70m
100 =
=
m
=m
Module 11
for t in the second
Substitute
equation.
Subtract
m from each side.
Subtract
from each side.
Divide each side by
495
.
Lesson 2
t = 30 + 70m Use one of the original equations.
t = 30 + 70
t=
(
,
( ) Substitute
for m.
) Write the sulotion as an ordered pair.
Both Internet providers will cost \$ after months.
Reflect
9.
If the variables in a real-world situation represent the number of months and cost, why must the values of
the variables be greater than or equal to zero?
10. A boat travels at a rate of 18 kilometers per hour from its port. A second boat is 34 kilometers behind the
first boat when it starts traveling in the same direction at a rate of 22 kilometers per hour to the same port.
Let d represent the distance the boats are from the port in kilometers and t represent the amount of time
⎧ d
= 18t + 34
in hours. The system of equations ⎨
can be used to represent this situation. How many
⎩ d = 22t
hours will it take for the second boat to catch up to the first boat? How far will the boats be from their
port? Use the substitution method to solve this real-world application.
Elaborate 11. When given a system of linear equations, how do you decide which variable to solve for first?
12. How can you check a solution for a system of equations without graphing?
13. Essential Question-Check-In Explain how you can solve a system of linear equations by substitution.
Module 11
496
Lesson 2
© Houghton Mifflin Harcourt Publishing Company
Evaluate: Homework and Practice
1.
• Online Homework
• Hints and Help
• Extra Practice
In the system of linear equations shown, the value of y is given. Use this
value of y to find the value of x and the solution of the system.
⎧
y = 12
⎨
⎩ 2x - y = 4
a. What is the solution of the system?
b. Graph the system of linear equations.
How do the solutions compare?
y
16
8
x
-16 -8
0
8
16
-8
-16
© Houghton Mifflin Harcourt Publishing Company
Solve each system of linear equations by substitution.
2.
⎧5x + y = 8
⎨
⎩2x + y = 5
3.
⎧ x - 3y = 10
⎨
⎩ x + 5y = -22
4.
⎧ 5x - 3y = 22
⎨
⎩ -4x + y = -19
5.
⎧
x + 7y = -11
⎨
⎩ -2x - 5y = 4
6.
⎧ 2x + 6y = 16
⎨
⎩ 3x - 5y = -18
7.
⎧ 7x + 2y = 24
⎨
⎩ -6x + 3y = 3
Module 11
497
Lesson 2
Solve each system of linear equations by substitution.
8.
⎧ x + y = 3
⎨
-4x - 4y = 12
⎩
⎧ 5x - y = 18
11. ⎨
⎩ 10x - 2y = 32
9.
⎧ 3x - 3y = -15
⎨
-x + y = 5
⎩
⎧ -2x - 3y = 12
12. ⎨
⎩ -4x - 6y = 24
x - 8y = 17
⎧
10. ⎨
-3x
+ 24y = -51
⎩
⎧ 3x + 4y = 36
13. ⎨
⎩ 6x + 8y = 48
Solve each real-world situation by using the substitution method.
used to represent this situation. If this trend continues, in how many months will the number
of DVDs sold equal the number of Blu-ray discs sold? How many of each is sold in that month?
15. One smartphone plan costs \$30 per month for talk and messaging and \$8 per gigabyte of data used each
month. A second smartphone plan costs \$60 per month for talk and messaging and \$3 per gigabyte of
data used each month. Let c represent the total cost in dollars and d represent the amount of data used in
⎧ c
= 30 + 8d
gigabytes. The system of equations ⎨
can be used to represent this situation. How many
⎩ c = 60 + 3d
gigabytes would have to be used for the plans to cost the same? What would that cost be?
Module 11
498
Lesson 2
© Houghton Mifflin Harcourt Publishing Company • Image Credits:
14. The number of DVDs sold at a store in a month
was 920 and the number of DVDs sold decreased by
12 per month. The number of Blu-ray discs sold in
the same store in the same month was 502 and the
number of Blu-ray discs sold increased by 26 per
month. Let d represent the number of discs sold and
t represent the time in months.
⎧ d = 920 - 12t
The system of equations ⎨
can be
⎩ d = 502 + 26t
16. A movie theater sells popcorn and fountain drinks. Brett buys 1 popcorn bucket and 3 fountain drinks for
his family, and pays a total of \$9.50. Sarah buys 3 popcorn buckets and 4 fountain drinks for her family,
and pays a total of \$19.75. If p represents the number of popcorn buckets and d represents the number of
⎧ 9.50 = p + 3d
drinks, then the system of equations ⎨
can be used to represent this situation. Find the
19.75
= 3p + 4d
⎩
cost of a popcorn bucket and the cost of a fountain drink.
17. Jen is riding her bicycle on a trail at the rate of 0.3 kilometer per minute. Michelle is 11.2 kilometers behind
Jen when she starts traveling on the same trail at a rate of 0.44 kilometer per minute. Let d represent the
distance in kilometers the bicyclists are from the start of the trail and t represent the time in minutes.
⎧ d = 0.3t + 11.2
The system of equations ⎨
can be used to represent this situation. How many minutes
d
⎩ = 0.44t
will it take Michelle to catch up to Jen? How far will they be from the start of the trail? Use the substitution
method to solve this real-world application.
© Houghton Mifflin Harcourt Publishing Company • Image Credits:
18. Geometry The length of a rectangular room is 5 feet more than its width. The perimeter of the room is
66 feet. Let L represent the length of the room and W represent the width in feet. The system of equations
⎧ L = W + 5
⎨
sed to represent this situation. What are the room’s dimensions?
can be u
⎩ 66 = 2L + 2W
19. A cable television provider has a \$55 setup fee and charges \$82 per month, while a satellite television
provider has a \$160 setup fee and charges \$67 per month. Let c represent the total cost in dollars and t
⎧ c = 55 + 82t
represent the amount of time in months. The system of equations ⎨
can be used to represent
c = 160 + 67t
⎩
this situation.
a.
In how many months will both providers cost the same? What will that cost be?
b.
If you plan to move in 12 months, which provider would be less expensive? Explain.
Module 11
499
Lesson 2
20. Determine whether each of the following systems of equations have one solution, infinitely many solutions,
or no solution. Select the correct answer for each lettered part.
x + y = 5 ⎧
a. ⎨
-6y
- 6y = 30
⎩
⎧ x + y = 7
b. ⎨
⎩ 5x + 2y = 23
⎧ 3x + y = 5
c. ⎨
⎩ 6x + 2y = 12
⎧2x + 5y = -12
d. ⎨
⎩ x + 7y = -15
⎧ 3x + 5y = 17
e. ⎨
⎩ -6x - 10y = -34
21. Finance Adrienne invested a total of \$1900 in two simple-interest money market accounts. Account A
paid 3% annual interest and account B paid 5% annual interest. The total amount of interest she earned
after one year was \$83. If a represents the amount invested in dollars in account A and b represents the
⎧ a + b = 1900
amount invested in dollars in account B, the system of equations ⎨
can represent
⎩ 0.03a + 0.05b = 83
this situation. How much did Adrienne invest in each account?
H.O.T. Focus on Higher Order Thinking
Hustace/Corbis
22. Real-World Application The Sullivans
are deciding between two landscaping
companies. Evergreen charges a \$79 startup
fee and \$39 per month. Eco Solutions charges
a \$25 startup fee and \$45 per month. Let c
represent the total cost in dollars and t represent
the time in months. The system of equations
⎧c
= 39t + 79
⎨
can be used to represent this
c = 45t + 25
⎩
situation.
a.
In how many months will both landscaping
services cost the same? What will that cost be?
b.
Which landscaping service will be less expensive in the long term? Explain.
Module 11
500
Lesson 2
23. Multiple Representations For the first equation in the system of linear equations below, write an
equivalent equation without denominators. Then solve the system.
x _
y
⎧ _
+ = 6
|
5
3
⎨
x
2y
=8
⎩
© Houghton Mifflin Harcourt Publishing Company
24. Conjecture Is it possible for a system of three linear equations to have one solution? If so, give
an example.
25. Conjecture Is it possible to use substitution to solve a system of linear equations if one equation
represents a horizontal line and the other equation represents a vertical line? Explain.
Module 11
501
Lesson 2
A company breaks even from the production and sale of a product if the total revenue
equals the total cost. Suppose an electronics company is considering producing two types of
smartphones. To produce smartphone A, the initial cost is \$20,000 and each phone costs \$150
to produce. The company will sell smartphone A at \$200. Let C(a) represent the total cost in
dollars of producing a units of smartphone A. Let R(a) represent the total revenue, or money
the company takes in due to selling a units of smartphone A. The system of
(a)= 20,000 + 150a
⎧C
can be used to represent the situation for phone A.
equations ⎨
R
⎩ (a)= 200a
To produce smartphone B, the initial cost is \$44,000 and each phone costs \$200 to produce.
The company will sell smartphone B at \$280. Let C(b) represent the total cost in dollars of
producing b units of smartphone B and R(b) represent the total revenue from
(b)= 44,000 + 200b
⎧C
selling b units of smartphone B. The system of equations ⎨
can be
⎩ R(b)= 280b
used to represent the situation for phone B.
Solve each system of equations and interpret the solutions. Then determine whether the
company should invest in producing smartphone A or smartphone B. Justify your answer.
Hustace/Corbis
Module 11
502
Lesson 2
```
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# vertical line test calculator
The vertical Line test. Get an answer to your question âWhen given a graph, the vertical line test can be used to determine functionality.Describe the vertical line test and explain the reasons ...â in ð Mathematics if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions. example. What properties of a written relation can tell us whether or not it is a function? Notice that graph touches the vertical line at 2 and -2 when it intersects the x axis at 4. If the line appears within five minutes of testing, then a faint line indicates a positive pregnancy. If a vertical line passes through two or more points on the graph, then the relations is not a function. This precalculus video tutorial provides a basic introduction into the vertical line test. Examples. Generating two different uniformly distributed points on a sphere using one uniform distribution: Cube V=8. Function? Function Worksheets (free pdf's with answer keys on all the topics normally covered in Algebra 1 and Algebra 2) One To One Functions. Look at the graph below. If no two different points in a graph have the same first coordinate, this means that vertical lines cross the graph at most once. It is a function. What does VERTICAL LINE TEST mean? Therefore, the vertical line test concludes that a curve in the plane 2010. By using this website, you agree to our Cookie Policy. This means that when x is 4, the y must be 5. How can we tell if a relation is a function by looking at its graph? 4th Edition. other vertical line. If each line crosses the graph just once, the graph passes the vertical line test. Your vertical line must only touch the graph at one point. Yes or no the vertical finest 13) 14. states that if a vertical line intersects the graph of the relation more than once, then the relation is a NOT a function. If a vertical line is drawn for each value in the domain of a relation and if that line only intersects the relation once then the relation is a function. https://mathworld.wolfram.com/VerticalLineTest.html. Describe the vertical line test in your own words. Yes or no the vertical fine test V 11. most once. If a vertical line can cross a graph more than once, then the graph does not pass the vertical line test. Practice online or make a printable study sheet. This is known as the vertical line test. Let's say your function has the ordered pair (4,5). Meaning of VERTICAL LINE TEST. (Unless you write y = ± something, which doesnât count.) The #1 tool for creating Demonstrations and anything technical. Is each input only paired with only one output? Walk through homework problems step-by-step from beginning to end. Lines: Point Slope Form. in two points. W. Weisstein. What properties of a written relation can tell us whether or not it is a function? Vertical Line Test. Le⦠Vertical Line Test. #y=3x+4# is a line and passes the vertical line test, it is a function. Let's look at our relation, b that we used in our relations example in the previous lesson.. Is this relation a function? Figure 12 A function can only have one output, y, for each unique input, x. Stewart, J. Calculus: Concepts and Contexts, 4th Edition. Applied Calculus for the Managerial, Life, and Social Sciences (8th Edition) Edit edition. The vertical line test determines if for every value of x there is only one output y corresponding to it. There are actually two ways to determine if a relation is a function. f(x) = 2x g(x) - 5x + 1 f(g(3)) = ? A test use to determine if a relation is a function. If the vertical line passes through at least two points on the graph, then an element in the domain is paired with more than 1 element in the range. In mathematics, the vertical line test is a visual way to determine if a curve is a graph of a function or not. New Blank Graph. "Vertical Line Test." Lines: Slope Intercept Form. Parabolas: Vertex Form. On the other hand, the vertical line test shows that the rightmost Hence equation of the required line is x = 3. example. The motivation for the vertical line test is as follows: A relation is a function precisely when each element intersect the curve in more than one point and, by observation, neither would any Function? Since the slope of the line passes which through the point (3, -5) is undefined, it should be vertical line. Explain your answer. is a way to determine if a relation is a function. \$\$ = \$\$ + Sign UporLog In. For any input x, a function can only have one corresponding y value. Vertical Line test : A relation is a function if and only if no vertical line intersects the graph of the relation at more than one point. The Vertical Line Test is a visual test that you can use to quickly check and see if a graph represents a function. example. Describe the vertical line test in your own words. Graphs that pass the vertical line test are graphs of functions. When you have connected all of the points, you will be told if your graph is a valid graph of a function. A plane curve which doesn't represent the graph of a function is sometimes said to have failed the vertical line test. It can be confusing when a woman sees two lines on her pregnancy testâone dark and another very faint. example. Further, there is a point of intersection if and only if is in the domain of the function. Horizontal Line Test. Free line equation calculator - find the equation of a line given two points, a slope, or intercept step-by-step This website uses cookies to ensure you get the best experience. The vertical line test is a method that is used to determine whether a given relation is a function or not. Now that you have two strategies under your belt, let's give it a whirl. The vertical line test is a simple method for determining if an equation is indeed a function. ordered pair function calculator, 3.1 Functions and Function Notation In this section you will learn to: ⢠find the domain and range of relations and functions ⢠identify functions given ordered pairs, graphs, and equations ⢠use function notation and evaluate functions ⢠use the Vertical Line Test (VLT) to identify functions ⢠apply the difference quotient https://mathworld.wolfram.com/VerticalLineTest.html. Hints help you try the next step on your own. The approach is rather simple. represents the graph of a function if and only if no vertical line intersects it Why does this work? Explore thousands of free applications across science, mathematics, engineering, technology, business, art, finance, social sciences, and more. Remember that it is very possible that a function may have an inverse but at the same time, the inverse is not a function because it doesnât pass the vertical line test. Composition of Functions. If you think about it, the vertical line test is simply a restatement of the definition of a function. Using the Vertical Line Test, determine whether each relation is a function. Join the initiative for modernizing math education. Rom version 2.0V: only three display test patterns (vertical lines, horizontal lines, and a combination checkerboard and vertical bar pattern) with no contrast test; the keyboard test starts with "49" displayed and decrements this count as each key is pressed in order starting with the "Y=" key. As a result, the relation is not a function. This remembers the word "VUX". Let's analyze our ordered pairs first. curve is a function on its domain: Indeed, none of the vertical lines drawn If we can draw any vertical line that intersects a graph more than once, then the graph does not define a function because a function has only one output value for each input value. Function? Input an equation, and use the slider to perform the vertical line test! The vertical line test is used to find out if a relation is a function. a result, any vertical line in the plane can intersect the graph of a function at more than once. The vertical line test can be used to determine whether a graph represents a function. Otherwise, there is no point of intersection. Yes or the vertical line 13. Stover. On the other hand, the sideways parabola x = 5y 2 + 4y â 10 isnât a function because thereâs no way to write it as y = something. See also. What Does an Evaporation Line Mean on a Pregnancy Test? Vertical Line Test is one of the Interactivate assessment explorers. Input an equation, and use the slider to perform the vertical line test! A relation is a function if there are no vertical lines that intersect the graph at more than one point. A function is basically anything you can graph on your graphing calculator in ây =â or graphing mode. So the equation is in the form x = a constant. Graphing Linear Inequalities Written in Standard Form (Quiz), Central Limit Theorem on Tossing a Fair Die. Therefore when x = 4 there are two different y-values (2 and -2). Yes or the vertical intest 12. If you are using the vertical line test, and your "line" touches the graph in more than one point, then it is not a function. One way is to analyze the ordered pairs, and the other way is to use the vertical line test. Improve your math knowledge with free questions in "Identify functions: vertical line test" and thousands of other math skills. Function? This is the only way to lock in the Vertical Line You should now see the draw screen Make sure that you are on the Graph Screen first Click the 2nd button Scroll down to 4:Vertical click Enter. Unlimited random practice problems and answers with built-in Step-by-step solutions. example. Stover, Christopher. Belmont, CA: Cengage Learning via Brooks/Cole, Vertical Line Test: Learn about the vertical line test for functions by trying to connect points in the plane to build a function. Since each input has a different output, this canbe classified as a function. to save your graphs! Boston, MA: Pearson, 2009. Parabolas: Standard Form. Information and translations of VERTICAL LINE TEST in the most comprehensive dictionary definitions resource on the web. The vertical line test is a graphical method of determining whether a curve in the plane represents the graph Lines: Two Point Form. Draw a few vertical lines spread out on your graph. So this function FAILS the vertical line test. In order to be a function, each x value can only be paired with exactly one y value. Definition of VERTICAL LINE TEST in the Definitions.net dictionary. Knowledge-based programming for everyone. The leftmost curve fails the vertical line test due to the fact that the single vertical line drawn intersects the curve Draw a vertical line cutting through the graph of the relation, and then observe the points of intersection. Vertical and Horizontal Shifts of Graphs ... \$\$ 3 \$\$ â A B C \$\$ \$\$ Ï \$\$ 0 \$\$. Function? The horizontal line test is a convenient method that can determine whether a given function has an inverse, but more importantly to find out if the inverse is also a function.. of a function by visually examining the number of intersections of the curve with vertical lines. How can we tell if a relation is a function by looking at its graph? Calculus: Concepts and Contexts, 4th Edition. The line y = 3x â 2 is a function, as is the parabola y = 4x 2 â 3x + 6. Explore anything with the first computational knowledge engine. Collection of teaching and learning tools built by Wolfram education experts: dynamic textbook, lesson plans, widgets, interactive Demonstrations, and more. The figure above shows two curves in the plane. The picture below shows the graph of the function and the line . Practice: Determining if a Relation is a Function (Vertical Line Test) Directions: Determine whether each relation is a function. The test states that a graph represents a function if and only if all vertical lines ⦠Vertical line test, Horizontal line test, One-to-one function. From MathWorld--A Wolfram Web Resource, created by Eric Lial, M.; Hornsby, J.; and Schneider, D. I. Precalculus Inverse of a Function. Create a Vertical Line using your Calculator A Few Tips: After you move your Vertical Line where you want it you MUST click Enter. The vertical line test says that any line parallel to the -axis, i.e., any line of the form , intersects the graph of the function at at most one point. 10) 30. 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• Level: GCSE
• Subject: Maths
• Word count: 2509
# Investigating the relationship between the total of a three-step stair on a number grid.
Extracts from this document...
Introduction
Tom Brown Number Stairs June 2003
10 D
Introduction:
For this piece of GCSE math’s coursework I have been asked to investigate the relationship between the total of a three-step stair on a number grid and it’s position on the grid. The stair will be on a 10x10 number grid like this:
91 92 93 94 95 96 97 98 99 100 81 82 83 84 85 86 87 88 89 90 71 72 73 74 75 76 77 78 79 80 61 62 63 64 65 66 67 68 69 70 51 52 53 54 55 56 57 58 59 60 41 42 43 44 45 46 47 48 49 50 31 32 33 34 35 36 37 38 39 40 21 22 23 24 25 26 27 28 29 30 11 12 13 14 15 16 17 18 19 20 1 2 3 4 5 6 7 8 9 10
I would need to find the total for stairs on the grid. For example the one shaded in red:
45+35+25+36+26+27=194
I would then tabulate this result and move the stair one square to the right so then I would get.
46+36+26+37+27+28=200
I would then continue moving it to the right until I am unable to anymore.
Part 1:
I will start of in the bottom left hand corner and work myself to the right of the grid.
Position 1
21 11 12 1 2 3
The total for this stair is:
21+11+1+12+2+3=50
Position 2
22 12 13 2 3 4
The total for this stair is:
22+12+2+13+3+4=56
Position 3
23 13 14 3 4 5
The total for this stair is:
23+13+3+14+4+5=62
I will now tabulate these results and others.
Position Number (Lowest number in grid) 1 2 3 4 5 6 7 8 Total 50 56 62 68 74 80 86 92
From looking at my results I can see that when I move the stair shape left by 1 column the total increases by 6 each time. I am going to try to find a formula to explain the relationship between the position and the total.
Middle
82
83
84
85
86
87
88
89
90
71
72
73
74
75
76
77
78
79
80
61
62
63
64
65
66
67
68
69
70
51
52
53
54
55
56
57
58
59
60
41
42
43
44
45
46
47
48
49
50
31
32
33
34
35
36
37
38
39
40
21
22
23
24
25
26
27
28
29
30
11
12
13
14
15
16
17
18
19
20
1
2
3
4
5
6
7
8
9
10
Position 1
The total for this stair is:
1+2+3+4+11+12+13+21+22+31=120
Position 2
The total for this stair is:
2+3+4+5+12+13+14+22+23+32=130
Position 3
The total for this stair is:
3+4+5+6+13+14+15+23+24+33=140
Position 4
The total for this stair is:
4+5+6+7+14+15+16+24+25+34=150
I will now tabulate these results and those for other 4 step stairs.
Position Number
1
2
3
4
5
6
7
## Stair
Total
120
130
140
150
160
170
180
I will see whether 6n+44 will work to find the stair total for 4 steps. I will use position 1 as an example.
n=1
=6n+44
=(6x1)+44
=6+44
=50
As you can see the formula does not apply to four step stairs. Now I will try to find a formula that does apply to four steps.
Since this time there is a difference of 10 between the totals I now times n by 10 in my equation. I will see what 10n looks like compared to the totals.
10n
10
20
30
40
50
60
70
Total
120
130
140
150
160
170
180
Difference
110
110
110
110
110
110
110
## The same way as with the three steps 10n does not equal the total it is larger. The difference is 110 so the formula for four step stairs is 10n+110. I will now test my new formula.
61+62+63+64+71+72+73+81+82+91=720
n=61
=10n+110
=(10x61)+110
=720
You can now see that my formula for the four step stairs is correct. I will now try five step stairs.
Position 1
The total for this stair is:
1+2+3+4+5+11+12+13+14+21+22+23+31+32+41=235
Position 2
The total for this stair is:
2+3+4+5+6+12+13+14+15+22+23+24+32+33+42=250
Position 3
The total for this stair is:
3+4+5+6+7+13+14+15+16+23+24+25+33+34+43=265
Position 4
The total for this stair:
4+5+6+7+8+14+15+16+17+24+25+26+34+35+44=280
Position 5
The total for this stair:
5+6+7+8+9+15+16+17+18+25+26+27+35+36+45=295
Position 6
The total for this stair:
6+7+8+9+10+16+17+18+19+26+27+28+36+37+46=310
Position Number
1
2
3
4
5
6
## Stair
Total
235
250
265
280
295
310
Conclusion
1
2
3
4
5
6
7
+11h3
11
88
297
704
1375
2376
3773
-11h
-11
-22
-33
-44
-55
-66
-77
Total
0
66
264
660
1320
2310
3696
Divide by 6
0
11
44
110
220
Total
1n+0
2n+11
6n+44
10n+110
15+220
So every time when 11h³-11h is divided by 6 it gives the right number in the second part of each formula. So I now have a complete general formula to find out any sized stair anywhere on a 10x10 number grid. The formula is:
n+(h²+h) + (11h³-11h)
2 6
However I will need to test this formula:
T: Total of the stair
n: Stair number (the number in the bottom left had corner of the stair)
h: The height of the stair
n=72 Three Step Stair:
T=72+73+74+82+83+92=476
## T=72((12)/2)+(264)/6
T=72(6)+44
T=432+44
T=476
For this one the answer is correct now I will try another sized stair
n=55 Five Step Stair:
## T=55+56+57+58+59+65+66+67+68+75+76+77+85+86+95
=1045
T=n((h²+h)/2)+(11h³-11h)/6
## T=55((30)/2)+(1320)/6
T=55(15)+220
T=825+220
T=1045
Now since I have tested it twice I am sure that my formula is correct and will work for any size stair anywhere on in the 10x10 grid. Now looking back I see what else I could have done to improve my investigation. Instead of changing stair height I could have instead changed the width of the grid or the height. I could also have changed the width and length simultaneously i.e. 5x5, 6x6, 7x7.
.
This student written piece of work is one of many that can be found in our GCSE Number Stairs, Grids and Sequences section.
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# 8.7: Looping and counting
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The following program counts the number of times the letter a appears in a string:
word = 'banana'
count = 0
for letter in word:
if letter == 'a':
count = count + 1
print(count)
This program demonstrates another pattern of computation called a counter. The variable count is initialized to 0 and then incremented each time an a is found. When the loop exits, count contains the result—the total number of a’s.
As an exercise, encapsulate this code in a function named count, and generalize it so that it accepts the string and the letter as arguments.
Then rewrite the function so that instead of traversing the string, it uses the three-parameter version of find from the previous section.
This page titled 8.7: Looping and counting is shared under a CC BY-NC 3.0 license and was authored, remixed, and/or curated by Allen B. Downey (Green Tea Press) via source content that was edited to the style and standards of the LibreTexts platform.
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# Question e7339
Apr 17, 2017
#### Answer:
It doesn't.
$\cos \left({\sin}^{-} 1 \left(\frac{2}{3}\right)\right) = \frac{\sqrt{5}}{3}$
#### Explanation:
Let's break this down:
color(red)(cos(color(blue)(sin^-1(2/3)))
The $\textcolor{red}{\cos}$ function is taking the cosine of the angle $\textcolor{b l u e}{{\sin}^{-} 1 \left(\frac{2}{3}\right)}$.
Let's say that $\textcolor{b l u e}{\theta = {\sin}^{-} 1 \left(\frac{2}{3}\right)}$. Remember that this is an angle.
In the triangle where this is true, we see that $\sin \left(\theta\right) = \frac{2}{3}$.
We can draw a triangle where $\sin \left(\theta\right) = \frac{2}{3}$: this means that the "opposite" side is $2$ and the hypotenuse is $3$.
Through the Pythagorean Theorem we see that the other leg is $\sqrt{5}$, since we have a right triangle.
What we want to find is color(red)(cos(color(blue)(sin^-1(2/3))), which is really color(red)(cos(color(blue)(theta))#.
That is, we just want cosine of theta in this triangle.
Cosine is "adjacent" over hypotenuse, so:
$\textcolor{red}{\cos \left(\textcolor{b l u e}{\theta}\right)} = \textcolor{red}{\cos \left(\textcolor{b l u e}{{\sin}^{-} 1 \left(\frac{2}{3}\right)}\right)} = \frac{\sqrt{5}}{3}$
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# singularity
Printable View
• May 12th 2011, 05:08 AM
hurz
singularity
$\displaystyle f(z)=\frac{1}{sinh(z)}$
It's ok that z=0 is a singularity, but why
$\displaystyle 2ik\pi$, where k is a integer, are singularities too? sinh(z)=0 only when z=0
Regards
• May 12th 2011, 05:24 AM
FernandoRevilla
$\displaystyle \sinh (2k\pi i)=\dfrac{e^{2k\pi i}-e^{-2k\pi i}}{2}=\dfrac{1-1}{2}=0$
• May 12th 2011, 05:29 AM
hurz
Quote:
Originally Posted by FernandoRevilla
$\displaystyle \sinh (2k\pi i)=\dfrac{e^{2k\pi i}-e^{-2k\pi i}}{2}=\dfrac{1-1}{2}=0$
$\displaystyle sinh(ki\pi)=0$
so, why the argument of the sinh must be an even multiple of $\displaystyle i\pi$ ?
• May 12th 2011, 07:51 AM
FernandoRevilla
Quote:
Originally Posted by hurz
$\displaystyle sinh(ki\pi)=0$ so, why the argument of the sinh must be an even multiple of $\displaystyle i\pi$ ?
You asked if $\displaystyle 2k\pi i$ were singularities of $\displaystyle f(z)$ and the answer is yes. If you want to find all of them:
$\displaystyle \sinh z=\dfrac{e^z-e^{-z}}{2}=0\Leftrightarrow\ldots\Leftrightarrow e^{2z}=1\Leftrightarrow\ldots\Leftrightarrow z=k\pi i\quad (k\in\mathbb{Z})$
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# Question: What Are Synonyms For Estimate?
## Is Estiment a word?
Correct spelling for ESTIMENT We think the word estiment is a misspelling.
It could be just an incorrect spelling of the words which are suggested below.
Review the list and pick the word which you think is the most suitable..
## What is a rough estimate?
A Rough Cost Estimate is an initial estimate that uses prior experience and other non-project data to estimate the cost of a project. It is also called a Rough Order of Magnitude (ROM) estimate, or a Conceptual Estimate.
## How do you estimate sums and differences?
The first step in estimating a sum or a difference is to round the numbers, by changing them to the nearest power of ten, hundred, thousand, etc. Round the numbers first, then use mental math to estimate an answer. When rounding, follow these rounding rules: If the number being rounded is less than 5, round down.
## What’s another word for estimate?
SYNONYMS FOR estimate 1 compute, count, reckon, gauge, assess, value, evaluate, appraise. 4 valuation, calculation, appraisal.
## What is a calculated estimate called?
As a verb, approximate means “to estimate.” Unlike the word guess, approximate implies the use of a logical or mathematical method. …
## What are two ways to estimate?
There are different methods for estimation that are useful for different types of problems. The three most useful methods are the rounding, front-end and clustering methods.
## What is an example of an estimate?
To find a value that is close enough to the right answer, usually with some thought or calculation involved. Example: Alex estimated there were 10,000 sunflowers in the field by counting one row then multiplying by the number of rows.
## Why is it good to estimate?
In real life, estimation is part of our everyday experience. … For students, estimating is an important skill. First and foremost, we want students to be able to determine the reasonableness of their answer. Without estimation skills, students aren’t able to determine if their answer is within a reasonable range.
## What are antonyms for estimate?
What is the opposite of estimate?suspectquestiongatherimagineassumefeelguesspresumereckonspeculate3 more rows
## How do you create an estimate?
What do I include in an estimate?Job description. Explain the work you’ll be doing. … Materials and labor. Provide a high-level view of the necessary materials and labor and the costs for each. … Total cost. Clearly and correctly tally up the total costs of the project.This is a big one. … Sales and company contact info.
## What’s another word for windy?
In this page you can discover 46 synonyms, antonyms, idiomatic expressions, and related words for windy, like: blustery, breezy, raw, airy, gusty, fresh, wind-shaken, stormy, wind-swept, blowing and drafty.
## What does the word of mean?
(Entry 1 of 3) 1 —used as a function word to indicate a point of reckoningnorth of the lake. 2a —used as a function word to indicate origin or derivationa man of noble birth. b —used as a function word to indicate the cause, motive, or reasondied of flu.
## Do synonyms English?
In this page you can discover 80 synonyms, antonyms, idiomatic expressions, and related words for do, like: deal with, accomplish, decipher, give, achieve, finish, doh, make out, effect, carry out and fulfill.
## What is the adjective of estimate?
Of or pertaining to an estimate or to estimation. Synonyms: foreseeable, anticipatable, calculable, conceivable, imaginable, likely, predictable, probable.
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# 4.3 Fitting linear models to data (Page 8/14)
Page 8 / 14
Find the linear function y , where y depends on $\text{\hspace{0.17em}}x,$ the number of years since 1980.
$y=-\text{3}00x+\text{11},\text{5}00$
Find and interpret the y -intercept.
For the following exercise, consider this scenario: In 2004, a school population was 1,700. By 2012 the population had grown to 2,500.
Assume the population is changing linearly.
1. How much did the population grow between the year 2004 and 2012?
2. What is the average population growth per year?
3. Find an equation for the population, P , of the school t years after 2004.
a) 800 b) 100 students per year c) $\text{\hspace{0.17em}}P\left(t\right)=\text{1}00t+\text{17}00$
For the following exercises, consider this scenario: In 2000, the moose population in a park was measured to be 6,500. By 2010, the population was measured to be 12,500. Assume the population continues to change linearly.
Find a formula for the moose population, $\text{\hspace{0.17em}}P.$
What does your model predict the moose population to be in 2020?
18,500
For the following exercises, consider this scenario: The median home values in subdivisions Pima Central and East Valley (adjusted for inflation) are shown in [link] . Assume that the house values are changing linearly.
Year Pima Central East Valley
1970 32,000 120,250
2010 85,000 150,000
In which subdivision have home values increased at a higher rate?
If these trends were to continue, what would be the median home value in Pima Central in 2015?
\$91,625
## Fitting Linear Models to Data
Draw a scatter plot for the data in [link] . Then determine whether the data appears to be linearly related.
0 2 4 6 8 10 –105 –50 1 55 105 160
Draw a scatter plot for the data in [link] . If we wanted to know when the population would reach 15,000, would the answer involve interpolation or extrapolation?
Year Population
1990 5,600
1995 5,950
2000 6,300
2005 6,600
2010 6,900
Extrapolation
Eight students were asked to estimate their score on a 10-point quiz. Their estimated and actual scores are given in [link] . Plot the points, then sketch a line that fits the data.
Predicted Actual
6 6
7 7
7 8
8 8
7 9
9 10
10 10
10 9
Draw a best-fit line for the plotted data.
For the following exercises, consider the data in [link] , which shows the percent of unemployed in a city of people 25 years or older who are college graduates is given below, by year.
Year 2000 2002 2005 2007 2010 Percent Graduates 6.5 7 7.4 8.2 9
Determine whether the trend appears to be linear. If so, and assuming the trend continues, find a linear regression model to predict the percent of unemployed in a given year to three decimal places.
In what year will the percentage exceed 12%?
2023
Based on the set of data given in [link] , calculate the regression line using a calculator or other technology tool, and determine the correlation coefficient to three decimal places.
$x$ 17 20 23 26 29 $y$ 15 25 31 37 40
Based on the set of data given in [link] , calculate the regression line using a calculator or other technology tool, and determine the correlation coefficient to three decimal places.
$x$ 10 12 15 18 20 $y$ 36 34 30 28 22
For the following exercises, consider this scenario: The population of a city increased steadily over a ten-year span. The following ordered pairs show the population and the year over the ten-year span (population, year) for specific recorded years:
Cos45/sec30+cosec30=
Cos 45 = 1/ √ 2 sec 30 = 2/√3 cosec 30 = 2. =1/√2 / 2/√3+2 =1/√2/2+2√3/√3 =1/√2*√3/2+2√3 =√3/√2(2+2√3) =√3/2√2+2√6 --------- (1) =√3 (2√6-2√2)/((2√6)+2√2))(2√6-2√2) =2√3(√6-√2)/(2√6)²-(2√2)² =2√3(√6-√2)/24-8 =2√3(√6-√2)/16 =√18-√16/8 =3√2-√6/8 ----------(2)
exercise 1.2 solution b....isnt it lacking
I dnt get dis work well
what is one-to-one function
what is the procedure in solving quadratic equetion at least 6?
Almighty formula or by factorization...or by graphical analysis
Damian
I need to learn this trigonometry from A level.. can anyone help here?
yes am hia
Miiro
tanh2x =2tanhx/1+tanh^2x
cos(a+b)+cos(a-b)/sin(a+b)-sin(a-b)=cotb ... pls some one should help me with this..thanks in anticipation
f(x)=x/x+2 given g(x)=1+2x/1-x show that gf(x)=1+2x/3
proof
AUSTINE
sebd me some questions about anything ill solve for yall
cos(a+b)+cos(a-b)/sin(a+b)-sin(a-b)= cotb
favour
how to solve x²=2x+8 factorization?
x=2x+8 x-2x=2x+8-2x x-2x=8 -x=8 -x/-1=8/-1 x=-8 prove: if x=-8 -8=2(-8)+8 -8=-16+8 -8=-8 (PROVEN)
Manifoldee
x=2x+8
Manifoldee
×=2x-8 minus both sides by 2x
Manifoldee
so, x-2x=2x+8-2x
Manifoldee
then cancel out 2x and -2x, cuz 2x-2x is obviously zero
Manifoldee
so it would be like this: x-2x=8
Manifoldee
then we all know that beside the variable is a number (1): (1)x-2x=8
Manifoldee
so we will going to minus that 1-2=-1
Manifoldee
so it would be -x=8
Manifoldee
so next step is to cancel out negative number beside x so we get positive x
Manifoldee
so by doing it you need to divide both side by -1 so it would be like this: (-1x/-1)=(8/-1)
Manifoldee
so -1/-1=1
Manifoldee
so x=-8
Manifoldee
Manifoldee
so we should prove it
Manifoldee
x=2x+8 x-2x=8 -x=8 x=-8 by mantu from India
mantu
lol i just saw its x²
Manifoldee
x²=2x-8 x²-2x=8 -x²=8 x²=-8 square root(x²)=square root(-8) x=sq. root(-8)
Manifoldee
I mean x²=2x+8 by factorization method
Kristof
I think x=-2 or x=4
Kristof
x= 2x+8 ×=8-2x - 2x + x = 8 - x = 8 both sides divided - 1 -×/-1 = 8/-1 × = - 8 //// from somalia
Mohamed
i am in
Cliff
hii
Amit
how are you
Dorbor
well
Biswajit
can u tell me concepts
Gaurav
Find the possible value of 8.5 using moivre's theorem
which of these functions is not uniformly cintinuous on (0, 1)? sinx
helo
Akash
hlo
Akash
Hello
Hudheifa
which of these functions is not uniformly continuous on 0,1
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# 9649 (number)
9649 is an odd four-digits prime number following 9648 and preceding 9650. In scientific notation, it is written as 9.649 × 103. The sum of its digits is 28. It has a total of one prime factor and 2 positive divisors. There are 9,648 positive integers (up to 9649) that are relatively prime to 9649.
## Basic properties
• Is Prime? yes
• Number parity odd
• Number length 4
• Sum of Digits 28
• Digital Root 1
## Name
Name nine thousand six hundred forty-nine
## Notation
Scientific notation 9.649 × 103 9.649 × 103
## Prime Factorization of 9649
Prime Factorization 9649
Prime number
Distinct Factors Total Factors Radical ω 1 Total number of distinct prime factors Ω 1 Total number of prime factors rad 9649 Product of the distinct prime numbers λ -1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ -1 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 9.17461 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 9649 is 9649. Since it has only one prime factor, 9649 is a prime number.
## Divisors of 9649
2 divisors
Even divisors 0 2 2 0
Total Divisors Sum of Divisors Aliquot Sum τ 2 Total number of the positive divisors of n σ 9650 Sum of all the positive divisors of n s 1 Sum of the proper positive divisors of n A 4825 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G 98.2293 Returns the nth root of the product of n divisors H 1.99979 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 9649 can be divided by 2 positive divisors (out of which none is even, and 2 are odd). The sum of these divisors (counting 9649) is 9650, the average is 4825.
## Other Arithmetic Functions (n = 9649)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ 9648 Total number of positive integers not greater than n that are coprime to n λ 9648 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π ≈ 1192 Total number of primes less than or equal to n r2 8 The number of ways n can be represented as the sum of 2 squares
There are 9,648 positive integers (less than 9649) that are coprime with 9649. And there are approximately 1,192 prime numbers less than or equal to 9649.
## Divisibility of 9649
m n mod m
2 1
3 1
4 1
5 4
6 1
7 3
8 1
9 1
9649 is not divisible by any number less than or equal to 9.
• Arithmetic
• Prime
• Deficient
• Polite
• Prime Power
• Square Free
## Base conversion 9649
Base System Value
2 Binary 10010110110001
3 Ternary 111020101
4 Quaternary 2112301
5 Quinary 302044
6 Senary 112401
8 Octal 22661
10 Decimal 9649
12 Duodecimal 5701
20 Vigesimal 1429
36 Base36 7g1
## Basic calculations (n = 9649)
### Multiplication
n×y
n×2 19298 28947 38596 48245
### Division
n÷y
n÷2 4824.5 3216.33 2412.25 1929.8
### Exponentiation
ny
n2 93103201 898352786449 8668206036446401 83639520045671323249
### Nth Root
y√n
2√n 98.2293 21.2893 9.91107 6.26464
## 9649 as geometric shapes
### Circle
Diameter 19298 60626.5 2.92492e+08
### Sphere
Volume 3.76301e+12 1.16997e+09 60626.5
### Square
Length = n
Perimeter 38596 9.31032e+07 13645.7
### Cube
Length = n
Surface area 5.58619e+08 8.98353e+11 16712.6
### Equilateral Triangle
Length = n
Perimeter 28947 4.03149e+07 8356.28
### Triangular Pyramid
Length = n
Surface area 1.61259e+08 1.05872e+11 7878.38
## Cryptographic Hash Functions
md5 7c2c48a32443ad8f805e48520f3b26a4 21e5bc36a0b1ed0fed2b3aa52fe3944cedfdb1c2 da780feb252d3b31a20c19b3928eb8fd75226d385af1e12676a412a9b6340179 92a7f213a6e9ed35b5ed381399626f328bfc21cb30c074413fdff936d4d52e72901f32b8164ea75d12e3240b9d817c8e1aca8558d51cdbb0069f8df990a62854 ea5cf9c252a6dc89cf6e5bfc521630024d63e364
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A278204 Number of nX4 0..1 arrays with every element both equal and not equal to some elements at offset (-1,-1) (-1,0) (-1,1) (0,-1) (0,1) (1,-1) (1,0) or (1,1), with upper left element zero. 1
0, 46, 666, 8242, 117088, 1674402, 23732454, 336380248, 4770344900, 67648850802, 959306211222, 13603641359640, 192909619956550, 2735599405114814, 38792794983910750, 550110139019848618, 7800963216380203384 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,2 COMMENTS Column 4 of A278208. LINKS R. H. Hardin, Table of n, a(n) for n = 1..210 FORMULA Empirical: a(n) = 10*a(n-1) +46*a(n-2) +157*a(n-3) +471*a(n-4) -202*a(n-5) -2198*a(n-6) -1400*a(n-7) -941*a(n-8) -1959*a(n-9) +2763*a(n-10) +374*a(n-11) -387*a(n-12) +3427*a(n-13) -1743*a(n-14) -335*a(n-15) +862*a(n-16) -1124*a(n-17) +584*a(n-18) -96*a(n-19) EXAMPLE Some solutions for n=4 ..0..0..1..1. .0..0..0..0. .0..0..0..0. .0..1..1..1. .0..1..0..0 ..0..1..0..0. .1..0..0..1. .1..0..1..1. .0..1..1..0. .0..0..1..1 ..1..1..1..0. .1..0..1..0. .1..1..1..0. .0..1..0..1. .1..0..1..0 ..0..0..0..0. .0..1..1..1. .1..0..0..0. .0..1..0..1. .1..0..0..0 CROSSREFS Cf. A278208. Sequence in context: A077732 A200898 A160151 * A223929 A224053 A224306 Adjacent sequences: A278201 A278202 A278203 * A278205 A278206 A278207 KEYWORD nonn AUTHOR R. H. Hardin, Nov 15 2016 STATUS approved
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Last modified July 21 17:20 EDT 2019. Contains 325198 sequences. (Running on oeis4.)
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# Search 3rd Grade Mathematic Mixed Operation Educational Resources
38 filtered results
38 filtered results
Mixed Operations
Mathematics
Sort by
More Mixed Minute Math
Worksheet
More Mixed Minute Math
Children work on their math automaticity and accuracy skills in this set of 48 mixed math problems.
Math
Worksheet
Multiplication & Division Word Problems Practice
Worksheet
Multiplication & Division Word Problems Practice
This resource gives your students practice with multiplication and division word problems. This worksheet can be used with the Stepping Through Multiplication & Division Word Problems lesson.
Math
Worksheet
Minute Math: Multiplication and Division
Worksheet
Minute Math: Multiplication and Division
Test your third graders math skills with this one minute exercise to see how fast they can complete multiplication and division problems.
Math
Worksheet
Workbook
Math
Workbook
Multiplication & Division: Picnicking Signs #1
Worksheet
Multiplication & Division: Picnicking Signs #1
Kids complete each equation on this third grade math worksheet by determining whether an equation is multiplication or division and writing in the correct sign.
Math
Worksheet
Practicing Commutative Property with Arrays
Worksheet
Practicing Commutative Property with Arrays
Children explore the concept of commutative properties with the visual aid of arrays.
Math
Worksheet
Properties of Multiplication: Distributive
Worksheet
Properties of Multiplication: Distributive
Give your third grader some extra practice with one of the key properties of multiplication: distributive property.
Math
Worksheet
Fill in the Missing Multiplication Factors
Worksheet
Fill in the Missing Multiplication Factors
Gear up for division with this missing factors multiplication worksheet.
Math
Worksheet
Division Facts: Nines
Worksheet
Division Facts: Nines
Learners work on their division facts for 9 in this practice worksheet.
Math
Worksheet
Properties of Multiplication: Commutative
Worksheet
Properties of Multiplication: Commutative
This worksheet introduces children to the commutative property, then uses a series of prompts to help them practice it.
Math
Worksheet
Division Facts: Fives
Worksheet
Division Facts: Fives
Learners work on their division facts for 5 in this practice worksheet.
Math
Worksheet
Word Problems: Division
Worksheet
Word Problems: Division
Use this worksheet to practice long division with one-digit divisors.
Math
Worksheet
Times Tables and Division
Worksheet
Times Tables and Division
It's time to start memorizing multiplication tables! Help your student master her times tables with this drill sheet, which also includes some division.
Math
Worksheet
Inverse Operations: Multiplication
Worksheet
Inverse Operations: Multiplication
Math
Worksheet
Input and Output Tables
Worksheet
Input and Output Tables
Find the mathematical rule in these math tables. Compare the input and output numbers to find the hidden pattern in each table.
Math
Worksheet
Inverse Operations: Division
Worksheet
Inverse Operations: Division
Learn about the relationship between division and multiplication with this inverse operations worksheet.
Math
Worksheet
Multiplication & Division: Beach Math
Worksheet
Multiplication & Division: Beach Math
Kids multiply or divide to complete each equation on this third grade math worksheet.
Math
Worksheet
Parking Lot Multiplication Word Problems #1
Worksheet
Parking Lot Multiplication Word Problems #1
Give your students a real-world visual of cars and parking lots as they get comfortable with arrays and the commutative property of multiplication.
Math
Worksheet
Distributive Property of Multiplication
Worksheet
Distributive Property of Multiplication
In this worksheet, children learn what the distributive property is, and how to use it to solve multiplication problems.
Math
Worksheet
Algebraic Thinking: Making 10
Worksheet
Algebraic Thinking: Making 10
Children learn basic algebraic thinking using multiplication fact families in this math worksheet.
Math
Worksheet
Multiplication & Division Clue Words Sorting
Worksheet
Multiplication & Division Clue Words Sorting
This resources teaches your students to distinguish between clue words that indicate multiplication problems versus those that signal division problems.
Math
Worksheet
Comparing Arrays
Worksheet
Comparing Arrays
Use this worksheet to get your students looking at arrays and noticing the similarities and differences.
Math
Worksheet
Input Output Math
Worksheet
Input Output Math
Use a little bit of logic to find out the rule that links "input" and "output" in these math puzzles.
Math
Worksheet
Multiplication Facts
Worksheet
Multiplication Facts
Give your fourth grader practice with an important concept: fact families. This one explores the multiplication facts that relate to division facts.
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# 13 4 tangent ratio lesson
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### Transcript of "13 4 tangent ratio lesson "
1. 1. Sec. 8 – 3Sec. 8 – 3 The Tangent RatioThe Tangent Ratio Objective:Objective: 1) To use tangent ratios to1) To use tangent ratios to determine side lengths indetermine side lengths in ΔΔ..
2. 2. This only for rightThis only for right ΔΔs!!s!! TrigonometryTrigonometry Greek WordGreek Word Trigon → TriangleTrigon → Triangle Metron → MeasureMetron → Measure Trigonometry RatioTrigonometry Ratio – Ratio of the lengths of– Ratio of the lengths of sides of a rightsides of a right ΔΔ..
3. 3. The tangent is just a button on your calculator!The tangent is just a button on your calculator! TanTan ** Make sure you calculator is in Degrees!!
4. 4. Tangent RatioTangent Ratio Tangent RatioTangent Ratio – Ratio of the length of the– Ratio of the length of the opposite leg from anopposite leg from an ∠∠ to the length of the legto the length of the leg adjacent to the sameadjacent to the same ∠∠.. A C B b a c * Can’t use the right ∠, ∠C Tangent ∠A = Length of leg Opposite of ∠A Length of leg Adjacent of ∠A Tangent ∠A = a b
5. 5. Writing tangent ratiosWriting tangent ratios Write the tangent ratio ofWrite the tangent ratio of ∠∠T andT and ∠∠U.U. T U S 6 8 10 Tangent ∠θ = Opposite Adjacent Tangent ∠T = 8 6 TS US = Tangent ∠U =TS US 6 8= ** Tangent ratio for ∠T & ∠U are reciprocals
6. 6. You can use the tangent ratio to findYou can use the tangent ratio to find the measure of a distance that isthe measure of a distance that is difficult to measure directly.difficult to measure directly. Example 1: Find w.Example 1: Find w. 10 w 54 Step 1: Set up the Tangent Ratio Tan 54 = opp adj Tan 54 = w 10 1.376 = w 10 13.76 = w
7. 7. Ex. 2: Solve for the variable usingEx. 2: Solve for the variable using the tangent ratio.the tangent ratio. 70° 8cm x Step 1: Set up the tangent ratio. Tan 70 = opp adj Tan 70 = 8 x 2.747 = 8 x Multiply both sides by the denominator, x 2.747x = 8 x = 2.9
8. 8. The Tangent Inverse: TanThe Tangent Inverse: Tan-1-1 Just another button on your Calculator!Just another button on your Calculator! Use it when you have the two sides of aUse it when you have the two sides of a ΔΔ andand are trying to find a missingare trying to find a missing ∠∠.. TanTan Tan-1Use the SHIFT (2nd ) Key to get to it. Once you press it, it should look like this: Tan-1 (
9. 9. Ex.3: Using the TanEx.3: Using the Tan-1-1 Use the TanUse the Tan-1-1 to solve for the missingto solve for the missing ∠∠.. 12mm 5mm y° Step 1: Set up the Tan Ratio Tan y = opp adj Tan y = 5 12 Tan y = .4167 At this point you will use the Tan-1 : 1) Hit shift Tan to get to Tan-1 ( 2) Type in the decimal and hit enter Tan-1 (.4167) = 22.6°
10. 10. Ex.4: Solve for mEx.4: Solve for m∠∠ZZ 8miles8miles 6miles6miles x Y Z Tan Z = opp adj Tan Z = 8 6 Tan Z = 1.333 Tan-1 (1.333) = m∠Z m∠Z = 53.1°
11. 11. What have I learned???What have I learned??? TanTan θθ == Use TanUse Tan-1-1 when looking for anwhen looking for an ∠∠ measure.measure. Opposite Side Adjacent Side
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• Resource ID: TEKS12_MATH_K_001
• Subject: Math
Representing Whole Number Quantities
This activity provides an opportunity for students to represent whole numbers with pictures.
• Resource ID: TEKS12_MATH_02_001
• Subject: Math
Modeling Addition and Subtraction of Two-Digit Numbers
In this activity students will model addition and subtraction of two-digit numbers.
• Resource ID: TEKS12_MATH_06_002
• Subject: Math
Area of Triangles, Parallelograms, and Trapezoids
These activities provide an opportunity for students to explore the area formulas for triangles, trapezoids, and parallelograms.
• Resource ID: M8M5L5
• Subject: Math
Selecting and Using Representations for Collected Data
Given a variety of data (including line plots, line graphs, stem and leaf plots, circle graphs, bar graphs, box and whisker plots, histograms, and Venn diagrams), the student will select and use an appropriate representation for presenting and displaying relationships among the collected data with and without the use of technology
• Resource ID: M8M2L6*
• Subject: Math
Determining Slopes from Equations, Graphs, and Tables
Given algebraic, tabular, and graphical representations of linear functions, the student will determine the slope of the relationship from each of the representations.
• Resource ID: M6M1L1
• Subject: Math
Applying the Order of Operations
Given a numerical expression including whole number exponents and prime factorization, students will use the order of operations to generate an equivalent expression.
• Subject: Math
11 OnTRACK Grade 8 Math: Proportionality
Students learn to to use proportional relationships to describe dilation; explain proportional and non-proportional relationships involving slope; and use proportional and non-proportional relationships to develop foundational concepts of functions.
• Resource ID: E_MSTAR001
• Subject: Math
ESTAR/MSTAR
The ESTAR/MSTAR Universal Screeners and Diagnostic Assessments are a formative assessment system administered to students in grades 2–8 to support instructional decisions in mathematics.
• Resource ID: T2T07
• Subject: Math
Rational Number Coefficients
Rational Number Coefficients
• Resource ID: T2T08
• Subject: Math
Equations: Combining Like Terms
Equations: Combining Like Terms
• Resource ID: T2T06
• Subject: Math
Measurement: Converting Units of Measure
Measurement: Converting Units of Measure
• Resource ID: T2T05
• Subject: Math
Building to Measurement with a Ruler
Building to Measurement with a Ruler
• Resource ID: T2T09
• Subject: Math
Equations: Vertical Alignment Discussion
Discussing the best way to help students understand the process of solving an equation.
• Resource ID: T2T03
• Subject: Math
Learn about addition, subtraction, and using number lines to solve two-step problems.
• Resource ID: T2T25
• Subject: Math
Integer Operations: Addition and Subtraction Models
Integer Operations: Addition and Subtraction Models
• Resource ID: T2T26
• Subject: Math
Integer Operations: Multiplication and Division Models
Integer Operations: Multiplication and Division Models
• Resource ID: T2T24
• Subject: Math
• Subject: Math
2 OnTRACK Grade 6 Math Module 1
This OnTRACK Grade 6 math module feature resources that touch upon student expectations for mathematical process standards, number and operations, proportionality, and personal financial literacy.
• Resource ID: T2TVideos
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RD Sharma Class 7 Maths Solutions Chapter 12 - Profit and Loss
Last updated date: 15th Sep 2024
Total views: 696.6k
Views today: 14.96k
RD Sharma Solutions For Class 7 Maths Chapter 12
The RD Sharma Class 7 Chapter 12 questions and solutions are prepared according to the NCERT curriculum. All these solutions are in a step by step manner to help the students to understand the concepts in easier ways. Here let us look into some important topics related to the Profit and Loss chapter.
• Definition of profit and loss
• Meaning of profit and loss with examples
• Definition of cost price
• Meaning of cost price with examples
• Definition of the selling price
• Meaning of the selling price with examples
• Profit percentage
• Loss percentage
• Finding profit or loss when cost price and selling price are given
• Finding cost price or selling price when profit or loss is given
• Finding profit or loss percentage
• Finding the selling price when cost price and profit or loss percentage are given
• Finding the cost price when selling price and profit or loss are given
Definition of profit and loss
Profits are gains, particularly the difference between the amount earned and the amount spent.
A loss occurs when a product is sold at a lesser cost. Loss is the difference between the price at which the good was bought and the price at which it was sold.
When a good is sold at a greater price then a profit occurs If the good is sold at a lesser price then there is a loss.
Meaning of profit and loss with examples
The meaning of profit and loss can be further illustrated with the help of some examples.
Example: If a student buys a mobile phone for Rs. 20000 and sells it for Rs. 25000 to his friend then the student makes a profit of Rs. 5000. Now, if the same student sells the mobile phone for Rs. 18000, then the student has a loss of Rs. 2000.
If a shopkeeper purchases milk for Rs. 80 and he sells it for Rs.70, he has a loss of Rs.10.If he sells it for Rs.100, he will make a profit of Rs.20.
Definition of cost price
The price at which the item is purchased by us is known as cost price. It is given by CP.
Meaning of cost price with examples
For you to understand cost price better, here are some examples.
For example, you bought a laptop for Rs 30,000. So the cost price is Rs. 30,000. Sometimes cost price may also include extra expenses such as transportation, repair costs, etc. So, if one buys a fridge at Rs 12,000 and spends Rs 500 for transport and Rs 500 for setting it up. Here the cost price is the sum of all the expenditure, that is, Rs 13,000.
Profit percentage
Profit percentage is calculated only when there is a profit. When the cost price of an item is less than the selling price, profit is considered.
This is the sole condition for profit on an item:
Selling Price > Cost Price
Loss percentage
Loss percentage is calculated only when there is a loss. When the selling price of an item is less than the cost price, loss is considered.
This is the sole condition for loss on an item:
Cost Price > Selling Price
Finding profit or loss when cost price and selling price are given
It’s very easy to find profit or loss when cost price or selling price is given. We just have to remember these simple formulae:
Net profit=SP- CP
Net loss=CP- SP
Finding cost price or selling price when profit or loss is given
When loss is given,
CP= Loss+ SP
SP=CP-Loss
When profit is given,
CP= SP- Profit
SP=Profit+CP
Finding profit or loss percentage
We can find profit or loss percentages from the following formulae.
Formula for calculating profit percentage:
Net profit = SP - CP
Profit = $\left ( \frac{(SP-CP)}{CP} \right )\times 100$
Profit =$\left ( \frac{(Net Gain)}{CP} \right )\times 100$
Formula for calculating loss percentage:
Net loss=CP- SP
Loss =$\left ( \frac{(CP-SP)}{CP} \right )\times 100$
Loss = $\left ( \frac{(Net loss)}{CP} \right )\times 100$
Finding the selling price when cost price and profit or loss
percentage is given
In order to find the SP when CP and profit or loss percentage is given, we need to follow the following steps.
When the loss percentage is given:
SP = CP-$\frac{(Loss \times CP)}{100}$
When the profit percentage is given:
SP = CP + $\left [ \frac{(Profit \times CP)}{100} \right ]$
Tips to Prepare for Exams Using RD Sharma Solutions For Class 7 Maths Chapter 12
• A basic understanding of converting decimals to percentages and vice versa should be known thoroughly before attempting any questions of profit and loss.
• Since most of the problems are directly related to the formulas of profit and loss, students should remember all the important formulas and conversion units to score good marks from the profit and loss chapter.
• Students are advised to practice as many problems on different concepts of profit and loss to score good marks in their exams.
Conclusion
The concepts covered in the RD Sharma Class 7 Chapter 12 are according to the CBSE syllabus and NCERT curriculum. All the important questions and solutions are created by the faculty from Vedantu who have a lot of teaching experience and good subject knowledge. These RD Sharma Solutions For Class 7 Maths Chapter 12 will definitely help students to secure good marks in their board exams.
FAQs on RD Sharma Class 7 Maths Solutions Chapter 12 - Profit and Loss
1. What is the formula of profit and loss?
It is very important for students to remember the formula of profit and loss for solving all the questions of profit and loss very easily. It is very important to familiarize ourselves with the concepts of profit and loss to get an idea about how the business world works. The term 'Profit and Loss' is very relevant in our lives.
The formulas are as follows:
The formula for profit is SP-CP.
The formula for loss is CP-SP.
Students are advised to remember these formulas for their exams.
2. What are some profit and loss examples?
There are many real-life examples to explain the concepts of profit and loss. Transactions happen everywhere. All businesses run on this concept thus it is of utmost importance for students to understand this concept very well, not only to score well in examinations but to live their life.
An example is: If you buy a pen for Rs.7 and sell it for Rs. 10, you make a profit of Rs. 3. Whereas loss is considered if you sell the pen for a price lesser than Rs. 5 say, Rs.3. Here you make a loss of Rs.2 then.
3. What is CP and SP in profit and loss?
CP and SP are the terms that are commonly used in profit and loss questions.
CP or Cost Price is the price at which you buy the item.
SP or Selling Price is the price at which you sell the item.
Finding the cost price when selling price and profit or loss are given
In order to find the CP when SP and profit or loss percentage is given, we need to follow the following steps.
When the loss percentage is given:
CP=$\frac{SP\times 100}{100-Loss percent}$
When the profit percentage is given:
CP=$\frac{SP\times 100}{Profit percent+100}$
Thus, they stand for the prices that are used in a transaction, business or exchange which involves profit or loss or none at all.
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Stats Final Exam
# Stats Final Exam - Statistics the study of how to collect...
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Statistics - the study of how to collect, organize, analyze and interpret info, Population - the collection of individuals or items of interest, Census - measurements from an entire pop are used, Sample - the subset of the pop on which we take measurements, Data - the measurements we collect, Individuals - the objects described by a set of data, Variable - any characteristic of and individual that can take different values for different individuals, Descriptive stats - methods of summarizing a set of data (no errors because you are summarizing the data you have) ex: sports stats, avg. GPA for students, car sales for the yr., # of students who live in Lex., Inferential stats - methods of making inference about a pop based on the info in the sample; collect data from a subset or sample of pop and use to estimate pop (is error because you don’t have all data, you have a subset of data) ex: presidential approval rating, Observational study - observes individuals and measures variables of interest but doesn’t attempt to influence responses ex: outcome of election predicted by polls, Sample survey- when data is collected by asking questions and recording answers; most common type of observational study ex: opinion polls, consumer surveys, teacher evals, Experiment - deliberately imposes some treatment on individuals to observe responses, Simulation - numerical facsimile of real world phenom; not used unless necessary, expensive) ex: crash tests, Bias - prejudice in one direction, Convenience sampling - uses results or data that are conveniently & readily obtained, can be extremely biased and shouldn’t be generalized to the overall pop or for inferential data, Voluntary response sample - common type of convenience sample that is selected by subjects volunteering, typically over represents people with strong opinions ex: comment cards, Random sample - sample determined completely by chance, is a representative sample, Simple random sample - sample of n measurements from a population selected in such a manner that every sample of size n from the population has equal probability of being selected, most common and all course is based on Stratified sampling - population is divided into at least two distinct strata or groups, a simple random sample is drawn from each, Systematic sampling - select a random starting point, select every fourth person in the sample, Cluster sampling -divide pop onto sections, randomly select the sections or clusters, Variability - describes the spread of values; we want the possible values of a statistic to not include bias and have a small amount of v.; to eliminate bias use random sampling; to control v. use a larger sample, Statistic - a numerical characteristic of sample, Parameter - a numerical characteristic of a population, Margin of error - can be calculated from the sampling variability; We can say with 95% confidence that the amount by which a proportion obtained from a sample will differ from the population proportion will not exceed Suppose you take a sample of 1600 adults and 800 enjoy amusement park rides. the sample proportion that
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## This note was uploaded on 07/12/2011 for the course STATS 200 taught by Professor Bradley,w during the Fall '10 term at KCTCS.
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Stats Final Exam - Statistics the study of how to collect...
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bounded by the given curves about the specified line.
y=12 y=x+(27/x) about the line x= -2
First we need to find the range of x for which the function is bound by the line y=12. We need the solution to 12=x+27/x. The solution to x^2-12x+27=0 is x=3 and 9. These will provide the limits of integration later. Now the region is defined. Next we need to find the volume of the bound region when rotated about a line parallel to the y axis, x=-2. It helps to imagine what the rotated region looks like. I see it as a circular gutter with a saucer shaped cross-section filled to the brim with water. Effectively, the volume of the rotated region will be equivalent to the volume of water in the gutter.
Consider the geometry. The radius of rotation will be x+2, because we are rotating around the axis x=-2. Consider a hollow cylinder of thickness dx and radius x+2. The volume of the cylinder will be 2(pi)h(x+2)dx where h is the height of the cylinder for the value of x. This height is the difference between the curve and the line y=12 so its value is 12-(x+27/x). Therefore the volume of the cylindrical shell is 2(pi)(x+2)(12-x-27/x)dx. This is the basis of the integral. If we expand this expression we get 2(pi)(12x+24-x^2-2x-27-54/x)dx=2(pi)(10x-x^2-3+54/x)dx. When this is integrated with respect to x between the limits 3 and 9 we get: 2(pi)[5x^2-x^3/3-3x+54ln(x)] for 3<=x<=9. Evaluating this we get 2(pi)(5(81-9)-(243-9)-3(9-3)+54ln(9/3))=2(pi)(360-234-18+54ln(3))=2(pi)(108+54ln(3))
=108(pi)*(2+ln(3))=1051.33 approx.
by Top Rated User (764k points)
edited by
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# Wikipedia:Reference desk/Archives/Mathematics/2008 April 2
Mathematics desk
< April 1 << Mar | April | May >> April 3 >
Welcome to the Wikipedia Mathematics Reference Desk Archives
The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages.
# April 2
## Infinities
I know infinity is pretty big, and you cant get bigger. But what if you raised infinity to the power infinity? What happens then/ —Preceding unsigned comment added by 79.76.250.241 (talk) 00:52, 2 April 2008 (UTC)
Well... there are different infinities. It's also rather deceptive. For instance, say we start with the counting numbers (1,2,3,etc). Now, it's quite clear that these are a strict subset of the integers (that is all whole numbers, positive or negative). But we can label the integers by the counting numbers. Let teh 1st integer be 0, the 2nd be 1, the 3rd be -1, the 4th be 2, the 5th be -2, etc. Then it is clear that we can just keep on counting to infinity, and that thus they have the same cardinality - essentially, there are the same number. We can do the same with the rationals. However, we cannot do this for the real numbers - see Cantor's diagonalization argument. This infinity is much bigger. Welcome to maths. -mattbuck (Talk) 01:14, 2 April 2008 (UTC)
Your question has already been treated by Georg Cantor in the 19th century. Generally, the concept of "infinity" has two different meanings: infinite ordinal number, and infinite cardinal number. If you mean "infinite ordinal number", then what you call "infinity to the power infinity" is traditionally symbolized: ${\displaystyle \omega ^{\omega }}$. See here for a deeper discussion on ${\displaystyle \omega ^{\omega }}$, as well as on its full meaning. A parallel concept of the desired power exists also for the infinite cardinal number. Eliko (talk) 01:28, 2 April 2008 (UTC)
I think it's a bit misleading to say that infinity has two meanings, since that suggests it has only two meanings. For instance, the phrase "infinity to the infinity" makes sense as a limit, in which context it's unambiguously infinity. On the other hand, in the nonstandard reals infinity to the infinity is different from both infinity and any interpretation in terms of cardinality or ordinality of sets. Black Carrot (talk) 08:45, 2 April 2008 (UTC)
When I wrote "two meanings" I meant "at least two meanings". Anyway, thank you for your comment. Eliko (talk) 13:03, 2 April 2008 (UTC)
Sorry, Kakarot, but that wouldn't be "infinity to the infinity, but rather "infinity to the infinity'th" or "[...] infinity'th power." Or you could be right as you stand. I don't really know. Anyway, here's a picture of what's being dealed with here:
${\displaystyle {\infty }^{\infty }}$ flaminglawyerc 21:32, 2 April 2008 (UTC)
"infinity to the infinity" and "infinity to the infinity'th (power)" sound like different ways of saying exactly the same thing to me. What distinction are you trying to draw? --Tango (talk) 00:42, 3 April 2008 (UTC)
It is linguistic nitpicking: just like we don't vocalize the written expression 33 as "three to the three", the fiery jurist objects to voicing ∞ as "infinity to the infinity". We do say "three to the power three". The intention of Black Carrot was clear enough. To the nitpicker: the past participle of the verb to deal is dealt. More importantly, recasting the question as ∞ completely ignores the earlier contributions pointing out that there are several notions of "infinity" in mathematics, for most of which the symbol ∞ is not appropriate.
In how many ways can we make a sequence of length 6 if each element of the sequence is picked from a set with 10 members (for example the set of the numbers 0 through 9)? The answer is 106. In general, there are NL different sequences of length L if each element belongs to a given set of size N. So an interpretation of "infinity to the power infinity" is: the number of different sequences of infinite length, each of whose elements belongs to a given set of infinite size.
In this interpretation we are talking about cardinalities, which measure set sizes. The simplest kind of infinite sequence is one whose elements can be indexed by natural numbers, and the simplest infinite set is again that of the natural numbers. So we are talking about the size of the set of infinite integer sequences with only nonnegative elements. The "size" (cardinality) of the natural numbers is denoted by ℵ0 (pronounced like "Aleph 0"), the smallest infinite cardinal. So then the question can be formulated as: how big is ℵ00?
For finite sequences, a somewhat surprising result (surprising until you get used to it) is that the length does not make a difference:
02 = ℵ03 = ℵ04 = ... = ℵ0.
Once you are used to this, the next surprising thing is that in the limit it does make a difference:
00 = ℵ1 > ℵ0.
1 ℵ00 is also the size of the set of real numbers, and is therefore known as the cardinality of the continuum. It is also denoted by ${\displaystyle {\mathfrak {c}}}$ or ℶ1. --Lambiam 08:32, 3 April 2008 (UTC)
Did it become standard to assume the continuum hypothesis when I wasn't paying attention? Matthew Auger (talk) 21:09, 3 April 2008 (UTC)
While we don't say "three to the three", we do say "e to the x", not "e to the x-th". Both ways are clearly used in some contexts. I think in the context of infinity, "infinity to the infinity" is the clearest way to phrase it. Exactly what it means depends very much on what you mean by "infinity". (And Matthew is right - you probably shouldn't assume the continuum hypothesis without saying you're doing so.) --Tango (talk) 21:17, 3 April 2008 (UTC)
Goody! I like linguistic quibbles. My rationalization would be that adding a -th suffix assumes we're dealing with the ordinal properties of infinity, which I was explicitly trying to move away from. I was treating infinity as an ideal element and exponentiation as a notational convenience, referring to an abstract binary operation. So, it was quite acceptable to phrase it that way. Black Carrot (talk) 09:39, 4 April 2008 (UTC)
I do say "three to the three".--196.209.177.241 (talk) 11:53, 5 April 2008 (UTC)
## Prime ideals in a ring
I'm struggling to prove something about prime ideals for an assignment: Given R, a nontrivial commutative ring with unity, use Zorn's lemma to show that the set of prime ideals has minimal elements with respect to inclusion, and that any prime ideal contains at least one such minimal prime ideal.
I've managed to show the existence of minimal prime ideals, but I'm struggling to figure out how I can show that every prime ideal contains a minimal prime ideal. I tried doing it by contradiction (assume that every prime ideal contains a smaller, nontrivial prime ideal) but I can't see any problems with that assumption that would give me a contradiction. The "turtles all the way down" notion seems a little counterintuitive, but not actually contradictory. Can anybody give me some pointers? Thanks, Maelin (Talk | Contribs) 01:34, 2 April 2008 (UTC)
Well, if R is also an integral domain, (0) is a minimal prime ideal. So that case is simple, and then assume you have zero divisors which gives you something else to work with, that may or may not be useful with what else you have. GromXXVII (talk) 11:03, 2 April 2008 (UTC)
What about applying Zorn’s lemma kind of backwards. That is, viewing the partial ordering as containment instead of inclusion. Then if there is an “upper bound” which in this case is a prime ideal contained in all prime ideals, Zorn’s lemma gives you precisely that every prime ideal contains a “maximal” prime ideal, which in this case is a minimal prime ideal. I’m not sure how to show that there is such an “upper bound” though, or if it even exists, because in general the intersection of prime ideals is not a prime ideal. GromXXVII (talk) 11:33, 2 April 2008 (UTC)
The intersection of a decreasing chain of prime ideals is prime, though, so it's indeed an application of Zorn's lemma. Bikasuishin (talk) 12:22, 2 April 2008 (UTC)
As a matter of interest, how did you manage to prove the existence of minimal prime ideals without proving the second part? To me they seem to both require essentially the same application of Zorn. Algebraist 14:54, 2 April 2008 (UTC)
## Tennis superiority
This should be slightly more complicated than it seems.
A tennis tournament has every participant playing every other participant. No ties are allowed. A player A is superior if for every other player B, either A beats B or there is a third player C such that A beats C and C beats B. Prove that if there is only one superior player then he or she beats every other player. Reywas92Talk 01:37, 2 April 2008 (UTC)
Here are a few steps towards the goal: Assume that A is the only superior player, but A' beats A. Then for any player B in the tournament, either A' beats A beats B, or there is another player C(B) such that A' beats A beats C(B) beats B (the (B) to explicitly state that the choice of C depends on the choice of B). From there, if you can prove that A' must be superior, you reach a contradition. Confusing Manifestation(Say hi!) 03:12, 2 April 2008 (UTC)
## Algebra
Okay, I help with this question here, but I really don't get it.
Given any three natural numbers, show that there are two of them, a and b, that a^3b-b^3a is divisible by thirty. I only understood to the even/odd part. Thanks, 76.248.244.196 (talk) 01:42, 2 April 2008 (UTC)
a^3 b - b^3 a = a b (a - b) (a + b), and to be divisible by 30 you need it to have factors of 2, 3 and 5. You say you understand the odd/even bit, so let's move on to divisible by 3. When Keenan talked about modulo arithmetic, what he means is talking about numbers in terms of their remainder when divided by various numbers. In the case of "modulo 2", there are two types of number - odd and even. In the case of "modulo 3", there are three - multiples of three, numbers one more than a multiple of three, and numbers one less than a multiple of three. We can write these as 3n, 3n + 1 and 3n - 1 respectively, where n is an arbitrary whole number.
This may be the long way of doing it, but consider all three cases for each of a and b, and what they mean for the factors a, b, a - b and a + b. For example:
If a = 3n + 1 and b = 3m - 1, then neither is divisible by 3, but a + b = 3n + 1 + 3m - 1 = 3n + 3m = 3(n + m), which is. If you go through all nine combinations, you will eventually see that in each case, at least one of a, b, a - b and a + b will be divisible by 3.
The "divisible by 5" part is what requires the pigeonhole principle. Again, for the sake of enlightenment, go the long way and consider all of the possibilities modulo 5 (so let a = 5n, 5n + 1, 5n + 2, 5n + 3 and 5n + 4, or if you're a fan of symmetry, 5n, 5n + 1, 5n + 2, 5n - 1 and 5n - 2 is equivalent). You'll see that there only some combinations are guaranteed to make one of the factors divisible by 5, so what you have to prove is that, given three numbers, there are always 2 that will fall into one of those combinations (for example, if at least one of them is divisible by 5 you're home and hosed). Hope that helps, Confusing Manifestation(Say hi!) 02:57, 2 April 2008 (UTC)
Thanks a lot! I fully understand now, but it will be a little difficult to put all the possibilites in a proof. Thanks, 76.248.244.196 (talk) 20:40, 2 April 2008 (UTC)
Listing all possible cases is considered a valid, if slightly ugly, method of proof. For an extreme case, check out the Four color theorem. Also, like I said, my method goes the long way but there are several shortcuts along the way depending on how comfortable you get with the modular arithmetic - for a start, you can simplify things by simply stating at various points "modulo 5" and then drop out all the 5n terms. And the 5 x 5 table in the last step comes down to a single statement involving one trivial case and one where there are two possible categories for the pigeonhole principle. Confusing Manifestation(Say hi!) 22:59, 2 April 2008 (UTC)
For showing that 5 can be made a factor by choosing a and b: consider the remainders modulo 5 of the three numbers, giving three numbers from the set {0,1,2,3,4}. If any of the three is 0, we are done (because of the factor ab). If any two are the same, we are also done (because of the factor a−b). Otherwise, we have a subset of three different numbers of the set {1,2,3,4}; only one of those four is missing. We need to show that we can pick a & b such that the final factor a+b is 5. If the missing fourth number is 1 or 4, take those giving remainders 2 & 3, otherwise take those giving 1 & 4. To complete the proof, show that for any pair of numbers a & b the number a3b − b3a is a multiple of 6. --Lambiam 09:09, 3 April 2008 (UTC)
## Geometry
In triangle ABC, draw angle bisectors AD and CE, where D is on BC and E is on AB. If angle B is 60 degrees, show that AC=CD+AE.
I've figure out that if the intersection of AD and CE is F, then <CFA and <EFD are 120 degrees and <DFC and <EFA are 60 degrees, but I'm not sure what's next. I'm reposting from here, but I'm not sure how to use the law of cosines on it, as I don't know any sides. Thanks, 76.248.244.196 (talk) 02:21, 2 April 2008 (UTC)
Also draw the third bisector. What point does it go through (other than B)? Figure out the sizes of some new angles that have appeared now. Do you spot any congruent triangles? --Lambiam 09:21, 3 April 2008 (UTC)
## Distribution of a maximum of multivariate normal
Does anyone know (or know a reference) on the distribution of max(X) if X is a multivariate normal random vector with various means (can be assumed to have constant, ie ${\displaystyle \sigma ^{2}I}$, variance). Simulation tells me it is gamma-like, but I haven't been able to prove it. Any ideas? Thanks, --TeaDrinker (talk) 02:28, 2 April 2008 (UTC)
Unless I have made a mistake or misunderstood your question, this is a straightforward calculation:
${\displaystyle F(x)=P(\max X\leq x)=P\left(\bigwedge _{i=1}^{n}X_{i}\leq x\right)=\prod _{i=1}^{n}P(X_{i}\leq x)=\prod _{i=1}^{n}\Phi \left({\frac {x-\mu _{i}}{\sigma }}\right)}$
${\displaystyle f(x)=\sum _{i=1}^{n}{\frac {1}{\sigma }}\Phi '\left({\frac {x-\mu _{i}}{\sigma }}\right)\prod _{j\neq i}\Phi \left({\frac {x-\mu _{i}}{\sigma }}\right)}$
It's possible that this is harder if the covariance matrix is not diagonal. -- Meni Rosenfeld (talk) 15:08, 2 April 2008 (UTC)
It is more complicated if the components are not independent. By the way, a standard notation for the pdf Φ' is φ. --Lambiam 09:27, 3 April 2008 (UTC)
Thanks! I tried this approach, but didn't see any slick simplifications (which I was hoping for--the iid case, of course simplifies nicely, but perhaps this one is just irreducibly ugly). Thanks again, --TeaDrinker (talk) 16:50, 2 April 2008 (UTC)
## Square digit chain
As in this problem. Why can't there be other loops? Can someone show me a proof that all possible chains will end in 1 or 89? 70.162.25.53 (talk) 03:43, 2 April 2008 (UTC)
It is not overly clever, but note that if you start with a four digit number, the largest number you can get is 324, and in fact, any number with four or more digits must decrease (and does so very rapidly until there are only three digits). Thus every number eventually becomes a number smaller than 325. It is therefore sufficient to simply check all the numbers 2 through 324 to verify that 1 and 89 are the only possible outcomes. It is not my area of expertise but I don't see a slick solution. --TeaDrinker (talk) 05:03, 2 April 2008 (UTC)
Actually, the bound can be brought down to 162. To see why, consider any x<324 that largest digit square sum you will find is 166, and then consider any number less than 166, that one that will produce the largest digit sum is 99, not 159 81+81=162, vs 1+25+81=26+81=107. so with that you can refine the bound to 162. Another hint, you can replace 89, by any of the other numbers after the first 89 in the sequence they provided, because those numbers are part of the loop. Additionally, any number who's sum converges to any of those number in that loop will go to 89, and every other number in that loop. A math-wiki (talk) 06:30, 2 April 2008 (UTC)
Going beyong squares, we can generalise this problem to iterating the sum of any power of a number's digits, and indeed generalise further to any base. Using similar arguments to the above, we can show that there is always some N such that the set of integers {1...N} is mapped to a subset of itself, and any integer greater than N is mapped to a strictly smaller integer. So all chains eventually enter the region {1...N} and once inside they never leave it. The long-term behaviour of any chain must therefore be to settle into a loop or to reach a fixed point (which is only a loop of length 1). Loops and fixed points can then be found by a finite number of trials.
Fixed points of these "powers of digits" functions are called narcissistic numbers. Considering sums of cubes of digits, for example, there are four fixed points (as well as the trivial fixed point 1), two loops of length 2, and two loops of length 3 [1]. Gandalf61 (talk) 11:21, 2 April 2008 (UTC)
Another loop is 00 (or is 0 no longer a number)? --Lambiam 09:39, 3 April 2008 (UTC)
But no other number has a sequence that reaches zero, so it's kind of useless. — Kieff | Talk 10:18, 3 April 2008 (UTC)
Must true facts in mathmatics are "kind of useless". The issue was that the statement "EVERY starting number will eventually arrive at 1 or 89" is false. Unless I made a mistake, the answer to the question "How many starting numbers below ten million will arrive at 89?", counting only nonnegative integers as starting numbers, is 8581146. This is completely useless knowledge, and only an idiot or a mathematician would pose that question. --Lambiam 11:29, 3 April 2008 (UTC)
So the ratio of numbers that arrive at 89 to numbers that arrive at 1 is approximately 8:1, and the 89 loop also has length 8. That's interesting ... Gandalf61 (talk) 22:59, 3 April 2008 (UTC)
The ratio is much closer to 6:1. --Lambiam 23:27, 3 April 2008 (UTC)
Ah, yes, 6:1. I wonder if that ratio converges as the upper limit increases ... Gandalf61 (talk) 08:31, 4 April 2008 (UTC)
Apparently, the behaviour of the ratio is rather erratic. See OEIS:A068571 which gives the number of happy numbers (numbers which arrive at 1) less than or equal to 10n for n=1 to 21. An optimist might see signs of convergence to some proportion around 12%, but I wouldn't put money on it ! Gandalf61 (talk) 10:36, 4 April 2008 (UTC)
(exdent) I'm pessimistic about there being a limit. Defining H(n) to be the fraction of happy numbers ≤ 10n, I find that H(43) = 0.1607595665888606352231265289855421563588180, while H(233) starts off like 0.13206866.... These are local extrema in the sequence. --Lambiam 22:55, 4 April 2008 (UTC)
## Area and Volume Ratios
I'm having trouble understanding Area and Volume ratios as the teacher didn't fully explain it at the time... does anyone have a basic summary of it? —Preceding unsigned comment added by Devol4 (talkcontribs) 06:46, 2 April 2008 (UTC)
A given 3 dimensional shape will have a surface area and a volume. The surface area to volume ratio is the surface area divided by the volume. That ratio is smallest for a fixed volume or surface area if the shape is a sphere. Does any of that help? If not, what specifically is confusing you? --Tango (talk) 12:39, 2 April 2008 (UTC)
Could it be about the fact that scaling a figure by a factor k will scale its area by a factor k2 and its volume by a factor of k3 ? -- Xedi (talk) 19:18, 2 April 2008 (UTC)
For a simple example, if you have a cube whose sides have a length of 1cm, then its surface area A = 6cm2, and its volume V = 1cm3. Then the ratio A/V = 6cm2/1cm3 = 6cm−1. If instead we take a block of dimensions 0.5cm × 1cm × 2cm, A = 7cm2 while V = 1cm3 as before, so now A/V = 7cm−1.
We also have an article Surface area to volume ratio, which however is not about the maths but about the significance in physical chemistry and biology. --Lambiam 09:58, 3 April 2008 (UTC)
## In 5-Card Draw Poker, what are the odds of a royal flush without a wild card when 2 are in play?
A friend and I were playing 5-card draw Poker, using 2 joker cards as wild cards. This was in addition to the standard 52-card deck. We each drew 3 cards on our turns, and I ended up with a royal flush (A,K,Q,J,10 all of the same suit), without a wild card. I don't recall what my friend ended up with, so I hope it doesn't matter much. I do remember that he didn't get a wild card either. I know the odds of a royal flush are about 650,000 to 1, but that's without a draw, in which case the odds are higher. So, i've been wondering ever since what odds I overcame to draw this hand without getting one of the 2 wild cards that were in play. Thanks so much to whoever can answer this for me. MoeJade (talk) 09:22, 2 April 2008 (UTC)
Because of the draw, there isn't a unique answer - it depends on your decision making process. I would think the decision making process that would give the highest odds of a royal flush (but would probably not be a good strategy in practise) would be to keep only cards 10 or higher, and only cards of the suit that would leave you with the most cards. If we assume that strategy, we can calculate the odds. It's not a quick calculation though, and I don't have time right now. If no-one else tackles it before I find time, I'll give it a go. My method would be to calculate it separately in the cases of drawing no cards, 1 card, up to 5 cards, and weight each of them by the chance of that case occurring. --Tango (talk) 12:32, 2 April 2008 (UTC)
I was afraid of that. Unfortunately, I have only a basic understanding of statistics, so I quickly became lost trying to determine the odds. Don't spend too much time on this problem; I was hoping it wouldn't be too complicated, but if it is, just give me a ballpark range if you can. MoeJade (talk) 14:41, 2 April 2008 (UTC)
How does "very unlikely" suit you? While it's going to be more likely than getting it straight away, it will still be very unlikely. We can find a lower bound by finding the chance of getting a royal flush when dealt 10 cards (since that's the most cards that can involved in one hand of 5-card draw, so you can only get a royal flush if there is one somewhere in those 10 cards). That's an easier calculation. There are 4 choices of Ace, and once that's decided, the next 4 cards are determined, you can then have anything as the remaining 5, and all of those can appear in any order. So that's (4*1*1*1*1*(10!/5!)*47*46*45*44*43)/(52!/(52-10)!)=0.00039, about 1 in 2578. So, we know the odds are somewhere between about 650,000 to 1 and 2500 to 1. Probably closer to the higher number, but it depends on your strategy. --Tango (talk) 17:43, 2 April 2008 (UTC)
If you have four 8's and a 9, will you throw all 5 cards away? If you don't, the probability of a royal flush is 0. If you do, you increase that probability. Someone who's trying for a royal flush at all costs will have a slightly better chacne of making one than someone who's using a sensible strategy. --tcsetattr (talk / contribs) 20:47, 2 April 2008 (UTC)
Exactly. The strategy I described above should maximise your chances of getting a royal flush, but it's a pretty stupid strategy in a real game. Trying to calculate the odds for a realistic strategy is pretty much impossible - you would have to define that strategy precisely, and that's going to be very complicated. --Tango (talk) 21:04, 2 April 2008 (UTC)
When you say "complicated," how complicated do you mean? I would like to, even if it takes me a couple weeks to memorize, know how to get a royal flush. I could pwn some n00bs in online poker... flaminglawyerc 21:20, 2 April 2008 (UTC)
If you're going to play draw poker by whatever strategy maximizes your probability of getting a natural royal flush, let me just say that I'd love to have you in my game :-) --Trovatore (talk) 22:42, 2 April 2008 (UTC)
I said "realistic strategy", I didn't say "strategy that will always get a royal flush". Such a strategy is obviously impossible. --Tango (talk) 00:37, 3 April 2008 (UTC)
It was a joke. The strategy is not very realistic for someone who does not want to lose most of the time. --Lambiam 11:36, 3 April 2008 (UTC)
The original poster only asked for an estimate of the odds. One thing that matters but wasn't specified is how many cards you are allowed to replace on the draw. To start with I'll assume that at most 3 can be replaced. (Even if you allow more, what you really want is the odds of getting a natural royal flush after drawing no more than 3 cards, right? Because that's a memorable aspect of the event.)
If the original hand had a pair or any higher combination, or any wild card, you'd keep it and therefore would be unable to improve your hand to a royal flush on the draw. I'll assume that in the absence of such a combination you would keep any holding that might be part of a royal flush, and discard anything else. (This is slightly wrong because you might keep a partial flush or partial straight; and also if your hand has two such combinations, like A-K of spades and Q-J of hearts, you might keep the wrong one — you keep A-K of spades only to draw A-K-10 of hearts — but these cases are pretty unlikely as long as you have to keep two cards, and won't affect the estimate much.)
Based on these assumptions we want the probability that 8 specific cards from the deck will include a natural royal flush, but the first 5 of them will not include a pair, straight, flush, or joker. I will make a further simplification and say that any 5 of them must not include a pair, straight, flush, or joker — since the cases where you will draw 3 cards are much more common than the cases where you will draw 2 or less, this won't affect the odds much. (That is, say the first 5 cards are two clubs and A-K-Q of spades; you keep the spades and draw J-10 of spades. The 8th card is the king of hearts. You'll never see it, but my simplification treats this case the same as if you got the king of hearts as the first case in place of one of the clubs and therefore would never draw enough cards to fill the straight flush.)
Okay, now, there are 54C8 = 1,040,465,790 combinations possible for 8 cards. Of these how many are suitable? First, there are 4 possible royal flushes, and for each one there are 47C3 = 16,215 combinations of other cards none of which are jokers. The first other card has a probability of about 15/47 that it will pair with a card in the royal flush, so 32/47 that it won't. Since we're only doing a rough estimate, we can pretend the probability is the same for the second and third other cards, and say that (32/47)³ of the cases are suitable. The case where the other three cards complete a natural flush has only 8C3 = 56 possibilities and can practically be ignored; for a straight it is not much higher, and I'll ignore that case too. So my estimate of the probability is (16,215 × (32/47)³) / 1,040,465,790 or say 1/200,000. Or in the form of odds, 200,000 to 1.
E&OE and HTH. :-)
--Anonymous, 00:27 UTC, April 5, 2008.
## statistics
Hi, I am a teacher trying to determine validity of test questions. I understand what validity is. My question is what percent of content must be measured in order to make a test valid. For example if there are 100 vocabulary words a test has only one question about them then the test could not be accurately measuring knowledge. If there are ten questions about them does that 10% represent enough of a statistical population to ensure the validity of the testing device. Is there a minimum percentage needed to do so? Thank you. 206.219.72.83 (talk) 15:20, 2 April 2008 (UTC) I am not sure what symbols to show?
Do you really mean validity? I believe you mean precision and/or accuracy. Accuracy is not the issue if you picked the question(s) randomly: even with one question, if I knew (say) 80 of the 100 answers, the average proportion of right answers would be 0.80, as I would get it right 80% of the time, and wrong 20% of the time. But this alludes to the real problem: precision. For one question, your assessment of me will be off by at least 20% all of the time. But if you quantify that (lack of) precision correctly, your results can still be valid, even if they are not useful.
You may wish to look at the binomial distribution article for a followup. Baccyak4H (Yak!) 18:00, 2 April 2008 (UTC)
I interpreted it as a question about statistical significance. See sample size - that article should get you started, at least. --Tango (talk) 18:46, 2 April 2008 (UTC)
Assuming all questions are equally difficult, and the student scores P correct answers out of a total of N randomly selected questions, the best estimate for the fraction of items known to the student in the total universe of possible questions is P/N. However, this is only an estimate. Let c be a measure of confidence we want to achieve, say c = 0.95. If the true fraction of items known to the student is x, the probability of observing P or more correct answers out of N is:
${\displaystyle \sum _{k=P}^{N}{\binom {N}{k}}x^{k}(1-x)^{N-k}\,.}$
As x gets smaller, this probability gets smaller, but if it goes below (1 − c)/2, it becomes unlikely small, so that means x is too small. Define xlow as the value of x for which the probability equals (1 − c)/2.
Likewise, the probability of observing P or fewer correct answers out of N is:
${\displaystyle \sum _{k=0}^{P}{\binom {N}{k}}x^{k}(1-x)^{N-k}\,.}$
As x gets larger, this gets smaller, and eventually 0 (unless P = N). Define xhigh as the value of x for which this probability equals (1 − c)/2. With some confidence we can say that x is in the range from xlow to xhigh. This depends on N, P, and c (but not on the size of the universe of questions). Given fixed N and c, the worst case (the widest range) is when P is about one half of N.
If N is fairly large and P = 1/2N, we may approximate the two probabilities above by
${\displaystyle 1-\Phi \left({\frac {{\tfrac {1}{2}}-x}{\sigma }}\right)}$
and
${\displaystyle \Phi \left({\frac {{\tfrac {1}{2}}-x}{\sigma }}\right)\,,}$
where
${\displaystyle \sigma ={\sqrt {\frac {x(1-x)}{N}}}\,}$
and Φ is the cumulative distribution function of the standard normal distribution.
How large N should be depends on how high c must be, and how narrow the range [xlow, xhigh] is required to be.
Mathematically this is not easy to handle, but since the critical area is around x = 1/2, we can slightly simplify things by just putting
${\displaystyle \sigma ={\frac {1}{2}}{\sqrt {\frac {1}{N}}}\,.}$
A 2-sided confidence interval for c = 0.95 means that xhighxlow is about 4σ. Suppose that we require this width to be at most 1/10. Then the simplified formula for σ tells us that N must be at least 400. --Lambiam 12:37, 3 April 2008 (UTC)
One other factor to consider is that you must give students enough time to complete the test (even the slowest students). It always seemed to me that tests in school put way too much emphasis on doing things quickly and not nearly enough emphasis on doing them correctly. In the real world, getting the correct answer 70% of the time is not a "passing grade". On the other hand, in the real world you're often expected to double check your work, and timed tests don't seem to encourage this. StuRat (talk) 18:48, 4 April 2008 (UTC)
## Afraid of sets
Since I realised that not every carelessly defined set is acceptable, I have been afraid that I will break something, and I don't know enough to understand the relevant axiom in ZFC (if that's even the axioms to use!). When I can figure out a bijection between my proposed set and something I know is a set, it's OK I guess, but how about "the set of all functions from ${\displaystyle \mathbb {R} }$ to ${\displaystyle \mathbb {R} }$"? What are your suggestions for figuring out what is an acceptable set and what is not? Thanks. —Bromskloss (talk) 16:39, 2 April 2008 (UTC)
ZFC is as good an axiomatization of set theory as any other, and I think the axioms guaranteeing the possibility to construct certain sets are easy enough to understand. Basically there are a few legal operations you can perform on sets, such as power sets, subsets and binary unions and Cartesian products. The collection of functions from ${\displaystyle \mathbb {R} }$ to ${\displaystyle \mathbb {R} }$ is a subset of ${\displaystyle P(\mathbb {R} \times \mathbb {R} )}$ and thus a set. -- Meni Rosenfeld (talk) 16:57, 2 April 2008 (UTC)
Ah, a subset of ${\displaystyle P(\mathbb {R} \times \mathbb {R} )}$. Of course! Thanks. Speaking of ZFC being as good as any other, what difference does the choice of axioms do? Is ZFC the one everyone is using today? With other axioms, would the resulting mathematics similar at all? Would it still work for physics? Bromskloss (talk) 17:07, 2 April 2008 (UTC)
The choice of axioms can make a lot of difference - they basically define what a set is. You define a set differently and it will behave differently. ZFC is pretty widespread these days. There are a few people working on alternative axioms (eg New Foundations). The axiom of choice (the C part) isn't always used (it's independent of ZF, that is, all the other axioms, so you can choose to include it or not - it usually is, though, in my experience). The things you might think are sets which actually aren't are called proper classes - they're usually very big (for an appropriate and vague definition of "big", at least) collections, like "all sets". A set of functions between two sets is quite small by comparison, and is a perfectly fine set. Most things you'll come across work perfectly well as sets - just stay away from anything like "the set of all sets", and you should be fine. --Tango (talk) 17:23, 2 April 2008 (UTC)
In practice I think a lot of working mathematicians use NBG. It's lighter than ZFC, and the set/class distinction is good enough for pretty much anything non set-theorists are ever going to do. If you're not doing set theory you just need to be able to steer clear of the most glaring pitfalls/paradoxes, and sets and classes, or a fixed universe of sets, or any similar light-weight type theory is good enough for that. -- Leland McInnes (talk) 18:05, 2 April 2008 (UTC)
In practice, working mathematicians don't use a formal axiomatic system at all. They work as Platonists, whether they claim to be or not. --Trovatore (talk) 18:09, 2 April 2008 (UTC)
A fair call - Ultimately I simply meant that the sets vs. classes distinction that NBG uses is the only point that I've seen crop up in non set theorist mathematics with regard to set theoreti paradoxes. Thus working mathematicians take NBG's sets and classes and pretty muh ignore the rest of axiomatic set theory.
As a side note, I'm more of a structuralist and never could entirely understand the platonists. I would prefer and welcome a category or topos theoretic foundation. -- Leland McInnes (talk) 01:13, 3 April 2008 (UTC)
I was using "Platonist" as an (admittedly imprecise) shorthand for "realist", which certainly includes structuralists -- in fact structuralism is the main branch of realism in contemporary foundational philosophy, in the sense of considering isomorphic structures to be the same, except when you need to distinguish between the way they're embedded in some larger structure. Note that structuralism is already a sufficiently strong form of realism to guarantee that, for example, the continuum hypothesis has a well-defined truth value. --Trovatore (talk) 02:39, 3 April 2008 (UTC)
That depends on the school of structuralism in question. The realist structuralists in mathematical philosophy, while hurdling the representation issue, still put themselves in what seems to me to be an untenable position. Count me in the same school as Bell, Mac Lane and McLarty I guess. -- Leland McInnes (talk) 12:29, 3 April 2008 (UTC)
Not directly, sure, but working mathematicians build on what's gone before, and that will be built on particular axioms (perhaps not historically, but someone will have formalised it at some point). Generally, mathematicians use whatever axioms are needed to do what they're trying to do (eg. if you need AC to prove your theorem, then you assume AC, otherwise you ignore it completely). The various (common) choices of axioms don't usually contradict each other in the kind of work most mathematicians do, so people don't generally worry about it. --Tango (talk) 18:36, 2 April 2008 (UTC)
"Someone will have formalized it at some point". Yes. But that does not mean that it is "built on axioms". The axioms are extracted from the mathematics, not the other way around. Oh, it's true that there's an interplay between them; the picture of the von Neumann hierarchy did not really come clear until after Zermelo's formalization, though in retrospect we can see that it was inherent in Cantor's work all along. --Trovatore (talk) 18:40, 2 April 2008 (UTC)
But until something is formalised, you can't be sure it's right (even then, Godel gets in the way, but lets not complicate things any further). While a lot of maths was (and, I guess, still is) invented/discovered without a formal axiomatic approach, it's only since it's been formalised that we've been able to be sure we're not talking nonsense (for example, see Russell's Paradox). So, while the maths itself may not be built on axioms, the proof that the maths is correct is. Whether you choose to distinguish between the two is a matter of definition of the word "built", I suppose. --Tango (talk) 18:54, 2 April 2008 (UTC)
You can never be "sure" it's right. Only "pretty sure". Essentially mathematics is an empirical science, subject to the limitations of human knowledge like any other; the differences are quantitative rather than qualitative. --Trovatore (talk) 19:02, 2 April 2008 (UTC)
Are you referring to Godel? If so, I already dismissed him as an unnecessary complication to this discussion. If not, then I disagree. By my understanding, assuming you're working in a consistent and complete framework, a mathematical proof is absolute. There is no need for any kind of experimentation. --Tango (talk) 19:18, 2 April 2008 (UTC)
Well, I could point out that it's a bit cavalier to dismiss as "an unnecessary complication" a proof that the thing you're assuming (depending, I guess, on what precisely you mean by a "framework") doesn't exist. But in a sense you're right; it is an unnecessary complication, because the error of Euclidean foundationalism should have been clear even in Euclid's time. It's an infinite regress -- you claim that your results are not subject to error because they are derived from your alleged foundation, but how do you know that the foundation itself, or your method of derivation, is not subject to error? --Trovatore (talk) 19:51, 2 April 2008 (UTC)
It's extremely cavalier, but you can't get bogged down in the details in every discussion. Our axioms are as good as it's possible to get them, so most of the time, it's easiest just to assume they are right. --Tango (talk) 20:55, 2 April 2008 (UTC)
That's fine by itself. But it's not a very convincing as part of an argument that formalization is some sort of absolute defense against error. My position is that the objects of study (natural numbers, real numbers, functions, sets) are the primary notion, and the axioms that describe them are the subordinate concept. You don't have to agree with that, but making the axioms primary on the grounds of certainty is not going to work. --Trovatore (talk) 21:15, 2 April 2008 (UTC)
How certain it is isn't really relevant. Without a formal axiomatic approach, you have very little certainty at all. You can't really prove anything about natural numbers until you know what a natural number is - the naive approach of "they're the numbers you count with - a 2 year old child could do it!" is just as dangerous as naive set theory and could well lead to equally paradoxical results. While we can't be completely certain we've got it right, what certainty we can have comes from the axiomatic approach. You can do an awful lot of maths without worrying about rigour, but there's always the chance you'll trip up like Russell did. An axiomatic approach doesn't eliminate that danger, but it gets pretty close. --Tango (talk) 23:44, 2 April 2008 (UTC)
No, the axiomatic approach is pretty much irrelevant to the question of eliminating the danger. Note that Russell's paradox was discovered in a completely formalized version of set theory, due to Frege, who thought he was formalizing informal Cantorian set theory but may well have been wrong about that (Wang Hao thought so -- it may depend on which moment in Cantorian thought you choose to look at). At best, axiomatics provides a way of, as it were, "localizing" the risk, saying, "if there's an inconsistency, it should show up here". When an apparent inconsistency shows up in informal argument, that's a useful thing to have. But confidence in the correctness of the results is fundamentally not about axiomatics. --Trovatore (talk) 00:04, 3 April 2008 (UTC)
Oh, I should say: irrelevant, except insofar as the formalization helps to find the errors. That is arguably what happened in the Frege-Russell case -- when Frege formalized his notion of set (which he thought was the same as Cantor's but he was probably wrong about that), he made it easier for Russell to find the error in the concept. This is a useful aspect of formalization. But the fact that a deductive system is formal, by itself, does not provide any extra confidence whatsoever in its results. --Trovatore (talk) 02:19, 3 April 2008 (UTC)
Certainly, a formal system can be just as wrong as any other system. However, it is possible to prove the consistency of a formal system (you need to use a different system to do it, of course, so it's far from perfect), it isn't possible to do so for an informal system. It's all just moving the risk around, as you say, but that can be quite handy. --Tango (talk) 12:43, 3 April 2008 (UTC)
It's an interesting thing to do for lots of reasons, but increasing confidence in the results is not really one of them, because to prove the consistency of a formal system you need to use a stronger theory (one even more "likely" to be wrong"). Tango, I really think you have the wrong idea about formal systems; you've let yourself be taken in by the errors of the formalists. Formal systems are more an object of study than they are a tool for deriving conclusions. --Trovatore (talk) 15:56, 3 April 2008 (UTC)
Are the ZFC axioms sufficient to rule out the possibility of a self-referential set, i.e., a set that contains itself as an element? -GTBacchus(talk) 17:50, 2 April 2008 (UTC)
Yes. It follows from the axiom of foundation that there is no such set. --Trovatore (talk) 17:58, 2 April 2008 (UTC)
From the discussion above, I seem to be able to extract that the most commonly discussed axiomatic set theories will eventually lead to the same mathematics, once you get past the most fundamental levels. I.e., most concepts in one theory would have its counterpart in another. They would all be useful for physics. Is that correctly understood? Is it possible that some completely different theory would lead to useful mathematics that is out of reach of what we are using now? —Bromskloss (talk) 19:43, 2 April 2008 (UTC)
Yeah, I think all the common axiomatic systems would agree on anything used in physics (as long as you can construct the real numbers in all of them, you're pretty much sorted, I would think). As for your last question - probably depends on what you mean by "useful". Different choices of axioms will certainly lead to new and interesting maths. Whether it will have any uses, it's hard to say, but people seem to be able to find real world uses for the strangest of mathematical concepts (which are generally created because a mathematician was curious as to what would happen, so thought "why not find out?", and just gave it a go), so I think there's a very good chance someone will make use of any maths we can come up with. --Tango (talk) 20:55, 2 April 2008 (UTC)
## Open loop transfer function
In my syllabus we've got study of control systems, and it says that that Open Loop transfer function for a normal feedback system is G(s)*H(s). Even if it is not a unity feedback system. I don't understand how it makes sense - open loops means that H(s) is disconnected from the system, yet how is it being considered? All my textbooks use G(s)*H(s), and I asked my professor, he could not come up with a satisfactory answer.
The example control system which I'm talking about is this - http://classes.engr.arizona.edu/ame558/public_html/topic1/bd2.gif That, incidentally, seems to be "standard" control system - almost all formulae and theorems are derived keeping that control system in mind.
Thanks! --RohanDhruva (talk) 19:50, 2 April 2008 (UTC)
"Open-loop", in this case, does not refer to the situation where you control your system without a feedback. Instead, you should unplug ${\displaystyle F(s)}$ from the circle (which calculates the difference) and consider the signal path from ${\displaystyle E(s)}$ to ${\displaystyle F(s)}$. That gives ${\displaystyle G(s)H(s)}$ (or ${\displaystyle KG(s)H(s)}$ as it seems to be written in the picture). They actually define it just like that, as ${\displaystyle F(s)/E(s)}$, on one of the accompanying pages. —Bromskloss (talk) 20:37, 2 April 2008 (UTC)
OK, if you did send in a signal ${\displaystyle R(s)}$ after unplugging ${\displaystyle F(s)}$ from the difference circle, you would be controlling the system without a feedback, but that's not the point. —Bromskloss (talk) 20:47, 2 April 2008 (UTC)
Thanks Bromskloss. I still don't understand the "logic" behind it being defined as ${\displaystyle F(s)/E(s)}$. In open loop, you mean to say, the signal ${\displaystyle F(s)}$ is not connected to a summing point, instead, it acts as a point to take the output from? If so, what happens to the output signal ${\displaystyle C(s)}$. The wikipedia page for Closed loop pole defines "Open loop transfer function" as -- The open-loop transfer function is equal to the product of all transfer function blocks in the forward path in the block diagram. Would that clarify anything? --RohanDhruva (talk) 02:10, 3 April 2008 (UTC)
I agree with you that it does not seem obvious that ${\displaystyle F(s)/E(s)}$ would be of any interest, but I think it can be useful when studying the porperties of the whole feedback system. I don't remember exactly how, though. In the mean time, ${\displaystyle C(s)}$ just sits there. —Bromskloss (talk) 07:03, 3 April 2008 (UTC)
## Dilemna!
I don't know how to write this using [itex] tabs, but I'll do my best to explain it. If you can, please write it in those "magic words."
The problem is:
There's a magic thing in algebra. Pick any positive whole number. Square it. Add [twice the original number]. Add 1. Now find the [square root] of that number. Subtract the original number, and the answer is 1.
Now that's fine and dandy. Even the reasoning isn't too hard.
${\displaystyle {\sqrt {x^{2}+2x+1}}-x=1}$
${\displaystyle {\sqrt {(x+1)^{2}}}-x=1}$
Meaning...
${\displaystyle x+1-x=1}$
Meaning...
${\displaystyle 1=1}$
That's fine and dandy. But if you don't find the quadratic roots - ah! A dilemma. See:
${\displaystyle {\sqrt {x^{2}+2x+1}}-x=1}$
We have to square every term in order to get rid of the [square root sign]. So:
${\displaystyle x^{2}+2x+1-x^{2}=1^{2}}$
Take out the zero pairs.
${\displaystyle 2x+1=1}$
Subtract...
${\displaystyle 2x=0}$
But alas! The number you are told to pick has to be a positive whole number. And zero isn't positive (or negative). So it shouldn't work. But try as hard as you like; as long as x is a positive whole number, it works. Can someone please explain this? flaminglawyerc 20:50, 2 April 2008 (UTC)
It's just an algebraic error: ${\displaystyle (a+b)^{2}=a^{2}+2ab+b^{2}\neq a^{2}+b^{2}}$. --Tango (talk) 21:01, 2 April 2008 (UTC)
(Edit conflict.) (Fixed [itex] notation.) You've made an error going from ${\displaystyle {\sqrt {x^{2}+2x+1}}-x=1}$ to ${\displaystyle x^{2}+2x+1-x^{2}=1^{2}}$. Note that ${\displaystyle (a+b)^{2}}$ does not simplify to ${\displaystyle a^{2}+b^{2}}$. —Bromskloss (talk) 21:04, 2 April 2008 (UTC)
(edit conflict squared) "We have to square every term in order to get rid of the [square root sign]. " Ummmm, no, it doesn't work that way ... Go ahead and add ${\displaystyle x}$ to both sides in place of that step, then square both sides (and most decidely not "every term"). --LarryMac | Talk 21:08, 2 April 2008 (UTC)
Oh, so you would square the sides. I thought it was every term. I get it now. (it was kind of new territory for me, since I am only taking Algebra 1) flaminglawyerc 21:13, 2 April 2008 (UTC)
## can u plz solve this?
(x+1)^10 —Preceding unsigned comment added by 64.229.54.4 (talk) 21:35, 2 April 2008 (UTC)
It's equal to (x+1)(x+1)(x+1)(x+1)(x+1)(x+1)(x+1)(x+1)(x+1)(x+1) --Carnildo (talk) 22:39, 2 April 2008 (UTC)
Check out Binomial expansion. And that can't be solved, because it isn't an equation, but it can, like I had linked, be expanded. Confusing Manifestation(Say hi!) 22:49, 2 April 2008 (UTC)
Thanks!!... it was easy:P...thanks! —Preceding unsigned comment added by 64.229.54.4 (talk) 23:24, 2 April 2008 (UTC)
${\displaystyle {x}^{10}+10\,{x}^{9}+45\,{x}^{8}+120\,{x}^{7}+210\,{x}^{6}+252\,{x}^{5}+210\,{x}^{4}+120\,{x}^{3}+45\,{x}^{2}+10\,x+1}$ --wj32 t/c 06:14, 3 April 2008 (UTC)
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# Math
posted by on .
2(a)Consider the arithmetic progression with terms A3=-2 and A12=23. Find the sum of A1+......+A40 (can do)
(b)In 1800 the population of England was 8 million. The economist Malthus (1766-1834) produced a hypothesis, suggesting:-that the population of england would increase, according to a G.P., by 2% per year
-that the english agriculture production,able to feed 10 million people in 1800,would improve according to an A.P. to feed an extra 400000 people every year
Po represents the english population in 1800 and Pn that population in the year 1800+n:
(i)express,according to Malthus' hypothesis Pn as a function n.
Ao represents the number of people that the english agriculture production can feed in 1800 and An that number in 1800+n:
(ii)express,according to Malthus' hypothesis,An as a function of n
(iii)Calculate the population of england in 1900 and the number of people that the english agriculture production can feed in 1900
(iv)Determine the year from which the english agriculture can no longer feed the english population according to Malthus' hypothesis(-use your calculator by graphing or creating the lists:n=L1;Pn=L2;An=L3 tp compare increases).
• Math - ,
b i)
a = 8 000 000
r = 1.2
Pn = 8 000 000 x 1.2^(n-1)
b ii)
a = 8 000 000
d = 400 000
Pn = 8 000 000 + (n-1)400 000
= 7 600 000 + 400 000n
This should help you to do the rest!
### Answer This Question
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# PIBRO Editorial
Tester: Kasra Mazaheri
Editorialist: Hussain Kara Fallah
### PROBLEM EXPLANATION
Given an array A[1 \ldots N] consisting of N binary digits (0,1) and a single integer K. You can select only one segment (subarray) of length K and change all zeroes inside this segment to 1.
Find the maximum number of consecutive ones inside the array you can achieve after applying the mentioned operation once.
Easy
### CONSTRAINTS
1 \leq N \leq 10^5
### EXPLANATION:
Let’s define 2 arrays head,tail such that:
head_i denotes the number of consecutive 1 (ones) to the left of i_{th} element (the element itself not included).
tail_i denotes the number of consecutive 1 (ones) to the right of i_{th} element (the element itself not included).
How to calculate each?
If A_{i+1}=0 then tail_i=0
If A_{i+1}=1 then tail_i=tail_{i+1}+1
Now how to pick the best segment?
Easy, let’s try all possible choices. Assume we pick the element with index st as the starting element then we will change all elements with indices in the range [st,st+K-1] into ones. So we would result in a segment made of ones of length head_{st}+K+tail_{st+K-1}
Try all possible st and pick the best result. Check implementation for details.
Editorialist's Solution
#include<bits/stdc++.h>
using namespace std;
int T , n , K , head[1<<17] , tail[1<<17];
string str;
int main(){
cin>>T;
while(T--){
cin>>n>>K>>str;
for(int j = 1 ; j < n ; j++) {
if (str[j - 1] == '1')
}
for(int j = n - 2 ; j >= 0 ; j--) {
if (str[j + 1] == '1')
tail[j] = tail[j + 1] + 1;
else tail[j] = 0;
}
int ans = 0;
for(int j = 0 ; j + K <= n ; j++)
ans = max(ans , K + head[j] + tail[j + K - 1]);
cout<<ans<<endl;
}
}
Tester Solution
// In The Name Of The Queen
#include<bits/stdc++.h>
using namespace std;
const int N = 100005;
int q, n, k, L[N], R[N];
char S[N];
int main()
{
scanf("%d", &q);
for (; q; q --)
{
scanf("%d%d%s", &n, &k, &S[1]);
for (int i = 0; i <= n + 1; i ++)
L[i] = R[i] = 0;
for (int i = 1; i <= n; i ++)
if (S[i] == '1')
L[i] = L[i - 1] + 1;
for (int i = n; i; i --)
if (S[i] == '1')
R[i] = R[i + 1] + 1;
int Mx = 0;
for (int i = 1; i + k - 1 <= n; i ++)
Mx = max(Mx, L[i - 1] + k + R[i + k]);
printf("%d\n", Mx);
}
return 0;
}
7 Likes
Got TLE one test :’(
https://www.codechef.com/viewsolution/28027923
Any way to optimize this?
Getting WA in 2 test case
What am i missing here …??
https://www.codechef.com/viewsolution/28014565
#include<bits/stdc++.h>
#include
#define ll long long int
using namespace std;
int main(){
ll t; cin>>t;
while(t–){
ll n,k; cin>>n>>k;
string s; cin>>s;
int ans = INT_MIN, tmp = 0, kk = 0;
if(n <= k){
ans = n;
cout<<ans<<endl;
}
else{
// ans = k;
for(int i=0;i<n-k;i++){
int j = i+k;
int l = i-1;
tmp = k;
for(;j<n;j++){
if(s[j] == '1'){
tmp++;
}
else{
break;
}
}
for(;l>=0;l--){
if(s[l] == '1'){
tmp++;
}
else{
break;
}
}
ans = max(ans,tmp);
}
cout<<ans<<endl;
}
}
return 0;
}
please inform me what’s wrong in this code:
from itertools import groupby
t = int(input())
while t>0:
n, k = input().split()
n = int(n)
k = int(k)
st = input()
ans = [0]
for x, y in groupby(st):
if x == '1':
ans.append(len(list(y)))
#print(x , )
a = min(n, max(ans)+k)
print(a)
t = t-1
What’s wrong with this approach.here i am trying to find max consecutive ones and then adding k to it taking care of boundary conditions.
https://www.codechef.com/viewsolution/28013706
can you please reveal a testcase for which my approach doesn’t work?
Can anyone please tell why tle in my code? According to me its time complexity is n*t which should be acceptable.
#include<bits/stdc++.h>
#define ll long long
#define vi vector
#define vll vector
#define pb push_back
#define rep(i,n) for(int i=0;i<n;i++)
#define repr(i,n) for(int i=n-1;i>=0;i–)
#define si set
#define sll set
#define ins insert
#define vpp vector<pair<int,int>>
#define mp make_pair
#define ft first
#define sc second
using namespace std;
struct interval{
int f,s;
};
int main(){
int t;
cin>>t;
while(t–){
int n,k;
cin>>n>>k;
string s;
cin>>s;
int i=0;
interval v[n];
int mx=INT_MIN;
while(i<n){
int j=i;
if(s[i]==‘0’){
i++;
if(mx<0)
mx=0;
continue;
}
while(j<n && s[j]==‘1’){
j++;
}
if(j-i >mx)
mx=j-i;
i=j;
}
i=0;
int p=0;
int ans=0;
while(i<n){
int j=i;
if(s[i]==‘0’){
i++;
if(mx==0){
ans=min(n,k);
break;
}
continue;
}
while(j<n && s[j]==‘1’){
j++;
}
if(j-i ==mx){
v[p].f=i;
v[p].s=j-1;
p++;
}
i=j;
}
i=0;
if(ans==0){
while(i<p){
int l=v[i].s-v[i].f+1;
if((v[i].s+k < n)||(v[i].f-k>=0)){
ans=l+k;
break;
}else{
int op1,op2;
if(v[i].s+k >= n){
op1=n-v[i].f;
}if(v[i].f-k<0){
op2=v[i].s+1;
}
if(ans<max(op1,op2)){
ans=max(op1,op2);
}
}
}
}
cout<<ans<<"\n";
}
return 0;
}
No,I think it is O(n) only
Why only one test case passes in this https://www.codechef.com/viewsolution/28054598
I used different approach.
No. It’s Linear.
can anyone please explain what is the problem in my code -
my code is understandable easily.
my code is -
#include <bits/stdc++.h>
using namespace std ;
int main()
{
int t ;
cin>>t ;
while(t--)
{
int n,k ;
cin>>n>>k ;
string str ;
cin>>str ;
if(n==k)
{
cout<<n<<"\n" ;
continue ;
}
vector<pair<int,int> > val ;
int kmax=0 ;
for(int i=0;i<n;i++)
{
if(str[i] == '1')
{
int j=i;
for(j=i;j<n;j++)
{
if(str[j] =='1')
kmax++ ;
else
{ break ;}
}
val.push_back(make_pair(kmax,i) ) ;
i=j-1 ;
kmax=0 ;
}
}
if(val.size()==0)
{
cout<<k<<"\n" ;
continue ;
}
sort(val.begin(),val.end()) ;
//cout<<"val is \n" ;
int ans=0,temp ;
//vector<int> ans ;
for(int i=val.size()-1;i>=0;i--)
{
if( val[i].second+1 > k )
{
temp = k + val[i].first ;
if(temp>ans)
{ans=temp;}
//break ;
}
else
{
if( val[i].first + val[i].second + k - 1 < n )
{
temp = val[i].first + k ;
if(temp>ans)
ans = temp ;
}
else
{
if(val[i].second + k < n)
{
temp = n - val[i].second ;
if(temp>ans)
ans=temp ;
}
}
}
}
cout<<ans<<"\n" ;
}
return 0 ;
}
Fails on the same testcase as this guy: Problem with code (PIBRO)
2 Likes
so I changed the code a little bit , having the data of the consecutive occurrences of 1’s in forward and backward direction, and then using it .
and here is my code but still I am getting W.A any guess why? please …
#include <bits/stdc++.h>
using namespace std ;
int main()
{
int t ;
cin>>t ;
while(t--)
{
int n,k ;
cin>>n>>k ;
string str ;
cin>>str ;
if(n<=k)
{
cout<<n<<"\n" ;
continue ;
}
vector<pair<int,int> > val ;
int forward[n] = {0},back[n]={0} ;
int kmax=0 ;
for(int i=0;i<n;i++)
{
if(str[i] == '1')
{
int j=i;
for(j=i;j<n;j++)
{
if(str[j] =='1')
{kmax++ ;forward[j]=kmax ;}
else
{ break ;}
}
val.push_back(make_pair(kmax,i) ) ;
i=j-1 ;
kmax=0 ;
}
}
kmax = 0 ;
for(int i=n-1;i>=0;i--)
{
if(str[i]=='1')
{
int j=i ;
for(j=i;j>=0;j--)
{
if(str[j]=='1')
{
kmax++ ;
back[j] = kmax ;
}
else
break ;
}
i=j+1 ;
kmax = 0 ;
}
}
if(val.size()==0)
{
cout<<k<<"\n" ;
continue ;
}
sort(val.begin(),val.end()) ;
//cout<<"val is \n" ;
int ans=0,temp ;
//vector<int> ans ;
for(int i=val.size()-1;i>=0;i--)
{
if( val[i].second+1 > k )
{
temp = k + val[i].first + forward[ val[i].second - k - 1 ] ;
if(temp>ans)
{ans=temp;}
//break ;
}
else
{
if( val[i].first + val[i].second + k - 1 < n )
{
temp = val[i].first + k + back[ val[i].second + k + 1 ] ;
if(temp>ans)
ans = temp ;
}
else
{
if(val[i].second + k < n)
{
temp = n - val[i].second ;
if(temp>ans)
ans=temp ;
}
if( val[i].second + val[i].first >= k )
{
temp = val[i].second + val[i].first + back[ val[i].second + val[i].first + 1 ] ;
if(temp>ans)
ans = temp ;
}
}
}
}
cout<<ans<<"\n" ;
}
return 0 ;
}
1
10 4
1111111111
every time I run it
2 Likes
thanks a lot , i figured out that trying to do it using only the consecutive occurrences of 1’s isn’t correct, as their can be cases of selecting the substring that may contain the consecutive 1’s in between , so have to traverse the entire array and check for the answer.
So , basically the answer to this question lies in the maximum length we can have between (i to i+k) + the forward 1’s till (i-1) + the backward 1’s till (i+k)
but still I am getting W.A why??
my code -
#include <bits/stdc++.h>
using namespace std ;
int main()
{
int t ;
cin>>t ;
while(t--)
{
int n,k ;
cin>>n>>k ;
string str ;
cin>>str ;
vector<pair<int,int> > val ;
int forward[n] = {0},back[n+2]={0} ;
int kmax=0 ;
if(n<=k)
{
cout<<n<<"\n" ;
break ;
}
for(int i=0;i<n;i++)
{
if(str[i] == '1')
{
int j=i;
for(j=i;j<n;j++)
{
if(str[j] =='1')
{kmax++ ;forward[j]=kmax ;}
else
{ break ;}
}
//val.push_back(make_pair(kmax,i) ) ;
i=j-1 ;
kmax=0 ;
}
}
kmax = 0 ;
for(int i=n-1;i>=0;i--)
{
if(str[i]=='1')
{
int j=i ;
for(j=i;j>=0;j--)
{
if(str[j]=='1')
{
kmax++ ;
back[j] = kmax ;
}
else
break ;
}
i=j+1 ;
kmax = 0 ;
}
}
/*cout<<"______the forward array is________\n" ;
for(auto f: forward)
cout<<f<<" " ;
cout<<"\n_____the backward array is_______\n" ;
for(auto f: back)
cout<<f<<" " ;
cout<<"\n" ;*/
int ans=0,temp ;
//vector<int> ans ;
ans = max(ans,k+back[k]) ;
for(int i=1;i+k<n;i++)
{
ans = max(ans, forward[i-1] + k + back[i+k] ) ;
}
//ans = max(ans,forward[n-1-k]+k) ;=
cout<<ans<<"\n" ;
}
return 0 ;
}
This solution fails for the following testcase:
1
5 4
10000
The answer should be 5:
Longest consecutive run of 1's = 5 obtained by changing the following marked range (of size 4) to all 1's:
10000
^ ^
Giving:
11111
just figured out how I used my last four hours but didn’t find out a bug that was in front of my eyes, all the time:rofl:
if(n<=k)
{
cout<<n<<"\n" ;
break ;
}
that break statement instead of continue .
1 Like
plz explain me the logic @kmaaszraa
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# Solitaire wins and losses
Let say a solitaire game can be solved 1 in 10 plays. (A play is defined as shuffling of cards and trying to solve the solitaire).
If after each successfully solved game we skip 8 games (we shuffle but don't try to solve it), what would be the average number of plays of non-skipped games per 1 solvable game and why?
• It should just be 10:1 right? Shuffling does not alter the probability of the next game being solvable if it did, then the number of times you shuffle the cards would matter. Jun 13, 2016 at 19:07
• OK, taking card game as an example probably is not the best choice, but the question could be "how the ratio would be changed if removing particular number of samples after next win sample?" Jun 13, 2016 at 19:20
• I still don't think there would be any difference. The probability distribution for the 1st sample after a win is exactly the same as the 9th sample after a win, and the 2nd sample has the same probability distribution as the 10th sample, and so on, and so forth. There is something kinda close to this which I think you may be getting at. The average run length between two Heads-Heads flips is longer than the average run length between two Heads-Tails flips. It would be wrong to interpret this as: "the probability of tails after heads is greater than the probability of heads after heads." Jun 13, 2016 at 19:36
• @TonyRuth Agreed. Jun 13, 2016 at 20:04
• Is the ratio (plays : solvable plays) or (plays : solvable shuffles)? Jun 13, 2016 at 20:39
Each game has two possible outcomes. We care about number of played games and number of winnable games (both played and unplayed).
Win
• $\frac{1}{10}$ chance
• Add 1 to played games
• Add 1 to solvable played games
• Add $\frac{8}{10}$ on average to solvable unplayed games
Loss
• $\frac{9}{10}$ chance
• Add 1 to played games
So on average once every 10 played games the number of unplayed solvable games rises by $\frac{8}{10}$ on average.
$10 : \frac{8}{10}$
We calculated ratio of played games to skipped solvable games. We have to add one to the right side to account for the one won played game.
$10 : \frac{8}{10} + 1$
$10 : \frac{18}{10}$
$100 : 18$
$50 : 9$
And that is the answer. Ratio of played games to winnable (both played and unplayed) is $50 : 9$. Average number of played games per solvable game would be $\frac{50}{9}$ or approximately $5.55$.
The average number of games played per solvable shuffle (given you always successfully solve if you play a solvable shuffle) would be
$\frac{50}9$
Because
If you shuffled for a play and it is not solvable you have made $1$ shuffle and it was unsolvable. (probability $\frac9{10}$: $0$ solvable shuffles)
Whereas if you shuffled for a play and it was solvable you have made a total of $9$ shuffles, the first of which was solvable and the other $8$, which you do not play, each have a probability of $\frac1{10}$ of being solvable. (probability $\frac1{10}$: $1+8\frac1{10}$ solvable shuffles)
The expected value of solvable shuffles per play is therefore $0\frac9{10}+(1+8\frac1{10})\frac1{10}=\frac{18}{100}=\frac9{50}$
The expected value of the number of plays per solvable shuffle is the inverse of this $=\frac{50}9$
In practice
It will be less if you decide ahead of time to play $n$ games and not to shuffle the other $8$ times if your last play is successful (why bother). In this case you expect $\frac9{50}+\frac1{10n}$ solvable shuffles per play and hence $\frac{50n}{9n+5}$ plays per solvable shuffle.
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## Lego Robot solves the Rubik’s Cube in 3 seconds!
Using a mobile phone to see the pieces, this Lego robot is able to beat all Rubik’s Cube records. In fact, we can check in the video that it solves the Rubik’s Cube in just 3.25 seconds.
Do you think that a human can beat that?
## Solve the Rubik’s Cube in 20 movements
One of the most famous questions related to Rubik’s Cube is: “How many movements are required to solve the Rubik’s Cube?” That is, what is the minimum number of moves I need to solve any Rubik’s cube position.
It may seem that this is an easy problem to solve using today’s computers. However, a Rubik’s Cube can be in so many different positions, that this problem can not be solved testing positions one by one.
## God number is 20
Because of the headaches that caused this issue, the number of moves needed to solve any Rubik’s Cube was called “god number”. In 1981 it was discovered that this number was bounded between 18 and 52. The margin of error was decreased in 1995, reducing the uncertainty between 20 and 29 movements.
Finally, in 2010, 29 years after the first estimation, it was found that the God number was exactly equal to 20. That is, any position of Rubik’s cube can be solved in 20 moves or less. To obtain this number were necessary Group Theory skills (remove symmetries and related positions), and the use of Google supercomputers.
## Can I solve the Rubik’s cube in 20 steps?
Because this information is not always given accurately, many people wonder if there is an algorithm they can use to solve the Rubik’s cube in 20 moves. Nope. Keep in mind that even if you can solve the Rubik’s cube in 20 moves, these moves will be completely each time. And because of the large number of possible positions, this knowledge can be useful to solve a Rubik’s cube with a computer, but not for a person.
So if you want to learn how to solve the Rubik’s Cube, you better start by our tutorial
| Categories: Uncategorized | 2 Comments
## You will figure out how to solve it
Just because you haven’t got it all figured out,
doesn’t mean you never will.
Some day you may even look back,
and wonder why you were ever worried.
| Categories: Uncategorized | 2 Comments
## Rubik’s Cube Solutions (without notation)
Years ago, most of the websites used strange notation to show how to solve a Rubik’s Cube. And, nowadays, a lot of the websites are still using them.
One of the worst things about notations is that you have to learn something before learning how to solve the Rubik’s Cube. As a consequence, many people think that solving the cube is something to difficult for them.
When I started to create this website, I wanted to show the simplest solutions to solve the Rubik’s Cube: no letters, no strange symbols, no notation at all. Only nice pictures.
Nowadays, there are more websites that also follows the ‘no-notaton’ philosophy. One of them is Rubik’s Solver, that offers a very simple guide to solve the Rubik’s Cube without notation.
And what do you prefer? Notation or no-notation solutions?
| Categories: Uncategorized | 1 Comment
## It took Erno Rubik one month to solve his own puzzle
Did you know that it took Erno Rubik, the creator of the Rubik’s Cube, more than one month to solve its own creation? In fact, we was not even sure if there was a method to solve it.
But for me, it was a code I myself had invented! Yet I could not read it.
It may seem surprising, but keep in mind that this website did not exist, hehe. Seriously, you have to put yourself in his shoes. Once he creates the Cube, he spends every day trying to solve it. No one had ever solved it. In fact, no one had attempted to solve it. Day after day, without achieving any progress most the days, must be exhausting.
Is like when you decide to go home after a nice walk in which you have seen many lovely sights. I decided it was time to go home, let us put the cubes back in order. And it was at that moment that I came face to face with the Big Challenge: What is the way home?
| Categories: Uncategorized | 1 Comment
## Solving a 11×11 Rubik’s Cube
Do you like big cubes? 4×4? 5×5? Let’s try 11×11 Rubik’s Cube!
| Categories: Uncategorized | 1 Comment
## Life is like a Rubik’s Cube
The Cube is an imitation of life itself – or even an improvement on life. The problems of puzzles are very near the problems of life, our whole life is solving puzzles. If you are hungry, you have to find something to eat. But everyday problems are very mixed – they’re not clear. The Cube’s problem depends just on you. You can solve it independently. But to find happiness in life, you’re not independent. That’s the only big difference.
Erno Rubik, Rubik’s Cube Creator
## The Cube and I (Rubik’s Cube Poem)
Last week I bought home a rubiks cube
To test my skill against its might
Well I said this silly toy
This plastic puzzle kids enjoy
Ill solve this night!
The night fled by the dawn arrived
The cube unsolved in disarray
Against each turn and twist
Gave me an aching wrist
As I worked through the day
At last I lined up one side orange
At first this seemed a great success
But it bought me little cheer
But the other five remained I fear
A multicolored mess
Two days flew past then three and four
My work neglected went undone
Though green lined up columns true
How come red was mixed in with the blue?
Does rubik call this fun?
A thousand times I turned the cube
My fingers gripped around its parts
Until the truth at last set in
I didn’t lack the will to win
I simply lacked the smarts
No more I said to rubiks cube
Shall you enslave me in your power?
Thus liberated from its hold
Gave it to a eight year old
Who solved it in a hour
Jarrod Wakefield
| Categories: Uncategorized | 2 Comments
## Interview to Patrick Bossert about Rubik’s Cube
Patrick Bossert was just 12 years old when he published the book You Can Do the Cube with the Rubik’s Cube solutions. It sold over 1.5 million copies and became the global best-seller of 1981.
In this nice video, he tells how a 12 years old child became a best-seller writer.
I’m sorry for that strange intro
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# Probability of server cluster failure
I have a problem where I'm trying to derive an expression for the availability of a server cluster.
Suppose I have 100 servers, where each one of them individually has a probability of failure 0.05. Now, I'm trying to find an expression for the probability that at most 3 of the servers fail.
Currently I have reasoned that:
$Pr$(at most 3 of the servers fail) = $Pr$(0, 1, 2, or 3 of the servers fail) = $Pr$(0 fail) + $Pr$(1 fails) + $Pr$(2 fail) + $Pr$(3 fail)
$Pr$(0 servers fail) = $0.95^{100} = 0.00592$
$Pr$(1 server fails) = $0.95^{99} \times 0.05^1 = 0.05623$
$Pr$(2 server fails) = $0.95^{98} \times 0.05^2 = 0.00906$
$Pr$(3 server fails) = $0.95^{97} \times 0.05^3 = 0.00702$
Thus, $Pr$(at most 3 of the servers fail) = $0.00592 + 0.05623 + 0.00906 + 0.00702 = 0.07823$.
Is that correct? That probability of 7.8% seems like a really small number for having at most 3 servers fail.
• You're forgetting some binomial coefficients. Oct 14, 2014 at 23:07
• Yes, you are very close., however you need to ask the question for each of $\textit{which}$ server(s) failed. Oct 14, 2014 at 23:09
When calculating the probabilities you should also take into account which of the servers will fail. Thus, the correct probabilities are
$\Pr$(0 servers fail) = $0.95^{100} = 0.00592$, (correct since $\binom{100}{0}=1$)
$\Pr$(1 server fails) = $\binom{100}{1}\times 0.95^{99} \times 0.05^1 = 0.03116$
$\Pr$(2 server fails) = $\binom{100}{2}\times 0.95^{98} \times 0.05^2 = 0.08118$
$\Pr$(3 server fails) = $\binom{100}{3}\times 0.95^{97} \times 0.05^3 = 0.13958$
• Thanks. So when I was earlier computing that $Pr$(2 server fails) = $0.95^{98}×0.05^2$, does that imply that I was computing the probability that the first 98 servers were good and the last 2 servers failed? Oct 14, 2014 at 23:21
• Yes, exactly. Actually you were computing the probability that two specific servers failed (f.e. as you say the last 2, or f.e. the servers in places 54 and 87 etc.) But you need to calculate the probability for any two. Oct 14, 2014 at 23:24
• Thank you. Now that I look further, I find it kind of extraordinary that $Pr$(0 servers fail) = $0.00592$. That seems really low when each of the servers seems pretty stable with each having probability $0.95$ of failing. But the math proves it, I guess. Oct 14, 2014 at 23:30
• 0.95 of not failing you mean. Yes, the point you mention is exactly the problem in inferential statistics, in case you want to make many simultaneous inferences. If you are "only" 95 percent sure that each of 100 statements is correct then there is 1-0.00592=0.99408 probability that at least one statement is wrong! Oct 14, 2014 at 23:34
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# Adding and Subtracting Rational Expressions where One Denominator is the LCD
## Combine fractions including variables, one-step LCM
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Progress
Practice Adding and Subtracting Rational Expressions where One Denominator is the LCD
Progress
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Adding and Subtracting Rational Expressions where One Denominator is the LCD
The length of a garden plot is $\frac{6x^2-5}{2x^2 + 4x - 6}$ . The width of the plot is $\frac{2x-7}{x+3}$ . How much longer is the garden plot than it is wide?
### Guidance
Recall when two fractions do not have the same denominator. You have to multiply one or both fractions by a number to create equivalent fractions in order to combine them.
$\frac{1}{2} + \frac{3}{4}$
Here, 2 goes into 4 twice. So, we will multiply the first fraction by ${\color{red}\frac{2}{2}}$ to get a denominator of 4. Then, the two fractions can be added.
${\color{red}\frac{2}{2} \cdot} \frac{1}{2} + \frac{3}{4} = \frac{2}{4} + \frac{3}{4} = \frac{5}{4}$
Once the denominators are the same, the fractions can be combined. We will apply this idea to rational expressions in order to add or subtract ones without like denominators.
#### Example A
Subtract $\frac{3x-5}{2x+8} - \frac{x^2-6}{x+4}$ .
Solution: Factoring the denominator of the first fraction, we have $2(x+4)$ . The second fraction needs to be multiplied by ${\color{red}\frac{2}{2}}$ in order to make the denominators the same.
$\frac{3x-5}{2x+8} - \frac{x^2-6}{x+4} &= \frac{3x-5}{2(x+4)} - \frac{x^2-6}{x+4} \cdot {\color{red}\frac{2}{2}} \\&= \frac{3x-5}{2(x+4)} - \frac{2x^2-12}{2(x+4)}$
Now that the denominators are the same, subtract the second rational expression just like in the previous concept.
$&= \frac{3x-5-(2x^2-12)}{2(x+4)} \\&= \frac{3x-5-2x^2+12}{2(x+4)} \\&= \frac{-2x^2+3x+7}{2(x+4)}$
The numerator is not factorable, so we are done.
#### Example B
Add $\frac{2x-3}{x+5} + \frac{x^2+1}{x^2-2x-35}$
Solution: Factoring the second denominator, we have $x^2-2x-35=(x+5)(x-7)$ . So, we need to multiply the first fraction by ${\color{red}\frac{x-7}{x-7}}$ .
$\overbrace{{\color{red}\frac{(x-7)}{(x-7)}} \cdot \frac{(2x-3)}{(x+5)}}^{FOIL} + \frac{x^2+1}{(x-7)(x+5)} &= \frac{2x^2-17x+21}{(x-7)(x+5)} + \frac{x^2+1}{(x-7)(x+5)} \\&= \frac{3x^2-17x+22}{(x-7)(x+5)}$
#### Example C
Subtract $\frac{7x+2}{2x^2+18x+40} - \frac{6}{x+5}$ .
Solution: Factoring the first denominator, we have $2x^2+18x+40=2(x^2+9x+20)=2(x+4)(x+5)$ . This is the Lowest Common Denominator, or LCD. The second fraction needs the 2 and the $(x+4)$ .
$\frac{7x+2}{2x^2+18x+40} - \frac{6-x}{x+5} &= \frac{7x+2}{2(x+5)(x+4)} - \frac{6-x}{x+5}{\color{red}\cdot \frac{2(x+4)}{2(x+4)}} \\&= \frac{7x+2}{2(x+5)(x+4)} - \frac{2(6-x)(x+4)}{2(x+5)(x+4)} \\&= \frac{7x+2}{2(x+5)(x+4)} - \frac{48+4x-2x^2}{2(x+5)(x+4)} \\&= \frac{7x+2-(48+4x-2x^2)}{2(x+5)(x+4)} \\&= \frac{7x+2-48-4x+2x^2}{2(x+5)(x+4)} \\&= \frac{2x^2+3x-46}{2(x+5)(x+4)}$
Intro Problem Revisit We need to subtract the width from the length.
$\frac{6x^2-5}{2x^2 + 4x - 6} - \frac{2x-7}{x+3}$
Factoring the first denominator, we have $2x^2+4x-6=(2x-2)(x+3)$ . So, we need to multiply the second fraction by ${\color{red}\frac{2x-2}{2x-2}}$ .
$\frac{6x^2-5}{2x^2 + 4x - 6} -\overbrace{{\color{red}\frac{(2x-2)}{(2x-2)}} \cdot \frac{(2x-7)}{(x+3)}}^{FOIL} &= \frac{4x^2-18x+14}{2x^2+4x-6} \\\frac{6x^2-5}{2x^2 + 4x - 6} - \frac{4x^2-18x+14}{2x^2+4x-6}\\\frac{6x^2-5-(4x^2-18x+14)}{2x^2+4x-6}\\\frac{6x^2-5-4x^2+18x-14}{2x^2+4x-6}\\\frac{2x^2+18x-19}{2x^2+4x-6}$
Therefore, the garden plot is $\frac{2x^2+18x-19}{2x^2+4x-6}$ longer than it is wide.
### Guided Practice
Perform the indicated operation.
1. $\frac{2}{x+1} - \frac{x}{3x+3}$
2. $\frac{x-10}{x^2+4x-24} + \frac{x+3}{x+6}$
3. $\frac{3x^2-5}{3x^2-12} + \frac{x+8}{3x+6}$
#### Answers
1. The LCD is $3x+3$ or $3(x+1)$ . Multiply the first fraction by $\frac{3}{3}$ .
$\frac{2}{x+1} - \frac{x}{3x+3} &= \frac{3}{3} \cdot \frac{2}{x+1} - \frac{x}{3(x+1)} \\&= \frac{6}{3(x+1)} - \frac{x}{3(x+1)} \\&= \frac{6-x}{3(x+1)}$
2. Here, the LCD $x^2+4x-24$ or $(x+6)(x-4)$ . Multiply the second fraction by $\frac{x-4}{x-4}$ .
$\frac{x-10}{x^2+4x-24} + \frac{x+3}{x+6} &= \frac{x-10}{(x+6)(x-4)} + \frac{x+3}{x+6} \cdot \frac{x-4}{x-4} \\&= \frac{x-10}{(x+6)(x-4)} + \frac{x^2-x-12}{(x+6)(x-4)} \\&= \frac{x-10+x^2-x-12}{(x+6)(x-4)} \\&= \frac{x^2-22}{(x+6)(x-4)}$
3. The LCD is $3x^2-12=3(x-2)(x+2)$ . The second fraction’s denominator factors to be $3x+6=3(x+2)$ , so it needs to be multiplied by $\frac{x-2}{x-2}$ .
$\frac{3x^2-5}{3x^2-12} + \frac{x+8}{3x+6} &= \frac{3x^2-5}{3(x-2)(x+2)} + \frac{x+8}{3(x+2)} \cdot \frac{x-2}{x-2} \\&= \frac{3x^2-5}{3(x-2)(x+2)} + \frac{x^2+6x-16}{3(x-2)(x+2)} \\&= \frac{3x^2-5+x^2+6x-16}{3(x-2)(x+2)} \\&= \frac{4x^2+6x-21}{3(x-2)(x+2)}$
### Practice
Find the LCD.
1. $x, \ 6x$
2. $x, \ x+1$
3. $x+2, \ x-4$
4. $x, \ x-1, \ x^2 - 1$
Perform the indicated operations.
1. $\frac{3}{x} - \frac{5}{4x}$
2. $\frac{x+2}{x+3} + \frac{x-1}{x^2+3x}$
3. $\frac{x}{x-7} - \frac{2x+7}{3x-21}$
4. $\frac{x^2+3x-10}{x^2-4} - \frac{x}{x+2}$
5. $\frac{5x+14}{2x^2-7x-15} - \frac{3}{x-5}$
6. $\frac{x-3}{3x^2+x-10} + \frac{3}{x+2}$
7. $\frac{x+1}{6x+2} + \frac{x^2-7x}{12x^2-14x-6}$
8. $\frac{-3x^2-10x+15}{10x^2-x-3} + \frac{x+4}{2x+1}$
9. $\frac{8}{2x-5} - \frac{x+5}{2x^2+x-15}$
10. $\frac{2}{x+2} + \frac{3x+16}{x^2-x-6} - \frac{2}{x-3}$
11. $\frac{6x^2+4x+8}{x^3+3x^2-x-3} + \frac{x-4}{x^2-1} - \frac{3x}{x^2+2x-3}$
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# What is lim x --> 1 [( 1- sqrt x)/ (1- x)]
justaguide | Certified Educator
calendarEducator since 2010
starTop subjects are Math, Science, and Business
Now we see that ( 1- sqrt x)/ (1- x)
=> (1 – sqrt x)/(1- sqrt x)*(1+ sqrt x)
=> 1/ (1+ sqrt x)
For lim x --> 1 [( 1- sqrt x)/ (1- x)]
we have lim x --> 1 [1/ ( 1+ sqrt x)]
=> 1 / (1 + sqrt 1)
=> 1/ (1+1)
=> 1/2
Therefore lim x --> 1 [( 1- sqrt x)/ (1- x)] = 1/2
check Approved by eNotes Editorial
giorgiana1976 | Student
We calculate the limit substituting x by 1
lim [( 1- sqrt x)/ (1- x)] = (1-1)/(1-1) = 0/0
We notice that we've get an indetermination case
We'll solve using L'Hospital rule:
lim [( 1- sqrt x)/ (1- x)] = lim ( 1- sqrt x)'/ (1- x)'
lim ( 1- sqrt x)'/ (1- x)' = lim (-1/2sqrtx)/-1
lim [( 1- sqrt x)/ (1- x)] = lim sqrt x/2
lim [( 1- sqrt x)/ (1- x)] = sqrt 1/2
lim [( 1- sqrt x)/ (1- x)] = 1/2
check Approved by eNotes Editorial
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97下複變期中考 電
# 97下複變期ä¸è€ƒ é›»
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Complex Analysis: Midterm Examination 10:20 AM - 12:00 PM, April 14, 2009. [1] (10 %) Find all values of ± - 8 - i 8 3 ² 1 / 4 in the form of a + ib . [2] (10 %) Prove f ( z )= e y e ix is nowhere analytic, where z = x + iy . [3] (15 %) True or false (If it is false, explain brie±y why it isn’t true) (a) (5 %) If f ( z ) is analytic on a closed contour C , then ³ C ± f ( z ) dz =0. (b) (5 %) If f ( z ) is differentiable at a point z 0 and at every point in some neighborhood of z 0 , then f ( z ) is an entire function. (c) (5 %) S = { z | Re( z ) ² =3 } is a domain ( open connected set ). [4] (15 %) Verify that u ( x, y )= e x ( x cos y - y sin y ) is harmonic. Find v ( x, y ), the conjugate harmonic function of u ( x, y ). [5] (10 %) Evaluate ³ C 1 z dz in the form of a + ib , where C is the arc of the circle z =4 e it with - π/ 2 t π/ 2. [6] (10 %) Evaluate ³ C ± ´ 3 z +2 - 1 z - 2 i µ dz , where C is the circle | z | =5. [7] (10 %) Expand f ( z )= 1+
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Unformatted text preview: z 1-z in the Taylor series centered at z = i , and give the radius of convergence of this series. [8] (10 %) Let f ( z ) = u ( x, y ) + iv ( x, y ) where the ²rst partial derivatives of u ( x, y ) and v ( x, y ) are continuous. Prove that f ( z ) is analytic at z if and only if ∂u ∂x = ∂v ∂y and ∂u ∂y =-∂v ∂x [9] (10 %) Assume f ( z ) is analytic in a domain D , and C is a closed contour lying entirely in D . Use the fact that f ± ( z ) = 1 2 πi ³ C ± f ( z ) ( z-z ) 2 dz , with z within C , to prove f ±± ( z ) = 2! 2 πi ³ C ± f ( z ) ( z-z ) 3 dz . Hint: f ±± ( z ) = lim Δ z → f ± ( z + Δ z )-f ± ( z ) Δ z . 1...
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Math = Love: Make It Even Brain Teaser
## Tuesday, September 29, 2015
### Make It Even Brain Teaser
Last Friday, I pulled a new-to-me brain teaser out for Figure It Out! Friday.
I found a puzzle called "Make It Even" in The Moscow Puzzles: 359 Mathematical Recreations by Boris A. Kordemsky.
The book is published by Dover, so it's super affordable at \$3.99! I've found numerous puzzles I want to try out with my students. This is the second week in a row that our figure it out puzzle has come from this book!
The original puzzle called for 16 objects. I decided to use bingo chips since I have a giant bag.
To save time and not have to bag up 16 bingo chips per student, I took a package of 5 plastic bowls I bought at Dollar Tree for \$1 and divided the bingo chips into these bowls. The bowls are cheap feeling, but at \$0.20/piece they are amazing!
Here they are, stacked and ready to go.
My Algebra 2 students and Stats students were much better at reading the instructions and attempting the puzzle on their own.
My Algebra 1 students struggled even when I read the instructions to them.
There were a ton of "This is too hard" or "This is impossible" comments. In each class, at least one student figured it out though!
I can't tell you how many times a student would raise their hand to have me check their solution, and I would walk over to find that they had a column or row with one in it. I think they were just too focused on making the rows even or making the columns even to even look at it the other way.
Part of me now wonders if students would perform differently on this task if it was presented differently. Maybe like this?
Some of my students tried rearranging the 10 leftover chips into 2 rows of 5. I think this second phrasing of the problem would prevent that. If I did the second version, I'd definitely give it to my kids inside a dry erase sleeve.
Or, maybe combine the two methods and give students the grid and only 10 chips?
1. This looks like a great resource!
1. I've used several ideas out of it!
2. I was just thinking to myself...what should I do for the Friday warmup? And I have that book :) I kind of like the idea of giving some groups chips & other groups the board in dry erase sleeves and then discussing which strategy was more effective or which one they liked more.
1. If you try it, you'll have to let us know how it goes!
3. Hey Sarah: I tried this activity with my Grade 6 math students who are withdrawn for math from their regular program as an introduction to arrays. First of all, they loved it. And second of all, they were thrilled when they figured it out! It required some grit and perseverance, but the smiles on their faces when they were successful was the highlight of my year so far. Thanks for sharing.
4. I can't figure it out! Can you stack the chips?
1. Nope. No stacking :) I promise it's possible.
5. Just to clarify. Is it even as in the same.. or even as in not odd?
6. Yes it is possible (y)
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Finding derivative of Inverse trigonometric functions
Chapter 5 Class 12 Continuity and Differentiability
Concept wise
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### Transcript
Misc 5 Differentiate π€.π.π‘. π₯ the function, (γπππ γ^(β1 ) π₯/2)/β(2π₯+7 ) , β 2 < π₯ < 2 Let π¦= (γπππ γ^(β1 ) π₯/2)/β(2π₯+7 ) Differentiating both sides π€.π.π‘. π₯ ππ¦/ππ₯ = π/ππ₯ ((γπππ γ^(β1 ) π₯/2)/β(2π₯+7 )) Using Quotient rule As (π’/π£)^β² = (π’^β² π£ β π£^β² π’)/π£^2 where u = cosβ1 π₯/2 & v = β(2π₯+7) ππ¦/ππ₯ = (π(γπππ γ^(β1 ) π₯/2)/ππ₯ . β(2π₯ + 7 ) β π(β(2π₯ + 7 ))/ππ₯ . γπππ γ^(β1 ) π₯/2 )/(β(2π₯ + 7 ))^2 ππ¦/ππ₯ = ((β1)/β(1 β(π₯/2)^2 ) . π(π₯/2)/ππ₯ β(2π₯ + 7 ) β 1/(2β(2π₯ + 7)) . π(2π₯ + 7)/ππ₯ . γπππ γ^(β1 ) π₯/2 )/(β(2π₯ + 7 ))^2 ππ¦/ππ₯ = ((β1)/β(γ(4 β π₯)/4γ^2 ) . 1/2 β(2π₯ + 7 ) β 1/(2β(2π₯ + 7)) . (2 + 0) . γπππ γ^(β1 ) π₯/2 )/((2π₯ + 7) ) ππ¦/ππ₯ = ((β2)/β(γ4 β π₯γ^2 ) . 1/2 β(2π₯ + 7 ) β 1/(2β(2π₯ + 7)) . 2 . γπππ γ^(β1 ) π₯/2 )/((2π₯ + 7) ) ππ¦/ππ₯ = ((β β(ππ + π ))/( β(γπ β πγ^π )) β (. γπππγ^(βπ ) π/π)/β(ππ + π) )/((2π₯ + 7) ) ππ¦/ππ₯ = (β β(2π₯ + 7 ) ( β(2π₯ + 7 )) β γπππ γ^(β1 ) π₯/2 β(γ4 β π₯γ^2 ))/((2π₯ + 7) (β(2π₯ + 7 )) (β(γ4 β π₯γ^2 )) ) ππ¦/ππ₯ = γβ(β(2π₯ + 7 ))γ^2/((2π₯ + 7) ( β(2π₯ + 7 )) (β(γ4 β π₯γ^2 )) ) "+ " (γπππ γ^(β1 ) π₯/2 β(γ4 β π₯γ^2 ))/((2π₯ + 7) ( β(2π₯ + 7 )) (β(γ4 β π₯γ^2 )) ) ππ¦/ππ₯ = (β(2π₯ + 7))/((2π₯ + 7) β(2π₯ + 7 ) β(γ4 β π₯γ^2 )) "+ " (γπππ γ^(β1 ) π₯/2 )/((2π₯ + 7) β(2π₯ + 7 ) ) ππ¦/ππ₯ = (β1)/(β(2π₯ + 7 ) β(γ4 β π₯γ^2 )) "+ " (γπππ γ^(β1 ) π₯/2 )/((2π₯ + 7) (2π₯ + 7)^(1/2) ) π
π/π
π = (βπ)/(β(ππ + π) β(γπ β πγ^π ))β(γπππγ^(βπ ) π/π )/((ππ + π)^(π/π) )
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Understanding the Vocabulary of Algebra How to Solve Algebra Problems with Grouping Symbols Recognizing Operational Symbols in Algebra
# Recognizing Relationship Symbols in Algebra
Algebraic relationship symbols show how numbers or terms of an equation relate to each other. The relationship symbols show if one value is larger, smaller, equal, or approximately equal to another value. These algebra symbols are straightforward — when you know what the symbols mean, algebra problems are much less intimidating.
• Equal (=) symbol: This symbol means that the first value is equal to, or the same as, the value that follows. For example:
• Not Equal (≠) symbol: This symbol means that the first value is not equal to the value that follows. For example:
• Approximately Equal (≈) symbol: This symbol means that one value is approximately the same, or about the same, as the value that follows; this is used when rounding numbers. For example:
• Less than or Equal to (≤) symbol: This symbol means that the value preceding the symbol is less than or equal to the value that follows. For example:
• Less than (<) symbol: This symbol means that the value preceding the symbol is less than the value that follows. For example:
• Greater than or Equal to (≥) symbol: This symbol means that the first value is greater than or equal to the value that follows. For example:
• Greater than (>) symbol: This symbol means that the first value is greater than the value that follows. For example:
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## Introduction to Chain Codes
##### Lionel (Lon) Shapiro: AuthorDavid L. Anderson: Author
Is it possible to design a computer system that can "perceive" objects in the world? If it is possible, one important ability that the machine must have is to determine the shape of an object and to determine when two objects have the same shape. How might a computer program accomplish this task? We are going to explore one type of program that can accomplish this task.
Imagine someone shows you a simple object. Later they ask you to pick out that object from a "line-up" of other objects. One strategy you might adopt would be to focus on remembering one or more salient attributes that you can pick out from viewing the object. These might include: shape, color, texture, etc.
Examining this figure we find an easily identifiable shape. That is, we can name this shape in a way that would convey meaning to others -- 'triangle.' By doing this, we are using a model (for the shape of triangles) that we can easily refer to with a word in our language.
When you learned about this shape early in your development, you may have been told that this shape has "three sides." Here our model is defined in terms of boundries between the object and its environment (in this case, the background). This model of the object can then be used for reconstruction (draw this object) or recognition (identify this shape).
Following this approach, we can construct a computer program to reconstruct or recognize shapes. In order to do so, we need to have a specific representation (encoding scheme) for our shape. For simple objects, whose construction and identification are based on boundries, we can use a chain code representation.
## Absolute chain codes
When drawing a triangle, you might start at one vertex (corner) and proceed to draw three straight, connected lines, such that you return to the starting vertex with the third line drawn. If you were to try to communicate what you were doing to someone unable to see your pad, but who is able to draw on their own pad, you might say something like, "starting with the pencil at one point on the pad, move in a straight line toward the top-center of the pad for about one inch, then continue toward the lower-right corner for another inch without lifting the pencil. etc." These instructions provide the listener with a loosely defined rule that can be used to reconstruct your shape.
Instructions like this, however, are not very precise. If our goal is to be able to draw a wide range of different shapes and if we want to be able to draw them on any piece of paper, then we need to be more systematic in the directions that we give. The directions ought to be clear and unambiguous and they must offer an adequate range of alternatives -- as many directions as we are likely to require to draw the shapes that interest us.
Conveniently, we are already familiar with a system of directions that begins to fit the previous description: The points of a compass. Here we have a range of eight different directions that we can use to tell people how they are to move. Say we want to communicate the shape, size and orientation of a two square mile section of a city that will be receiving a new water system. First, we pick a starting point and then we tell them were to go to traverse the perimeter of this designated section. The directions might go something like this:
Start at the corner of 5th and Main. Go west on 5th St. for 6 blocks. Go north on Maple St for 12 blocks. Go east on 17th St for 6 blocks. And go south on Main St for 12 blocks, which brings you back to the corner of 5th and Main.
The points of the compass give us an absolute set of reference points, because each direction is fixed by the Earth's magnetic fields. Thus, the "North" direction always points to a specific point in the frame of reference and is independent of which way we might be facing. Chain codes that are created using absolute points of direction are called absolute chain codes.
Absolute chain codes do not need to be fixed to anything as large as the whole earth, however. We can also fix the direction points to something smaller (like a piece of paper for example, or your computer monitor). If we do that, it is natural to make the top of the paper (or monitor) "N", the bottom "S", and so on. Once we've designated the offical "North" end of the paper, then it doesn't matter which direction we are facing as we look at the paper. "N" will always be in a fixed (absolute) direction in relation to the dimensions of the paper.
If the shape we are encoding is a large one, then a city block might be the appropriate unit of measure. But if we are to encode a smaller shape, like the triangle at the top of this page, we must use a smaller unit of measure. Let's use one inch as our unit of measure and the points of the compass as our directions. "North" will be fixed at the top of the paper / monitor. The size and shape of the triangle, then will be encoded with a string of directional letters. To interpret such a string, we will follow the following rule:
For each directional letter(s) that is written down (N, SE, W, etc.), move the pencil in that direction for a distance of one inch*. If the pencil is to be moved in the same direction for more than one inch (let's say, 3 inches), then the directional letter must be written more than one time (write: "w, w, w,"). Directional letters can either be written in a vertical column or on a horizontal line with commas separating the letter(s).
*[Note: For diagonal directions, the distance will be slightly longer than one inch -- for reasons that will be explained below.]
Using this convention, we can then represent the size and shape of an object as a chain code -- that is, a set of directional codes, with one code following another like links in a chain. Let's return to our triangle. The vertices (or corners) we will label v1 - v3. To begin we determine a starting point. Let's start at v1, going in a clockwise direction. First we move one inch in a NE direction, moving from v1 to v2. Then we move SE one inch to v3. Finally, we return to v1 by moving W one inch. (On many monitors the sides of the triangle will look longer than one inch.)
The chain code that we just produced is an algorithm that gives a symbolic representation of the triangle's shape, size and orientation. This information also provides precise rules for reconstructing that figure in a drawing. We will typically write the chaincode in one of either two ways. Either as
NE SE W or as NE, SE, W
Now it is your turn to create a chain code for the figure below. Starting in the lower right hand corner, and moving clockwise, write the chain code for this shape on a piece of scratch paper.
After you've written down the complete chain code, click the button that says "Completed Chain Code" below, and see if you got it right.
### Using numbers in writing chain codes
Since chain codes provide an algorithm for representing the size and shape of an object, it is possible to build a machine that is capable of encoding that information. But algorithms cannot be implemented in a computer program using just any coding system. It must be a language that the computer can recognize. Since digital computers typically use a numerical system for storing information (everything is coded in "O's" and "1's") it will be a step in the right direction to modify our directional system to a numerical system rather than a letter system. Computer programmers who write chain codes based on an eight-way directional system, typically represent those eight directions using the numbering scheme at the right.
From now on, we will use this numbering system for writing absolute chain codes. The "North" (or the "top" of the page or monitor) will be represented by the number "2", "South" is "6", "West" is "4", "East" is "0", and so on.
It is now time to use an interactive java program where you will get to work with chaincodes. When you go to the Java program, you will find that the objects are all located on a grid. This is to make it as convenient as possible to apply the units of measure to the objects. With the eight-direction system that we are using, a chain code can capture the shape of any object that can be drawn by connecting together intersection points on the grid. Each move will take us from one intersection point on the grid, to another. Each horizontial and vertical move is of 1 unit length and each diagional move is slightly bigger than 1 unit in length (its length is the square root of 2, to be precise).
You should now be familiar with absolute chain codes. But there is another type of chain code that also has important applications. We turn to it next.
## Relative chain codes
There will be times when absolute chain codes won't do the job. Either, because no absolute point of reference is available or because absolute chain codes are not capable of doing the job that needs to be done. In such cases, using a relative perspective might be preferable.
Suppose, for example, that you had no working compass or other system for naming directions in a global frame of reference. This is easily imagined by closing your eyes and listening for a specific sound (say footsteps of an approaching friend). You may be inclined to describe the direction with respect to your body, that is, you may hear your friend approaching from your right. Note, this system of direction naming is based on positions relative to your current orientation. If you were to do a quick "about-face," your friend would appear to be approaching from your left in this example.
It is possible to base an eight-directional coding system on relative directions. Each of the eight directions will be defined in relation to a moving perspective. Think of it this way. Imagine that you are driving a car around the perimeter of the object, marking a line behind you as you go. From any given point you have eight options. You can continue to go forward ("F") in the same direction that you have been going, you can go backward ("B") in the opposite direction, you can turn to the left ("L") or to the right ("R"). These are the four main points of the compass. The other four directions are derived from the four basic ones, as can be seen in the compass to the right.
Now take a try at producing a relative chain code for the same figure that you worked with above. Remember this time, the "directional compass" does not stay fixed in one position. The "F" does not always point "up". The "F" points "forward" in whatever direction YOU are moving as you mentally traverse around the perimeter of the object. Starting in the lower right hand corner, and moving clockwise, write the relative chain code for this shape on a piece of scratch paper. Keep in mind: The directional code representing any particular section of line is relative to (and thus dependent upon) the directional code of the preceding line segment.
Write down the relative chain code for this figure on a piece of scratch paper. When you've finished, click the button that says "See Completed Chain Code" below, and see if you got it right.
Did you get it right? It is common for people to get the very first letter wrong (even if they get all the others right). There is a reason for this. When you begin making the chain code, you locate yourself (mentally) in the lower righthand corner and you tend to orient yourself in the direction that the first line segment is moving. So it feels natural to many people to put an "F" for the first line segment. However natural that might be, you aren't allowed to do it that way. You don't get to choose your orientation. Remember the hint above:
The directional code representing any particular section of line is relative to (and thus dependent upon) the directional code of the preceding line segment.
What this means is that relative chain codes have one rather tricky element. The directional code for the first line segment cannot be determined until you know what the directional code is for the last line segment. That means that you either have to look back to the preceding line or you have to leave the space for the first code blank and wait until you have finished the rest of code and then fill in the direction for the first line segment. Given the direction that the last line segment was moving, a right turn ("R") was required to place the first line segment moving in the proper direction.
Did this all make sense? If not, try to do the chain code again. When you've come full circle and return to the "start" button, don't stop there. Make one more move as you turn the corner. That move around the corner (a move to the "right") is the first directional code for the figure. Since the direction of any line segment is relative to the orientation of the line preceding it, we must consult the preceding line segment.
Now let's try a relative chain code for the triangle.
Write the down the relative chain code for this figure on a piece of scratch paper. When you've finished, click the button that says "See Completed Chain Code" (below) and see if you got it right.
Just as we did with absolute chain codes, we can modify the directional system that we use with relative chain codes, using a numerical system rather than a letter system. Computer programmers who write relative chain codes based on an eight-way directional system, typically represent those eight directions using the numbering scheme at the right. The "forward" direction that was represented with an "F" is now represented with the number "2," "Backwards" is "6," "Left" is "4," "Right" is "0," and so on.
We now have two different methods for recording the shape of an object, one using an absolute point of reference, the other using relative reference points. The next step is to consider how a computer program might encode this information and make use of it to "recognize" objects.
## Machines that do chain codes
One of the topics that we've been exploring is that of "machine vision." How might a machine be made to process visual information about the world? Given our present discussion of chain codes, we might ask: What kinds of visual information could a machine acquire using chain codes? As you may have already deduced, each type of chain code has its own advantages.
One of the strengths of the absolute chain code is that it carries information not only about the size and shape of the object but also about its rotation. Suppose we want to encode the information about the size and shape of the triangle, but also about the position of the figure on the page. An absolute chain code can distinguish between the triangle on the left (below) and the one on the right, that has been rotated 180 degrees.
The absolute chain code for the triangle on the left is: 4,1,7. For the one on the right, it is: 0, 5, 3. They don't share even one number in common. This is because the rotation of the triangle as well as the size and the shape are being represented in the code. So, if we want to build a machine that can preserve the orientation of an object (its rotation) as well as its shape and size, then the machine would need to represent the object with an absolute chain code.
Relative chain codes do not behave in the same way. The relative code for the first triangle is 7, 7, 0. The code for the triangle on the right will also be 7,7,0 if the starting point remains the same (as shown by the words "start" above). Things only change slightly if we move the starting point to the right-angled corner of the triangle. The code then becomes: 0, 7, 7 regardless of whether the base of the triangle is pointed up or down.
The similarity in these chain codes has a beneficial feature, one that our machine can take advantage of. Since the numbers that are used are the same, a computer can be programmed to compare the chain codes of two different objects and determine if they are the same size and shape. Here is how it works.
## Comparing chain codes
Since the starting point for a relative chain code is completely arbitrary, we can shift the numbers around to reflect different possible starting positions. Now, we must be careful not to change the serial order of the numbers, because that is an essential part of the way the shape and size of the object is represented. One thing we can change, however, is which number we start with as we generate the series. So, let's say that our machine is trying to determine if the following two chain codes represent objects that are the same size and shape. Let's take the two relative chain codes that we created above. Here are the triangles and the chain codes:
Triangle #1 Chain Code #1 Chain Code #2 Triangle #2 7 0 7 7 0 7
If our computer program compares the numbers in these two lists, it will not produce a match. The top and bottom numbers don't match.
However, the fact that they don't match up could be the result of nothing more than the codes having been started at different vertices of the triangle. So, the only way to tell for certain whether or not the two objects are the same size and shape, is to change the order of the numbers on one of the chain codes, trying every possible starting point. You can visually see how this would work by pulling a number off the bottoom of the chain and moving it to the top.
This preserves the serial order of the numbers. You can cycle through the entire code this way, shifting the starting point one line segment at a time. If you cycle through each number in the code, you will be able to see if the two codes eventually "line up" so that you can visually see (as in column 3 above) that the two codes are exactly the same.
If we wanted to give a machine the ability to compare chain codes, a fairly simple program can be written that can determine whether or not two chain codes are exactly the same. This would give the machine the ability to judge that two shapes were the same, even if they were rotated quite differently.
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Anonymous
Anonymous asked in Science & MathematicsMathematics · 9 years ago
# What is a fast way of adding all the numbers from 1 to 100 ?
i really need to find out the trick ASAP
thans :) xx
Relevance
• 9 years ago
Imagine a square 4 * 4 dots in size
/_._._.
,_/_._.
,_,_/_.
,_,_,_/
You can split it diagonally into two triangles of the same size and a diagonal
The number of dots in the square is 4*4 = 16
The number of dots along the diagonal is 4
The remaining dots are 16 - 4 = 12 and since the triangles are the same size we can find the number by dividing 12 by 2 = 6
Say we want a slightly bigger triangle, one that includes the diagonal then the number of dots would be 6 + 4 = 10
Why do these dots and triangles matter?
The triangle has 1 dot in the first row, 2 in the second, 3 in the third etc. so knowing the number of dots in the triangle here tells us the sum of 1, 2 and 3 = 6
More generally if when had a square with n dots by n dots there would be n² dots in total.
Along the diagonal there would be n dots, so both triangles would have (n² - n)/2
To create the bigger triangle:
(n² - n)/2 + n = (n² - n)/2 + 2n/2 = (n² - n + 2n)/2 = (n² + n)/2
This can be written as n(n+1)/2
In the case of n = 100 rows then there are 100*101/2 = 5050 dots which is also the sum from 1 to 100.
• 9 years ago
You can use the sum formula for arithmetic series
n=100 (No. of values)
d=1 (Difference between two consecutive values)
a1=1 (First value of series)
Sum=S= n/2 (2a1 + (n-1)(d))
S=100/2 (2 + 99(1))
S=50(101)
S= 5050
• 9 years ago
arithematic progression
tn-1=a+d(n-1) this is the formula
100=1+1(n-1)
100=1+n-1
100=n
n/2(first term+last term)
100/2(1+100)
50(101)=5050
• 9 years ago
hmmmm...hmmmm...
all the numbers? ALL the numbers? Every bloomin' number?!!
yeah right.
would you be meaning: whole numbers, natural numbers, rational numbers, integers, squared numbers...???
need just a tad bit more information to provide a constructive answer. based on the information in the question, the 'fast way' would be pencil and paper.
good luck.
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You are reading an older version of this FlexBook® textbook: CK-12 Texas Instruments Trigonometry Student Edition Go to the latest version.
# 2.3: Trigonometric Patterns
Difficulty Level: At Grade Created by: CK-12
This activity is intended to supplement Trigonometry, Chapter 1, Lesson 8.
ID: 12434
Time required: 15 minutes
## Activity Overview
Students will use the unit circle to examine patterns in the six trigonometric functions.
Topic: Trigonometry
• Unit Circle
• Right Triangle Trigonometry
Teacher Preparation and Notes
Associated Materials
## Trigonometric Patterns
Students will move the triangle on the unit circle to find the angle measures listed in the table on the student worksheet. Students will record the values and then answer questions about patterns in the results. Because the Cabri Jr. file only measures the angle less than $90^\circ$, there is opportunity for some further student learning. This means that when the angle being displayed is $30^\circ$ and the point is in the second quadrant, the angle being observed is really $150^\circ$ $(180^\circ - 30^\circ)$.
Students can write the ratios on their worksheet and then use the Home screen to do their calculations.
Note that due to rounding in Cabri Jr., the answers are approximate, not exact.
Discussion Questions
• What happens at $90^\circ$, $180^\circ$, $270^\circ$, and $360^\circ$?
• Why is the tangent function undefined for some angle measures?
## Extension – Patterns in reciprocal functions
In this problem, students will repeat the activity for the co-trigonometric functions. Students should notice that these functions are the reciprocals of the functions from the first part of the activity by looking at the given formulas. This will help them in calculating these functions because they can use simply find the reciprocals on the Home screen instead of finding new ratios.
Discussion Questions
• Are any of the functions undefined? For what values?
Do any of the patterns match the patterns from the functions in the activity?
## Solutions
$\theta$ $\sin\theta$ $\cos\theta$ $\tan\theta$
$30^\circ$ 0.5 0.866 0.577
$45^\circ$ 0.707 0.707 1
$60^\circ$ 0.866 0.5 1.732
$90^\circ$ 1 0 Undefined
$120^\circ$ 0.866 -0.5 -1.732
$135^\circ$ 0.707 -0.707 -1
$150^\circ$ 0.5 -0.866 -0.577
$180^\circ$ 0 -1 0
$210^\circ$ -0.5 -0.866 0.577
$225^\circ$ -0.707 -0.707 1
$240^\circ$ -0.866 -0.5 1.732
$270^\circ$ -1 0 Undefined
$300^\circ$ -0.866 0.5 -1.732
$315^\circ$ -0.707 0.707 -1
$330^\circ$ -0.5 0.866 -0.577
$360^\circ$ 0 1 0
1. $0^\circ < \theta < 180^\circ$
2. $180^\circ < \theta < 270^\circ$
3. Positive $0^\circ < \theta < 90^\circ, 180^\circ < \theta < 270^\circ$ because sine and cosine have same sign Negative $90° < \theta < 180^\circ, 270^\circ < \theta < 360^\circ$ because sine and cosine have different signs
4. $\cos(330^\circ)$
5. Possible response: $\cos(45^\circ) = \cos(315^\circ); \cos(60^\circ) = \cos(300^\circ)$
6. $\tan(225^\circ)$
7. Possible response: $\tan(60^\circ) = \tan(240^\circ), \tan(30^\circ) = \tan(210^\circ)$
8. Answers will vary. Sample response: The values in the first quadrant are repeated in the other quadrants, but have different signs.
9. Answers will vary. Sample response: The values of sine and cosine switch within a quadrant, such as $\sin(30^\circ) = \cos(60^\circ)$, but the signs may be opposite in some quadrants.
Extension
$\theta$ $\sin \theta$ $\cos \theta$ $\tan \theta$
$30^\circ$ 1.155 2 1.732
$45^\circ$ 1.414 1.414 1
$60^\circ$ 2 1.155 0.577
$90^\circ$ Undefined 1 0
$120^\circ$ -2 1.155 -1.732
$135^\circ$ -1.414 1.414 -1
$150^\circ$ -1.155 2 -1.732
$180^\circ$ -1 Undefined Undefined
$210^\circ$ -1.155 -2 1.732
$225^\circ$ -1.414 -1.414 1
$240^\circ$ -2 -1.155 0.577
$270^\circ$ Undefined -1 0
$300^\circ$ 2 -1.155 -0.577
$315^\circ$ 1.414 -1.414 -1
$330^\circ$ 1.144 -2 -1.732
$360^\circ$ 1 Undefined Undefined
## Date Created:
Feb 23, 2012
Nov 04, 2014
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# Algorithm Complexity(Big-Theta, Big-Omega, Big-O Notations) …Fcukthecode
For Big O Notation, On Another Page 👉 👉 Big O Notation (With Examples) 🙂
Unlike Big-O notation, which represents only upper bound of the running time for some algorithm, Big-Theta is a tight bound; both upper and lower bound. Tight bound is more precise, but also more difficult to compute.
The Big-Theta notation is symmetric: f(x) = Ө(g(x)) <=> g(x) = Ө(f(x))
An intuitive way to grasp it is that f(x) = Ө(g(x)) means that the graphs of f(x) and g(x) grow in the same rate, or that the graphs ‘behave’ similarly for big enough values of x.
The full mathematical expression of the Big-Theta notation is as follows:
Ө(f(x)) = { g: N0 -> R and c1, c2, n0 > 0, where c1 < abs(g(n) / f(n)), for every n > n0 and abs is the absolute value }
An example
If the algorithm for the input n takes 42n^2 + 25n + 4 operations to finish, we say that is O(n^2) , but is also O(n^3) and O(n^100) . However, it is Ө(n^2) and it is not Ө(n^3) , Ө(n^4) etc. Algorithm that is Ө(f(n)) is also O(f(n)) , but not vice versa!
Formal mathematical definition
Ө(g(x)) is a set of functions.
Ө(g(x)) = {f(x) such that there exist positive constants c1, c2, N such that 0 <= c1g(x) <= f(x) <= c2g(x) for all x > N}
Because Ө(g(x)) is a set, we could write f(x) ∈ Ө(g(x)) to indicate that f(x) is a member of Ө(g(x)) . Instead, we will usually write f(x) = Ө(g(x)) to express the same notion – that’s the common way.
Whenever Ө(g(x)) appears in a formula, we interpret it as standing for some anonymous function that we do not care to name. For example the equation T(n) = T(n/2) + Ө(n) , means T(n) = T(n/2) + f(n) where f(n) is a function in the set Ө(n) .
Let f and g be two functions defined on some subset of the real numbers. We write f(x) = Ө(g(x)) as
x->infinity if and only if there are positive constants K and L and a real number x0 such that holds:
``````K|g(x)| <= f(x) <= L|g(x)| for all x >= x0
The definition is equal to:
f(x) = O(g(x)) and f(x) = Ω(g(x))
A method that uses limits
if limit(x->infinity) f(x)/g(x) = c ∈ (0,∞) i.e. the limit exists and it's positive, then f(x) = Ө(g(x))``````
Common Complexity Classes
## Comparison of the asymptotic notations
Let f(n) and g(n) be two functions defined on the set of the positive real numbers, c, c1, c2, n0 are positive real constants.
The asymptotic notations can be represented on a Venn diagram as follows:
## Big-Omega Notation
Ω-notation is used for asymptotic lower bound.
Formal definition
Let f(n) and g(n) be two functions defined on the set of the positive real numbers. We write f(n) = Ω(g(n)) if
there are positive constants c and n0 such that: 0 ≤ c g(n) ≤ f(n) for all n ≥ n0 .
Notes
f(n) = Ω(g(n)) means that f(n) grows asymptotically no slower than g(n) . Also we can say about Ω(g(n)) when algorithm analysis is not enough for statement about Θ(g(n)) or / and O(g(n)) .
From the definitions of notations follows the theorem:
For two any functions f(n) and g(n) we have f(n) = Ө(g(n)) if and only if f(n) = O(g(n)) and f(n) = Ω(g(n)) .
Graphically Ω-notation may be represented as follows:
For example lets we have f(n) = 3n^2 + 5n – 4 . Then f(n) = Ω(n^2) . It is also correct f(n) = Ω(n) , or even f(n) = Ω(1) .
Another example to solve perfect matching algorithm : If the number of vertices is odd then output “No Perfect Matching” otherwise try all possible matchings.
We would like to say the algorithm requires exponential time but in fact you cannot prove a Ω(n^2) lower bound using the usual definition of Ω since the algorithm runs in linear time for n odd. We should instead define f(n)=Ω(g(n)) by saying for some constant c>0 , f(n)≥ c g(n) for infinitely many n . This gives a nice
correspondence between upper and lower bounds: f(n)=Ω(g(n)) iff f(n) != o(g(n)) .
Formal definition and theorem are taken from the book “Thomas H. Cormen, Charles E. Leiserson, Ronald L. Rivest, Clifford Stein. Introduction to Algorithms”.
For Big O Notation, On Another Page 👉 👉 Big O Notation (With Examples) 🙂
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Week 3 Course Notes
# Week 3 Course Notes - MATH 137 WEEK 3 NOTES PROF DOUG PARK...
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MATH 137 WEEK 3 NOTES PROF. DOUG PARK 1. Limit laws Limit Laws . Suppose f ( x ) L and g ( x ) M , as x a . Then Sum Law: f ( x ) + g ( x ) L + M Difference Law: f ( x ) - g ( x ) L - M Constant Multiple Law: cf ( x ) cL , where c = constant. Product Law: f ( x ) · g ( x ) L · M Quotient Law: f ( x ) g ( x ) L M , as long as M 6 = 0. Composition Law: g ( f ( x )) g ( L ), as long as g ( x ) is continuous at x = L . We will prove the Sum Law using the precise definition of limit. Proofs of other laws are found in Appendix F of Textbook. Proof of Sum Law . We need to prove lim x a ( f ( x ) + g ( x )) = L + M if lim x a f ( x ) = L and lim x a g ( x ) = M . Given any 1 > 0, there is δ 1 > 0 such that 0 < | x - a | < δ 1 = | f ( x ) - L | < 1 Given any 2 > 0, there is δ 2 > 0 such that 0 < | x - a | < δ 2 = | g ( x ) - M | < 2 Given any > 0, let δ = min { δ 1 , δ 2 } . Then 0 < | x - a | < δ = | f ( x ) - L | < 1 and | g ( x ) - M | < 2 = | f ( x ) - L | + | g ( x ) - M | < 1 + 2 = | ( f ( x ) - L ) + ( g ( x ) - M ) | ≤ | f ( x ) - L | + | g ( x ) - M | < 1 + 2 = | ( f ( x ) + g ( x )) - ( L + M ) | < 1 + 2 We are done if 1 + 2 = . Thus we set 1 = 2 = 2 . 2. Computing limits of 0 0 form Ex . Evaluate L = lim x a x 3 - a 3 x - a ( a = constant). Sol’n . If we plug in x = a , we get 0 0 form. In such cases we try to cancel the common zero factor. x 3 - a 3 = ( x - a )( x 2 + ax + a 2 ) Date : September 28, 2009. 1
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2 DOUG PARK Thus L = lim x a x - a x - a · ( x 2 + ax + a 2 ) = lim x a ( x 2 + ax + a 2 ) = a 2 + a 2 + a 2 = 3 a 2 Ex . Evaluate L = lim x 2 x - 3 x - 2 x + 2 - 2 . Sol’n . 0 0 form so we cancel ( x - 2) terms. The ( x - 2) terms are found when we multiply by conjugate expressions! Hence multiply both top and bottom by ( x + 3 x - 2)( x + 2 + 2). We get ( x 2 - 3 x + 2)( x + 2 + 2) ( x + 2 - 4)( x + 3 x - 2) = ( x - 1)( x - 2)( x + 2 + 2) ( x - 2)( x + 3 x - 2) = ( x - 1)( x + 2 + 2) x + 3 x - 2 -→ (2 - 1)( 2 + 2 + 2) 2 + 6 - 2 = 4 4 = 1 3. One-sided limits One-sided limits will usually be of constant 0 form, or limits involving | | symbol. Theorem . lim x a f ( x ) = L ⇐⇒ lim x a - f ( x ) = lim x a + f ( x ) = L Ex . Evaluate the following limits if they exist. (i) lim x 1 1 ( x - 1) 3 (ii) lim x 1 1 | x - 1 | (iii) lim x 1 x - 1 | x - 1 | Strategy . Note that 1 0 + = + , 1 0 - = -∞ . Sol’n . (i) lim x 1 + 1 ( x - 1) 3 = 1 (1 + - 1) 3 = 1 (0 + ) 3 = 1 0 + = + lim x 1 - 1 ( x - 1) 3 = 1 (1 - - 1) 3 = 1 (0 - ) 3 = 1 0 - = -∞ Hence the limit does not exist (DNE). (ii) lim x 1 + 1 | x - 1 | = 1 | 0 + | = 1 0 + = + lim x 1 - 1 | x - 1 | = 1 | 0 - | = 1 0 + = + Hence the limit is + , but DNE as a real number.
MATH 137 3 (iii) If x 1 + , then x > 1, so x - 1 > 0. Thus x - 1 | x - 1 | = x - 1 x - 1 = 1 lim x 1 + x - 1 | x - 1 | = lim x 1 + 1 = 1 If x 1 - , then x < 1, so x - 1 < 0. Thus x - 1 | x - 1 | = x - 1 - ( x - 1) = - 1 lim x 1 - x - 1 | x - 1 | = lim x
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# Binomial Expansion to the power of a non-natural number
What is the reasoning that $$|x|$$ has to be less than $$1$$ for $$(1+x)^n$$ when $$n$$ is not a natural number?
• radius of convergence for taylor series. – N. S. Oct 23 '18 at 0:09
• Were |x| > 1, the terms in the expansion would grow without bound. – ncmathsadist Oct 23 '18 at 0:11
If you want an intuitive explanation, then consider this scheme.
The Binomial expansion is a Taylor series at $$x=0$$.
The Taylor series is a polynomial that you can view as the polynomial that interpolates "a number" of points close to the origin.
Now, when $$n$$ is an integer, a polynomial of degree $$n$$ interpolating near the origin will also interpolate $$(1+x)^n$$ perfectly for $$x \to \infty$$.
If $$n$$ is not an integer, than for large $$x$$ $$(1+x)^n \approx x^n$$ and no polynomial of integral degree can approximate a non integral power of x.
The radius of convergence $$R$$ of the power series $$\sum_{k \geq 0} \binom{\alpha}{k} x^k$$ is, by the ratio test,
$$R = \lim_{k \to \infty} \left\lvert \frac{\binom{\alpha}{k}}{\binom{\alpha}{k+1}}\right\rvert = \lim_{k \to \infty} \left\lvert\frac{k+1}{\alpha - k}\right\rvert = 1$$
And so we are only guaranteed convergence for those $$x \in \mathbb{C}$$ satisfying $$|x| < 1$$.
Edit: I should add the caveat above that $$\alpha \in \mathbb{C} \setminus \{0, 1, 2, \ldots\}$$. If $$\alpha$$ is a nonnegative integer, then the sequence $$\binom{\alpha}{0}, \binom{\alpha}{1}, \ldots$$ is eventually zero, and the power series is just a polynomial, which converges everywhere.
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# Forces
• Forces are usually a push or a pull.
• Forces are measured using a newtonmeter and the unit of force is Newtons.
• Forces work in pairs - as you push down on a table the table pushes up with equal force.
• When there is no change in velocity or shape of an object then the forces are balanced e.g. book on table
• When there is a change in velocity or shape of an object then the forces are not balanced e.g. person falling through ice.
• An example of a force is gravitational force. To work out the weight of an object (with gravity working on it) you use the following formula:
• Weight (N) = mass (kg) x gravitational force (N/kg)
• The main formula to use to work out forces is:
• Force (N) = mass (kg) x acceleration (m/s/s)
• OR F=ma
Forces and motion
#### Acceleration
##### Acceleration
Forces and motion
#### Forumulae
##### Forumulae
Forces and motion
#### Hooke's Law
##### Hooke's Law
Forces and motion
#### Momentum
##### Momentum
Forces and motion
#### Pressure
##### Pressure
Forces and motion
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# 8*7+8%5+6/2 to me this == 59.6. Apparently I am wrong.
Apparently I am wrong. Please explain?
Normally the % (modulo) operator has the same level of precedence as multiplication and division. So in this statement we can evaluate it from left to right.
8*7 = 56 8%5 = 3 <- the remainder is the answer to a modulus expression 6/2 = 3 56+3+3 = 62
## You're forgetting about operator precedence.
All multiplicative operations (which include multiplication, division, and modulus) are performed before any additive opertions ( + or - ).
I'll bet you can figure it out now without a spoiler.
MOD
Ok so for Computer Science the Order of Operations is:
1. P ---> Parentheses
2. E ---> Exponents
3. M ---> Multiplication
4. D ---> Division (This is for programming only, division holds the same weight as multiplication
5. M ---> Modulos (This is for programming only, and it can has the same order weight as division and multiplication)
7. S ---> Subtraction
So by following these rules we know that our addition will be done last and our multiplication, division, and Modulos will be done first, so this could have effectively been written as `(8*7) + (8%5) + (6/2)`
We know that `8 * 7 = 56` We know that `8 % 5 = 3` We know that `6 / 2 = 3`
So when we add them all together we get `56 + 3 + 3 ----- or ----- 56 + 6 = 62`
There's no common rule for all of "computer science", operator precedence varies with language. For example, in C#, multiplication does not take precedence over division — so 12/2*3 is 18, not 2.
That is because of associativity, so you are correct there! I realized that I have not specified that division and modulos hold the same weight as multiplication so I will fix that.
By the C# rules, the associativity of multiplicative operations is left-to right. That's why 12/2*3 evaluates to 18, instead of 2.
I recommend following the rules of the language you are programming in. If there are "general rules" that the language is "breaking", it doesn't seem like they would be useful to the developer and they could lead to confusion and unexpected program behavior.
I see where you are coming from, I have a math background so I always see that there are general rules (keeps us sane) but in programming I have noticed that some languages seem to break this (for some reason or another).
PEMDMAS does apply in Python where it's more like PE[MDM][AS]
```>>> 12 / 2 * 3
18.0 # not 2
>>> 12 / 4 % 2 * 3
3.0 # / then % then *
>>> 25 % 13 / 2 * 3
18.0 # % then / then *
>>> 15 - 3 + 2
14 # not 10
```
Normally the % (modulo) operator has the same level of precedence as multiplication and division. So in this statement we can evaluate it from left to right.
8*7 = 56 8%5 = 3 <- the remainder is the answer to a modulus expression 6/2 = 3 so... 56+3+3 = 62
Excellent answers, thanks all. Sometimes learning the write software it's difficult to see the wood for the trees but this has been explained well!
PLUS
I like to put paranteses between numbers which are going to be multiplied, divided or modulated to seperate additive and multiplicative operations to make it easier:
(8*7)+(8%5)+(6/2)
This way you can easily solve it.
(56)+(3)+(3) = 62
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# Easy Time, Speed & Distance Solved QuestionAptitude Discussion
Q. Two identical trains $A$ and $B$ running in opposite direction at same speed tale 2 min to cross each other completely. The number of bogies of $A$ are increased from 12 to 16. How much more time would they now require to cross each other?
✖ A. 40 sec ✖ B. 50 sec ✖ C. 60 sec ✔ D. 20 sec
Solution:
Option(D) is correct
Total initial bogies is $12+12 = 24$
Additional bogies $= 16−12 = 4$
24 bogies take 2 minutes.
4 bogies will take:
$=\dfrac{2\times 60}{24}\times 4$
20 sec.
## (3) Comment(s)
Chirag Goyal
()
4 bogies increase in one train is similar to 2 bogies increase in 2 trains each.
so 12 bogies takes 2 mins and 2 bogies will take 1/6 of 2 mins
i.e. 20 sec more.
Ash
()
How did we get 60? In the solution??s
Anita
()
total boggies 24 take 2 min i.e; 120 sec
1 boggie will take 5 sec
therefore 4 boggies will take 5 * 4=20 sec
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# Biology Help - Fish/gills/respiration/maths watch
1. Does anybody know how to answer this?😊
Volume of water absorbed by the gills from each dm3 of water / cm3 = 7
Mass of fish /kg = 0.4
Oxygen required by fish / cm3 kg-7 hour-1 = 90
Calculate the volume of water that would have to pass over the gills each hour to supply the oxygen required by the fish. Show your working.
Attached Images
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2. 1. The volume of O2 per dm3 of water is 7cm3
2. The O2 required per hour is 90cm3
3. If one dm3 of water passed over the gills per hour, this would mean the fish gets 7cm3 of O2 per hour
4. I found it easiest to get down to 1cm3 of O2 per hour and work from there.
5. 7cm3 of O2/7 = 1cm3 of O2. In order to work out the volume of water you'd need to get this amount you need to do the same thing:
6. 1dm3 of water/7 = 0.143dm3 of water
7. The fish needs 90 cm3 of O2. To get this, you need to multiply the 1cm3 from part 5 by 90. Again, you need to do the same thing with the water.
8. 0.143dm3 of water x 90 = 12.86dm3 of water.
Therefore, you need 12.86 dm3 of water per hour.
Hope this makes sense (and that it's right).
3. Thanks so much this is a massive help😊x
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What is the special of product of the incidence matrix and its transpose for an undirected graph
Can some one tell me what is the special of product of the incidence matrix and its transpose for an undirected graph ?. I try get product of a incidence matrix and its transpose for an undirected graph with three vertices, two edges but i don't get any clue
The $$i,j$$ entry of this matrix product is $$\tag1b_{i,j}=\sum_k a_{i,k}a^T_{k,j}=\sum_ka_{i,k}a_{j,k}.$$ The product $$a_{i,k}a_{j,k}$$ is $$1$$ if vertices $$i$$ and $$j$$ are both incident with edge $$k$$, and $$0$$ otherwise. Thus for $$i\ne j$$, the entry $$b_{i,j}$$ counts the number of edges from vertex $$i$$ to vertex $$j$$ (so for simple graphs, $$b_{i,j}$$ is $$1$$ or $$0$$, depending on whether there is an edge between the two vertices or not). For $$i=j$$ something different happens: each edge having vertex $$i$$ as one of its endpoints contributes $$1$$ to the sum in $$(1)$$, hence we see that $$b_{i,i}=\rho(i)$$ is the degree of vertex $$i$$.
Note that one needs to be careful: The details depend on how the incidence matrix is defined (does $$a_{i,j}$$ denote the incidence of vertex $$i$$ with edge $$j$$ or that of edge $$i$$ with vertex $$j$$?) and likewise whether one computes $$AA^T$$ or $$A^TA$$. When doing it "wrong", one obtains that $$b_{i,j}$$ tells us whether edges $$i$$ and $$j$$ have an endpoint in common and $$b_{i,i}=2$$ for the diagonal entries - a much less interesting result.
The product of incidence matrix Q by its transpose is known as Laplacian matrix. The Laplacian matrix has many properties linked to graphs.
The Laplacian Matrix can be obtained with
L=D-A
with D diagonal matrix of degree/valences of nodes, and A the adjacency matrix.
• Just for the sake of posterity, I'm pretty sure this cannot be true. In particular, the incidence matrix has only nonnegative entries, so so much the product of it with its transpose, while the Laplacian matrix of a graph with at least one edge has a negative component. Notice also that the Laplacian does not fit the description given by Hagen above. Commented Sep 16, 2021 at 3:16
• @CadeReinberger It is true. If you use oriented incidence matrix $B$, the $L=B^TB$. Read more at: mbernste.github.io/posts/laplacian_matrix Commented Sep 4, 2022 at 21:21
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Hello, learners welcome to The Engineering Projects. We are working on MATLAB, and in this tutorial, you are going to learn a lot about matrices in MATLAB. We are going to learn them from scratch, but we will avoid unnecessary details about the topic. So, without wasting time, have a look at the topics that you will learn in detail.
• What is an array?
• What is the matrix?
• How can we declare a matrix in MATLAB?
• What are the different types of matrices?
• Can we find the unknown values of two equal matrices?
• How can we solve the simultaneous equation in MATLAB?
## What is an Array?
In this world of technology, the use of data is everywhere, and therefore, we can say there is a need for arrays in every field. You will find the reason soon. But before this, look at the introduction of an array.
An array is a simple data structure that contains a collection of data presented in contiguous memory locations.
So, the term “contiguous” used in the definition tells us that the data is in a continuous format, so we are not required to search here and there because the data is in a structured format. Moreover, arrays are of many kinds, such as
• Two-dimensional arrays
• Three-dimensional arrays
In different types of cases, the suitable array is picked up so that we may get the best result with limited memory occupancy. With this type of foundation concept, we can now move forward toward our main topic, which is matrices.
## What is a Matrix?
In real-life applications and in higher studies, matrices are used in plenty in different forms, and therefore, we have decided to talk about them from a very basic level since it is important to understand the key features of the topics we are studying. Moreover, matrices are introduced in early classes, and it is important to refresh the basics in our minds so that we may proceed to the more complex problems. Here is the definition of "matrix":
A matrix is a two-dimensional array in the form of an ordered rectangular array enclosed by a square bracket that has entries of the same kind in it in the form of real or complex data.
The plural of the matrix is matrices, and sometimes the rectangular bracket is replaced by the parentheses according to the case. Just look at the image given below:
This is a matric that contains nine elements, and you can also name this matric anything you want. In this way, it becomes easy to deal with more than one matrix, and you will see this action soon.
### Order of a Matrix
To proceed forward, you must know the types of matrix and, for this, it is important to know the order of the matrix.
The matrix given above is a square matrix and the horizontal lines are called columns, whereas the vertical entries are termed the rows of that particular matrix.
If we represent the rows with the name m and the columns as n, then the order of the matrix is given as:
mxn
In this way, it is clear that the matrices given above have the order 3x4. If it seems to be an unnecessary thing to you, think again, because, with the help of order, we can know the type of a matrix and then perform different types of operations on it. But before this, have a look at some code in MATLAB to design matrices of different kinds.
### Code for the Simple Matrix
The Matrix is easily used in MATLAB, and you can start working with it by following the simple steps given below:
• Go to the command window.
• Start writing the following code:
A=[23 14 -8 33; 17 -102 0 37;3 -31 98 4];
• Press enter.
In the image given overhead, these are the same entries that we have seen in the image given above, and in MATLAB, you will see the following result:
The square bracket is not shown on the sides of the array in MATLAB. As you can see, the semicolon after every three entries indicates that the row is completed and the MATLAB compiler has to start the other row.
Here, A shows the name of the matrix that is compulsory, and you can name your matrix any word. If you do not follow the exact format and provide the number of entries different in rows, you will get the error. Once you know how to get started, you are ready to learn about the types of matrices.
## Types of Matrices
There are several different types of matrices, and you can perform different arithmetic operations on the matrices only if they are of the same kind. This condition is not applied to all the operations, but most of them follow these rules. Here are some important types of matrices.
### Row Matrix
A row matrix contains only one row and it is one of the simplest forms of a matrix. In this way, we get the matrix with a horizontal shape. The order of this matrix is:
mxn=1xn
Where n may be any number.
### Column Matrix
As you can guess, the column matrix is a type of matrix containing only one column and one or multiple rows. In this way, we get a matrix that has a vertical shape. Have a look at the order of a column matrix:
mxn=mx1
Where m may be any number, but the value of n is always one.
### Square Matrix
A square matrix always has the number of rows and columns equal. It means, that no matter what the total number of entries is, the number of entries in each row and column must always be equal. In other words,
m=n
When you examine the example of a square matrix, you will get the reason why it is called so. The shape of this type of matrix is always square.
### Rectangular Matrix
A rectangular matrix is one that has the arrangement of elements in such a way that the number of rows of the matrix is not equal to the number of columns. The same statement can be represented in the equation given next:
m!=n
Therefore, the matrix formed is in a rectangular shape, either in vertical format or horizontal format, according to the number of rows and columns.
### Diagonal Matrix
We all know that the diagonal is the line or area that joins the upper left area with the lower right area of a rectangular or square. By the same token, a diagonal matrix is the one that contains all the diagonal values equal to zero and s in such a way that all the values other than the diagonal are zero. It will be clearer when you see the example of the diagonal matrix. We have set the examples of all the types of matrices that we have defined previously into a single MATLAB screen so you may have the best idea of each of them.
Code and Output
Moreover, here you can observe that instead of naming the matrices A, B, and so on, we have used the real names of the matrices for a clear declaration. Your homework is to make examples of each of them by yourself for the sake of practicing.
## Finding the Unknown Values Between Two Matrices
Do you remember when we said the order of the matrix matters? This is one of the uses of an order of a matrix. Suppose we have two matrices named A and B, declaring that both are equal. This means that each corresponding value of a matrix A at position row 1 column 1 is equal to the corresponding value of the same position of matrix B. This is true for all the remaining values q of both matrices. Let me be clear with one example. Have a look at the picture given below:
So, the value of r and, in return, the value of all r variables in each entry can be easily obtained by following the rules of the equation. It is one of the simplest examples of doing so, but in real life, we face complex problems. So, we use MATLAB for simplicity and accurate results. Have a look at the MATLAB code where we are going to show you an application of you can easily solve the simultaneous equation in MATLAB as well.
## Solving Simultaneous Equations in MATLAB
By using the property of the matrix of equality in more than one matrix, we can easily solve the simultaneous equations that are difficult and time taking if we solve them by hand. So let's see how we can declare and solve the simultaneous equation in MATLAB.
Code:
syms x y
equa1= 6*x + 9*y==13;
equa2= 9*x + 6*y==12;
[A,B]= equationsToMatrix([equa1,equa2],[x,y])
z=linsolve(A,B)
Output:
### Understanding the Code
To understand this code, you have to learn the basic definition of the function we have used in the code. It is the equationsToMatrix function.
#### equationsToMatrix Function
The equationsToMatrix is a pre-defined function of MATLAB that converts the linear equation into a matrix so that we can use different operations on it more efficiently. It does it in the same way as we do in real life while solving the simultaneous equation with pen and paper. There are three types of syntax if this particular function. The one that we have used has the following syntax:
[A,b] = equationsToMatrix(eqns,vars)
Here, a minimum of two equations are required and the variables have the same condition. You must keep all the functions in mind and have to follow the exact syntax. Otherwise, it will show an error.
#### linsolve Function in MATLAB
In MATLAB, to solve the linear equation, we use this pre-defined function as it works in two ways:
1. LU factorization with partial pivoting when in equation AB=X, A is a square.
2. QR factorization, otherwise.
In our case, it has used the QR factorization. Now, you are able to understand the code clearly.
• First of all, the syms sign tells MATLAB that we are defining the variables. These may be one or more. But, we wanted two variables here, and we named them x and y.
• Now, we simply provide the values of the equation to MATLAB and store both of them into variables named equa1 and equa2 respectively.
• The values of variables and equations are fed into the eqautionToMatrix function to convert the linear simultaneous equation into a matrix for easy solving.
• In the end, we simply named a matrix z and told MATLAB that we wanted the value of variables x and y.
### Simultaneous Equation in MATLAB: Method 2
By the same token, we can use the other method that is similar to it but the way it solves the equation is a little bit different.
Code:
syms x y
equa1= 6*x + 9*y==13;
equa2= 9*x + 6*y==12;sol=solve([equa1,equa2],[x,y])
asol=sol.x
bsol=sol.y
Output:
Here, the only pone this is to understand. sol.x and sol.y are the functions that are used by the compiler to find the value of variables x and y respectively. You can use any variable with this sol function, after naming them at the beginning. After that, a variable is used to store and present the value of the answer obtained.
It was an interesting lecture about the matrix, and we worked a lot from scratch to the end on many topics. We have defined the arrays and seen the introduction of the matrix. We also found information about the types of matrices. Once we have a grip on the basics, we learn that a matrix can be used to find the unknown value of two matrices, and as an application of this method, we found the values of the variable by using linear equations and learned how to declare, use, and solve the linear equation with the help of matrices in MATLAB.
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SolSec8.6
# SolSec8.6 - Problems and Solutions Section 8.6(8.50 through...
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Unformatted text preview: Problems and Solutions Section 8.6 (8.50 through 8.54) 8.50 Consider the machine punch of Figure P8.15. Recalculate the fundamental natural frequency by reducing the model obtained in Problem 8.16 to a single degree of freedom using Guyan reduction. Solution: From the results of 8.16 K = 4-2-2 2 × 10 8 , Μ = .052 .013 .013 .026 Φρομ(8.10429 Θ Τ ΜΘ = .052 + .013 + .013 + .026 = .104 Φρομ(8.10529 Θ Τ ΚΘ = (4 - 229 ×10 8 = 2 ×10 8 ϖ = 2 ×10 8 .104 = 43852.9ραδ/σ which is a poor prediction of the first natural frequency. If we reorder K and M (reducing to coordinate 2) we get Q T MQ = .026 + .013 + .013 = .052 Θ Τ ΚΘ = (2 -129 ×10 8 = 1 ×10 8 ϖ = 43852.9ραδ/σ which is the same result as reducing to coordinate 1. 8.51 Compute a reduced-order model of the three-element model of a cantilevered bar given in Example 8.3.2 by eliminating u 2 and u 3 using Guyan reduction. Compare the frequencies of each model to those of the distributed model given in Window...
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## This note was uploaded on 11/28/2010 for the course ME 4440 taught by Professor Hill during the Winter '09 term at Detroit Mercy.
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ΙΒ μαθηματικά HL- Combinatorics – Probability
Permutations Combinations – IB Mathematics HL
1. Find the number of possible choices for choosing a sub-committee of five persons consisting of the chairperson, secretary and three ordinary membersfrom a committee of 40 people.
Solution
The chairperson can be chosen by$$^40\mathrm{C}_1$$ ways
The secretary can be chosen by $$^39\mathrm{C}_1$$ ways
And the three ordinary members can be chosen by $$^38\mathrm{C}_3$$ ways
So, the members of this sub-committee can be chosen by
$$^40\mathrm{C}_1 \cdot ^39\mathrm{C}_1 \cdot ^38\mathrm{C}_3=$$
$$=40 \cdot 39\cdot8436=13,160,160$$ ways.
2. In a class of 12 students, 4 are male and 8 are female. How can we find the number of committees of 5 members can be formed containing 2 males and 3 females?
Solution
The males can be chosen by $$^4\mathrm{C}_2$$ ways
The females, can be chosen by $$^8\mathrm{C}_3$$ ways
So, the members of this committee can be chosen by
$$^4\mathrm{C}_2 \cdot ^8\mathrm{C}_3=$$
$$=6 \cdot 56=336$$ ways.
3. Find the number of different arrangements of the letters of the word “LONDON”? and how can we calculate the probability of the event that the middle two letters both are N’s.
Solution
The number of different arrangements of the six letters from which there are two pairs of same letters is given by the following formula
$$\frac{6!}{2!2!}=\frac{720}{4}=180$$
Since the two N’s are located in the middle of the word there are $$\frac{4!}{2!}=\frac{24}{2}=12$$ ways to arrange the remaining 4 letters from which there are two O,s.
So, the probability equals to $$\frac{12}{180}=\frac{1}{15}$$
4. Find the probability $$P(A\cap B’)$$ if A and B are two events such that $$P(A)=0.5$$, $$P(B)=0.3$$ and $$P(A \cup B)=0.7$$.
Solution
We know that
$$P(A \cup B)=P(A)+P(B)-P(A \cap B) \Rightarrow$$ $$P(A \cap B)=0.5+0.3-0.7=0.1$$
Finally, we have
$$P(A \cap B’)=P(A)-P(A \cap B) =0.5-0.1=0.4$$
Conditional Probability – IB Mathematics HL – IB maths HL
5. Find the probability $$P(A/B’)$$ if A and B are two events such that $$P(A)=0.4$$, $$P(B)=0.2$$ and $$P(A/B)=0.8$$.
Solution
From conditional probability formula we have that
$$P(A/B)=\frac{P(A \cap B)}{P(B)}\Rightarrow$$ $$0.8=\frac{P(A \cap B)}{0.2}\Rightarrow P(A \cap B) =0.8 \cdot 0.2=0.16$$
Finally, we have that
$$P(A/B’)=\frac{P(A \cap B’)}{P(B’)}=\frac{P(A)- P(A \cap B)}{1-P(B)}=$$ $$=\frac{0.4-0.16}{1-0.2}=\frac{0.24}{0.8}=0.3$$
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# Excel Point and Click Method: Simplifying Data Entry and Calculation
While many are familiar with Excel’s core functionalities, one often underutilized feature is the “point and click” method. This approach streamlines the process of entering cell references into formulas, offering numerous advantages, including reduced errors, saved time, and more meaningful data.
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## Understanding the Point and Click Method in Excel
The point and click method is a technique that leverages the mouse pointer or arrow keys to input cell references into Excel formulas. This method simplifies the process of constructing formulas and performing calculations, making it a valuable tool for both beginners and experienced users.
## Step-by-Step Guide
To utilize the point and click method effectively, follow these straightforward steps:
1. Initiate the Formula: Start your formula by typing an equal sign (=), plus sign (+), or minus sign (-) in the cell where you want the result to appear.
2. Select the First Cell Reference: Click on the cell you wish to reference first in your formula, or navigate to it using the arrow keys. Excel will highlight the selected cell with a dashed blue line, and the cell reference will automatically appear in your formula.
3. Define the Operator: Enter the mathematical operator you want to use, such as +, -, *, or /.
4. Select the Next Cell Reference: Continue by clicking on the next cell you want to reference or navigating to it using the arrow keys. As before, Excel will highlight the selected cell with a dashed red line, and the cell reference will be added to your formula.
5. Repeat as Needed: Repeat steps 3 and 4 until you have included all the cell references and operators necessary for your formula.
6. Finalize and Calculate: Once your formula is complete, press Enter or click the Enter button on the Formula bar. Excel will automatically calculate the result and display it in the designated cell.
## Practical Example
Let’s illustrate the point and click method with a simple example. Suppose you want to calculate the sum of values in cells B2 and C2. Here’s how you can do it:
1. In cell D2, type the equal sign (=) to initiate the formula.
2. Click on cell B2 or use the arrow keys to navigate to it. The cell reference B2 will appear in the formula, and a dashed blue line will surround cell B2.
3. Enter the addition operator by typing +.
4. Click on cell C2 or use the arrow keys to move to it. The cell reference C2 will appear in the formula, with a dashed red line around cell C2.
5. To complete the formula and calculate the sum, press Enter or click the Enter button on the Formula bar. Excel will display the result in cell D2, and the formula will be saved as =B2+C2.
## Advantages of the Point and Click Method
The point and click method in Excel offers several notable advantages:
• Reduced Errors: By allowing users to select cell references visually, the method minimizes the risk of typographical errors in formulas.
• Time Savings: This method speeds up the formula creation process, making it a time-efficient way to work in Excel.
• Improved Data Interpretation: The visual representation of cell references enhances the clarity of your data, making it easier to understand and troubleshoot.
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# Is this variant of ATM decidable?
Ok so I understand how $\mathrm{ATM} = \{\langle M,w \rangle \mid \text{$M$is a TM and$M$accepts$w$}\}$ is undecidable.
Is this because $w$ is a variable?
What if the parameter is fixed?
Consider $\mathrm{BTM} = \{\langle M,w \rangle \mid \text{$M$is a TM and$M$accepts the string 101}\}$.
BTM is decidable right? The diagnolization problem here doesn't seem to apply because it would seem trivial to build a Turing machine that is 100% capable of accepting only the input "101" and rejecting every other possible input, correct?
And our machine would always reject itself as an input, since it only accepts "101", right?
• Note our reference questions. Commented Sep 29, 2014 at 7:47
• Just quote Rice's theorem! Commented Sep 29, 2014 at 13:54
• What's the use of $w$ in BTM? Commented May 25, 2018 at 12:08
You can show that $\mathrm{BTM}$ is undecidable with a simple reduction from $\mathrm{ATM}$.
Let $\langle M,w\rangle \in ATM$. We define the reduction function $f$, $\mathrm{HP} \leq \mathrm{BTM}$ as follow:
$f(\langle M,w\rangle) = M_w$, Where $M_w$ on every input $x$, runs $M$ with input $w$. if $M$ stopped, then $M_w$ accepts $x$. if $M$ rejects, then $M_w$ rejects. Clearly, if $\langle M,w\rangle \in ATM$ then $L(M_w) = \Sigma^*$, and in particular $101 \in L(M_w)$. if $\langle M,w\rangle \notin ATM$ then $L(M_w) = \emptyset$, and $101 \notin L(M_w)$.
Because $\mathrm{ATM}$is undecidable, so is $\mathrm{BTM}$.
It should be intuitive that $\mathrm{BTM}$ is undeciadble, because given a turing machine $M$, you can't tell if $M$ will halt with input $101$
• Hmm thank you I am getting the concept of the behaviour of one turing machine versus a language describing how a set of turing machines behave - still a bit confused. Commented Sep 29, 2014 at 6:44
BTM is also undecidable, with a similar diagonalization proof. Suppose the Turing machine $M$ decided BTM. Define a Turing machine $T$ that, on input $x$ an encoding of a Turing machine, it computes the encoding $y_x$ of a Turing machine which runs the Turing machine encoded by $x$ on input $x$; if $M(y_x)=1$ then $T$ gets into an infinite loop, and otherwise it halts. Then $T$ halts on input $\#T$ (where $\#T$ is the encoding of $T$) iff $M(y_{\# T})=0$ iff $y_{\# T}$ doesn't halt on 101 iff $T$ doesn't halt on input $\#T$, contradiction.
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Simplifying radicals | MAT222 | Ashford University
In this discussion, you will simplify and compare equivalent expressions written both in radical form and with rational (fractional) exponents. Read the following instructions in order and view the MAT222 Week 3 Discussion Example to complete this discussion. Please complete the following problems according to your assigned number. (Instructors will assign each student their number.)
On pages 575 – 577, do the following problem: 22
On pages 584 – 585, do the following problem: 28
• Simplify each expression using the rules of exponents and examine the steps you are taking.
• Incorporate the following five math vocabulary words into your discussion. Use bold font to emphasize the words in your writing. Do not write definitions for the words; use them appropriately in sentences describing the thought behind your math work.
• Principal root
• Product rule
• Quotient rule
• Reciprocal
• nth root
Refer to Inserting Math Symbols for guidance with formatting. Be aware with regards to the square root symbol, you will notice that it only shows the front part of a radical and not the top bar. Thus, it is impossible to tell how much of an expression is included in the radical itself unless you use parenthesis. For example, if we have √12 + 9 it is not enough for us to know if the 9 is under the radical with the 12 or not. Therefore, we must specify whether we mean it to say √(12) + 9 or √(12 + 9), as there is a big difference between the two. This distinction is important in your notation.
Another solution is to type the letters “sqrt” in place of the radical and use parenthesis to indicate how much is included in the radical as described in the second method above. The example above would appear as either “sqrt(12) + 9” or “sqrt(12 + 9)” depending on what we needed it to say.
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# Convert an array to a Linked List
Problem Statement: Given an array arr[] of size N. The task is to create a linked list from the given array and return the head of the linked list.
Examples
```Example 1:
Input Format: N=3, arr[]=[0,1,2]
Result 0->1->2
Explanation: After converting the array into a linked list, the head of the linked list would be the start of the array, and thus the output will be 0.
Example 2:
Input Format: N=6, arr[]=[1,2,3,4,5,6]
Result 1->2->3->4->5->6
Explanation: Again, after converting the linked list, the head would be the first element of the array, and thus the output will be 1.
```
### Solution
Disclaimer: Don’t jump directly to the solution, try it out yourself first.
### Approach:
In order to convert an array to a linked list, start by defining a reference point. In this case, try to fix the start of the array, which will be the head of the linked list.
After fixing the head, iterate through the entire array, convert every element into a node of the linked List, and keep linking them.
### Algorithm:
• Initially, assign the 0th index element by creating a new node which will be the head of the linked list.
• Once we have the head(store a reference in the mover node, this mover will always be the last node of the linked list created to date), which has the 0-th index element, the next task is to connect the next set of elements to the LinkedList. In order to do that, iterate through the entire array, and for every element, do the following:
1. Create a new node and assign the i-th index element value to it.
2. Link the previous nodes next to this new node.
Assign this node to be the mover node for the next iteration.
Code:
## C++ Code
``````class Node{
public:
int data;
Node* next;
public:
Node(int data1, Node* next1){
data=data1;
next=next1;
}
public:
Node(int data1){
data=data1;
next=nullptr;
}
};
Node* convertarr2LL(vector<int>& arr){
for (int i=1;i<arr.size();i++){
Node* temp=new Node(arr[i]); //creating new node every time
mover->next=temp; //pointing mover to temp
mover=mover->next; //moving mover to the next node
}
}
int main(){
vector<int> arr={2,5,8,7};
}
``````
Output: 2
Time Complexity: O(N), since we are iterating over the entire array, where N is the number of elements in the array.
Space Complexity: O(1), as we have not used any extra space.
## Java Code
``````class Node{
int data;
Node next;
Node(int data1, Node next1){
this.data=data1;
this.next=next1;
}
Node(int data1){
this.data=data1;
this.next=null;
}
};
private static Node convertarr2LL(int[] arr){
for (int i=1;i<arr.length;i++){
Node temp= new Node(arr[i]);
mover.next=temp;
mover=mover.next;
}
}
public static void main(String[] args){
int[] arr={2,5,8,7};
}
}
``````
Output: 2
Time Complexity: O(N), since we are iterating over the entire array, where N is the number of elements in the array.
Space Complexity: O(1), as we have not used any extra space.
## Python Code
``````class Node:
def __init__(self, data1, next1=None):
self.data = data1
self.next = next1
def convert_arr_to_ll(arr):
for i in range(1, len(arr)):
temp = Node(arr[i]) # creating new node every time
mover.next = temp # pointing mover to temp
mover = mover.next # moving mover to the next node
arr = [2, 5, 8, 7]
``````
Output: 2
Time Complexity: O(N), since we are iterating over the entire array, where N is the number of elements in the array.
Space Complexity: O(1), as we have not used any extra space.
## JavaScript Code
``````class Node {
constructor(data1, next1 = null) {
this.data = data1;
this.next = next1;
}
}
function convertArrToLL(arr) {
for (let i = 1; i < arr.length; i++) {
let temp = new Node(arr[i]); // creating new node every time
mover.next = temp; // pointing mover to temp
mover = mover.next; // moving mover to the next node
}
}
let arr = [2, 5, 8, 7];
``````
Output: 2
Time Complexity: O(N), since we are iterating over the entire array, where N is the number of elements in the array.
Space Complexity: O(1), as we have not used any extra space.
In case you are learning DSA, you should definitely check out our free A2Z DSA Course with videos and blogs.
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