thedudemanguyperson

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 $n=\frac{24\cdot 864}{216}$.

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

Drop $C$ to $AB$ at $D$. Because of congruence, $AD=BD=3$. Then $CD=\sqrt{8^2-3^2}=\sqrt{55}$. So the area of the triangle is $3\sqrt{55}$.

Then the semiperimeter is 11. So $11r=3\sqrt{55}\implies r=\frac{3\sqrt{55}}{11}$, which is the inradius. The incircle area is then $\frac{45}{11}\pi$

The area is also $\frac{abc}{4R}$. So $\frac{96}{R}=3\sqrt{55}$ so $R=\frac{32}{\sqrt{55}}$. The area of the circumcircle is then $\frac{1024}{55}\pi$.

So the answer is $\frac{1024}{55}\pi - \frac{45}{11}\pi=\boxed{\frac{799}{55}\pi}$.

1:- slope is rate of change

2:- Notice that $\overline{PQ}=26$ by distance formula, hence the point is the midpoint, or $(15,24)$.

Just take $P(x)=(x-1)^3-4$.

First find a polynomial that $x^2=7+2\sqrt{10}$ is a root of by using the quadratic formula equation.