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According to the Vieta's Formula, we know that -b/a in the question is equal to the sum of the roots.

Plug in-> -7/2 is a+b which is the answer

The chance of the 1st selection greater than 3 is 6/10 or 3/5

The chance of the 2nd selection greater than 3 is 5/9

The chance of the 3rd selection greater than 3 is 4/8 or 1/2

The chance of the 4th selection greater than 3 is 3/7

The chance of 5th selection greater than 3 is 2/6 or 1/3

Multiply all chances together we get: 1/42

So, the answer is 1/42 chance the 5 numbers selected are greater than 3

Question:

My attempt:

\(\begin{array}{|rcll|} \hline \vec{M} &=& \frac{\vec{A}+\vec{C}}{2} \\ \vec{N} &=& \frac{\vec{B}+\vec{D}}{2} \\ \hline \vec{M}\vec{N} &=& \frac{(\vec{A}+\vec{C})}{2}* \frac{(\vec{B}+\vec{D})}{2} \\ 4\vec{M}\vec{N} &=& (\vec{A}+\vec{C}) (\vec{B}+\vec{D}) \\ 4\vec{M}\vec{N} &=& \vec{A}\vec{B} +\vec{A}\vec{D}+\vec{C}\vec{B}+\vec{C}\vec{D} \\ \hline \end{array}\)

Hi, can you explain the property of the telescoping sum, I don't really understand it. Thanks! 

GS

\(\sqrt{2*5^0},\sqrt{2*5^1},\sqrt{2*5^2},\sqrt{2*5^3},\\ \sqrt{2*5^4}, \sqrt{2*5^5}, \sqrt{2*5^6}=5^3\sqrt{2}\)

In base 10 there are 10 distinct digits    0,1,2,3,4,5,6,7,8,9

In base 3 there are 3 distinct digits  0,1,2

In base 27 there are 27 distinct digits

Now you can answer yourself

You're welcome. :))

=^._.^=

Plug in x as 2, and solve for k. Once you get k, factor the equation to find the other 2 roots. :))

=^._.^=

Given that the polynomial x^2-kx+160 has only positive integer roots, find the average of all distinct possibilities for k.

Hello Guest!

\(x^2-kx+160\\ x=-\frac{(-k)}{2} \pm \sqrt{(\frac{k}{2})^2-160}\\ [(\frac{k}{2})^2-160]\in\{Square\ numbers \}\\ [(\frac{k}{2})^2-160]\in\{  1 , 4,  9, 16,  25,  36 ,  49 , 64 , 81\\ , 100,  121 , 144 , 169, 196 , 225 ,  256 , 289 , 324 ,  361 , 400 , 441 , 484 , 529 ,  576 , 625 \}\)

\([(\frac{k}{2})^2-160]\in \{\)9,    36,   65,   96,   129, 164, 201,  240, 281,  324, 369,  416,  465\(\}\)  

      \(k\in\{\)              \(2\sqrt{196},\)                                                       \(2\sqrt{484}\ \}\)

      \(\color{blue} k\in\{\)                    28,                                                                44   \(\}\)

  \(x=-\frac{(-k)}{2} \pm \sqrt{(\frac{k}{2})^2-160}\\ x1=\frac{28}{2} \pm \sqrt{(\frac{28}{2})^2-160}=14 \pm6 \\ \color{blue} x_1=8\\ \color{blue} x_2=20\\ x_3=\frac{44}{2} \pm \sqrt{(\frac{44}{2})^2-160}=22\pm 18\\ \color{blue} x_3=4\)

  \(x_4=40\)

  !

since its a `like terms` thing and its only asking you for $x$ in the power of 2; and how do we get $x^2$? by multiplying two $x$'es right?

thus, you have yo multiply all the variables who have got an $x$ in the power of one in them:

therefore, the figure you are looking for is $-42x^2$

also you have got the $-5x^2\times 1$ as well as the $1\times -x^2$ -- add those to $-42x^2$ and you will have an answer.

first of all we are given that the roots $\in \mathbb{Z}^{++}$. going easy on this, by using vietas formula we know that the sum of the roots of that polynomial is $\frac{16}{a}$, and the product of them is $\frac{-(-k)}{a}$ where $a=1$ right?

so, this polynomial is such that it has got these three solutions: $(16,1) \ \wedge\ (8,2) \ \wedge\ (4,4)$ , therefore the sum of all those would give us $k \{ 17 \ \wedge\ 10 \ \wedge\ 8 \}$

adding them alltogether we get $35$ which when divided by three gives us $11. \overline{6}  $ or more correctly $11\frac{2}{3}$rds

By the Pythagorean Theorem, the area of the circular base is 356*pi square feet.

.25 (4) +1.0(x)= .5(4+x)       solve for x gallons of pure

Thank you so much. I've been trying to figure this out for a while.

2x + 7 does not equal 3x + 1, however for some "x" it does. 

2x + 7=3x +1

2x + 6 = 3x

6 = x

=^._.^=

We can write -4 in exponential notation as 4e^(pi*i), so the equation is z^4 = 4e^(pi*i).

By Hamilton's Theorem, the solutions are z = 4^{1/4}*e^(pi*i/4), 4^{1/4}*e^(pi*i/4 + pi/4), 4^{1/4}*e^(pi*i/4 + 2*pi/4), and 4^{1/4}*e^(pi*i/4 + 3*pi/4).  Since 4^{1/4} = sqrt(2) and e^(pi*i/4) = (1 + i)/sqrt(2), the first solution is 1 + i.  Then the other roots work out as

4^{1/4}*e^(pi*i/4 + pi/4) = 1 - i,

4^{1/4}*e^(pi*i/4 + 2*pi/4) = -1 - i, and

4^{1/4}*e^(pi*i/4 + 3*pi/4) = -1 + i.

By the property of telescoping sum, we have quite easily that \dfrac{1}{a_n} = 2-\sum_{k=0}^{n-1}\dfrac{1}{n+a_k}.

First we show that a_n<1 which is analogous to proving that \sum_{k=0}^{n-1}\dfrac{1}{n+a_k}<1. Now, we note that \{a_k\} is an increasing sequence, hence a_k\geq a_0 for all k\geq0. This gives \sum_{k=0}^{n-1}\dfrac{1}{n+a_k}\leq\dfrac{n}{n+a_0}=\dfrac{2n}{2n+1}<1 and thus we are done.

Now we prove the other part of the inequality. We note that \sum_{k=0}^{n-1}\dfrac{1}{n+a_k}\geq\dfrac{n}{n+a_n} because of increasing property of the sequence a_n. Now using the fact that \dfrac{1}{a_n}=2-\sum_{k=0}^{n-1}\dfrac{1}{n+a_k} we have, after some algebra that 2a_n^2+(n-1)a_n-n\geq0 implying, and keeping in mind that \{a_k\} is a positive sequence, a_n\geq\dfrac{-(n-1)+\sqrt{(n-1)^2+8n}}{4}. It remains to show that this huge quantity is greater than 1-\dfrac{1}{n}. By squaring both sides, and cancelling out terms, we come to 3n>2 which is true for any n. Hence, 1/(a_1 + 1) + 1/(a_2 + 1) + ... + 1/(a_n + 1) < 2.

p(1,-1) = 1 - (-2) = 1 + 2 = 3

p(-5, -2) = -5 - (-4) = -5 + 4 = -1

p(3, -1) = 3 - (-2) = 3 + 2 = \(\boxed{5}\)

Here's the link to the same problem that's been answered:

By the Binomial Theorem

(3 + sqrt(5))^4

= 3^4 + C(5,1)*3^3*sqrt(5) + C(5,2)*3^2*sqrt(5)^2 + C(5,3)*3*sqrt(5)^3 + sqrt(5)^4

= 360 + 144*sqrt(5)

Let the number of friend that Braydon have = x

Case 1: When he given 9 stickers to each friend.

Since he gave 9 stickers to each friend,

∴ Number of distributed stickers = 9x

While distributing 9 stickers to each friend he will be short for 69 stickers.

∴ The number of stickers Braydon actually have = 9x-69

Case 2 : When he given 4 stickers to each friend.

Since he gave 4 stickers to each friend,

∴ Number of distributed stickers = 4x

While distributing 4 stickers to each friend he will have 46 more tickets.

∴ The number of stickers Braydon actually have = 4x+46

Since Braydon have equal number of tickets in each case,

∴ 9x-69 = 4x+46

⇒ 9x-4x = 46+69

⇒ 5x = 115

⇒ x = 23

Hence there are 23 friends.

Number of stickers = 4x+46 = 4(23)+46 = 138

Now we know that there are 138 tickets in total and 23 friends.

Hence the number of tickets each friend will get such that no sticker remains = \({138 \over 23}{}{} = 6\)

Hence Braydon must given 6 tickets to each of his friend such that no sticker is left.

All you need to do is this:

140 = 12.5( 0.2^t -1) + 20t            can you solve this for 't' ?

I am having trouble understanding your post..... did part of it get left out?