All Answers 355949 Answers

The probability that you get those exact three is $\frac{1}{\binom{20}{3}}$. The probability that you get the SuperBall is $\frac{1}{10}$. Multiply them.

Factoring the denominator:

$$\frac{x^2+5x-\alpha}{(x+11)(x-4)}$$

Thus $(x+11)\mathrm{ or }(x-4)|(x^2+5x-\alpha)$.

Hence either $x^2+5x-\alpha=(x+11)(x+t)$ or $x^2+5x-\alpha = (x+t)(x-4)$.

The first case, sum of roots is $-5=-11-t\implies 5=11+t\implies t=-6$ and hence $(x+11)(x-6)=x^2+5x-66$ thus $\alpha = 66$. 

The second case, sum of roots is $-5=4-t\implies t=9$ and hence $\alpha=36$.

The answer is then $66+36=102$.

Using binomial distribution it's $\frac{\binom{5}{4}}{2^5}$

We want:

$$\log \binom{10}{0}\binom{10}{1}\cdots \binom{10}{10}=\log \binom{10}{0}^2\binom{10}{1}^2\binom{10}{2}^2\binom{10}{3}^2\binom{10}{4}^2\binom{10}{5}=\log (1\cdot 10^2\cdot 45^2\cdot 120^2\cdot 210^2\cdot 252)=\log(2^12\cdot 3^10\cdot 5^8\cdot 7^3)\approx 16.51$$

The question is wrong, no integer solutions exist to $2^n\equiv 5\pmod{15}$.

$2x+5y$ isn't a line.

Start by squaring both sides. 

2sqrt(t-2) = sqrt(7-t)

sqrt(4t - 8) = sqrt(7-t)

4t - 8 = 7 - t

5t = 15

t = 3

=^._.^=

I'll solve for $a,b,c$?

Multiplying them all together, $a^4b^4c^4=2^2\cdot 3^4\cdot 7^2\implies a^2b^2c^2=2\cdot 9\cdot 7$. So $c^2=\frac{a^2b^2c^2}{a^2b^2}=\frac{9}{2}$, now plug into eqn 3 to find a and then plug into eqn 1 to find b.

this question seems to be incomplete. what's the missing text?

$a\cdot 2^r=1$, $a\cdot 32^r=16$. Dividing the two, $\frac{32^r}{2^r}=16^r=16\implies r=1$.

$\omega$ is a cube root of unity so write it in exponential form everywhere. This problem is easy to guess as well.

$(1111111+909091)(1111111-909091)=(2\cdot 10^6+2\cdot 10^4+2\cdot 10^2+2\cdot 10^0)(2\cdot 10^5+2\cdot 10^3+2\cdot 10)=4(10^6+10^4+10^2+10^0)(10^5+10^3+10)=4(10^{11}+10^{9}+10^{7}+10^9+10^7+10^5+10^7+10^5+10^3+10^5+10^3+10)=4(10^{11}+2\cdot 10^9+2\cdot 10^7+2\cdot 10^5+2\cdot 10^3+10)=4(102020202010)=408080808040$.

Don't be scared.

It's $1+i\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...\right)=1+i$.

$x^2-3x-80=0$. You know that $r=-\frac{3}{2}$ so to find $s$ subtract $\left(x-\frac{3}{2}\right)^2$ from the LHS.

It's $8x^2-8x-5$.  So $a=8$, then factor it out and complete the square to find $h,k$.

Let the side length be $s$. Draw the height and draw the bottom point connected to two vertices of the base. Because the height hits the bottom at the centroid, the length of each of those legs are half the height of the triangles, which is $\frac{s\sqrt{3}}{4}$. This forms a right angle with the height, and the hypotenuse is $s$. Using the Pythagorean theorem: $\sqrt{\left(\frac{s\sqrt3}{4}\right)^2+20^2}=s$. The end is near.

Consider the following scenario: You have 5 baskets. At each basket you can pick anywhere from 0 to 4 apples. How many ways can you pick 3 apples? The function given is the generating function for this question, and so it suffices just to solve the scenario.

$(0,0,1,1,1), (0,0,0,1,2), (0,0,0,0,3)$ are the only ones that work. So the answer is $\frac{5!}{2!3!}+\frac{5!}{3!1!1!}+\frac{5!}{4!1!}=10+20+5=\boxed{35}$.

Alternatively:

See if you can find a pattern in @heureka's answer -- prove it.

Use the fact that $\gcd(a,b)lcm(a,b)=ab$. So $\gcd(n,216)lcm(n,216)=216n=24\cdot 864$. So $n=\frac{24\cdot 864}{216}$.

$\begin{align}e^{17\pi i/60}+e^{27\pi i/60}+\cdots+e^{57\pi i/60}&=e^{17\pi i/60}\left(1+e^{\pi i/6}+\cdots+e^{4\pi i/6}\right)\\&=e^{17\pi i/60}\frac{1-e^{5\pi i/6}}{1-e^{\pi i/6}}\\&=e^{17\pi i/60}\left(\frac12\left(2+\sqrt3\right)+\frac12\left(3+2\sqrt3\right)i\right)\\&=\left(2+\sqrt3\right)e^{17\pi i/60}\left(\frac12+\frac{\sqrt3}2i\right)\\&=\left(2+\sqrt3\right)e^{17\pi i/60}e^{\pi i/3}\\&=\left(2+\sqrt3\right)e^{37\pi i/60}.\end{align}$

2(4) + 2 - 16 = -6

So the remainder when 2a + b - c is divided by 19, is -6, or 13. (19-6)

=^._.^=

.0000    LOL         Show WHAT to 4 decimal places? 

3 x 5 = 15      so all multiles of 15 are marked with two  = 10

3*7    = 21          all multiples of 21 (105 already marked)    = 6

5x7 = 35            35 70 105  and 140  (but 105  already marked)  = 3            total 19

What is the coefficient of \(\color{red}x\) in \((x^4 + x^3 + x^2 + x + 1)^5\)?

What is the coefficient of \(\color{red}x\) in \((x^4 + x^3 + x^2 + x + 1)^5\)?
What is the coefficient of \(\color{red}x^3\) in \((x^4 + x^3 + x^2 + x + 1)^5\)?

\((x^4 + x^3 + x^2 + x + 1)^5 =(1 + x^2 + x^3 + x^4)^5\)

\(\begin{array}{l} \mathbf{(1 + x^2 + x^3 + x^4)^2} \\ =(1 + x^2 + x^3 + x^4)(1 + x^2 + x^3 + x^4) \\ \end{array}\\ \begin{array}{|lcccccccccc|} \hline &=& &1&+&x &+&x^2 &+&x^3+\dots \\ && & &+&x &+&x^2 &+&x^3+\dots \\ && & & & &+&x^2 &+&x^3+\dots \\ && & & & && &+&x^3+\dots \\ \hline &=& &1&+&2x&+&3x^2&+&4x^3+\dots \\ \hline \end{array} \)

\(\begin{array}{l} \mathbf{(1 + x^2 + x^3 + x^4)^3} \\ =(1 + 2x+3x^2+4x^3 +\dots)(1 + x^2 + x^3 + x^4) \\ \end{array}\\ \begin{array}{|lcccccccccc|} \hline &=& &1&+&x &+&x^2 &+&x^3+\dots \\ && & &+&2x &+&2x^2 &+&2x^3+\dots \\ && & & & &+&3x^2 &+&3x^3+\dots \\ && & & & && &+&4x^3+\dots \\ \hline &=& &1&+&3x&+&6x^2&+&10x^3+\dots \\ \hline \end{array}\)

\(\begin{array}{l} \mathbf{(1 + x^2 + x^3 + x^4)^4} \\ =(1 + 3x+6x^2+10x^3 +\dots)(1 + x^2 + x^3 + x^4) \\ \end{array}\\ \begin{array}{|lcccccccccc|} \hline &=& &1&+&x &+&x^2 &+&x^3+\dots \\ && & &+&3x &+&3x^2 &+&3x^3+\dots \\ && & & & &+&6x^2 &+&6x^3+\dots \\ && & & & && &+&10x^3+\dots \\ \hline &=& &1&+&4x&+&10x^2&+&20x^3+\dots \\ \hline \end{array}\)

\(\begin{array}{l} \mathbf{(1 + x^2 + x^3 + x^4)^5} \\ =(1 + 4x+10x^2+20x^3 +\dots)(1 + x^2 + x^3 + x^4) \\ \end{array}\\ \begin{array}{|lcccccccccc|} \hline &=& &1&+&x &+&x^2 &+&x^3+\dots \\ && & &+&4x &+&4x^2 &+&4x^3+\dots \\ && & & & &+&10x^2 &+&10x^3+\dots \\ && & & & && &+&20x^3+\dots \\ \hline &=& &1&+&{\color{red}5}x&+&15x^2&+&{\color{red}35}x^3+\dots \\ \hline \end{array}\)