thedudemanguyperson

Draw a venn diagram. Let there be $x$ students taking only chemistry and $y$ taking only biology. Then $(x+8)=3(y+8)$ and $x+y+8=43$. Now solve this, and the answer will be whatever $x+8$ is.

I am assuming you want $\theta$ not $0$.

We have that $\cos\theta=\frac{-2\sqrt{2}}{4}=-\frac{\sqrt{2}}{2}$. Use the fact that $\cos^{-1}\left(-\frac{\sqrt{2}}{2}\right)=\frac{3\pi}{4}$ and the fact that the period of $\cos x$ is $2\pi$ to finish.

I will rephrase it:

Given $x+y+z=5, x,y,z\ge 0$, find the probability that $x,y,z\le 2$.

Put $x=2-a, y=2-b, z=2-c$ Then put $a=t_1+\frac13$, etc.

It is now, $t_1+t_2+t_3=0$, $t_1, t_2, t_3\in [-10/3, 5/3]$, find probability that $t_i\in [-1/3, 5/3]$ for all $i$. (this is because $a,b,c\in [-3,2]$ and we want $a,b,c\in [0,2]$ given $a+b+c=1$, it's simply bounding). 

Now we do casework:

Case 1: There are 2 negatives and 1 positive. WLOG $t_1\le t_2\le t_3$. Then $t_1+t_2=-t_3$. and $t_3$ is positive. So we need $t_1+t_2\ge -\frac53$ given that they are in $[-10/3, 0]$. We can graph this letting $t_1=x, t_2=y$. Find it here: https://www.desmos.com/calculator/32dlklyibx. We then note that the area of the blue region is $\left(\frac{10}{3}\right)^2$ and the area of the red region in the blue region is $\frac{1}{2}\left(\frac{5}{3}\right)^2$. Using the principle of geometric probability, the probability here is $\frac{1}{8}$.

Case 2: There is 1 positive and 2 negatives. This is the same as the last case (note this by letting $t_i=-\alpha_i$).

All other cases are negligible since they have a probability of $0$ of occurring.

The answer is then $\frac{1}{8}+\frac{1}{8}=\boxed{\frac{1}{4}}$.

Beautiful problem. +1

Surface area is $2\pi r(h+r)=12\implies rh+r^2=\frac{6}{\pi}$.

We want to maximize $\pi r^2 h$. 

Note that $\lcm(a,b)\cdot \gcd(a,b)=ab$. Note $ab$ has less than $10^6\cdot 10^6=10^{12}\to 12$ digits. And $\frac{\text{12 digits}}{\text{10 digits}}$ has less than $\frac{10^{11}}{10^{8}}=10^{3}$ digits. The answer is then $2$.

I don't think there is one.

It's much easier just to visualize it, as you did. With a square root in the denominator, there are always sign issues. 

For your first question, they did it for the sake of convenience (so that we could have signs on $\sin(2\alpha)$ and $\cos(2\alpha)$.)

Graph it: https://www.desmos.com/calculator/9rbnk7rx1s

The trapezoid is a triangle + a rectangle. The probability is $\frac{\text{area of trapezoid}}{3*6}$

Make it a rectangle and a triangle and use Pythagorean theorem to find that the top part is $7$. The perimeter is $7+13+24+13+25=\boxed{82}$.

Forget about the girl/boy condition. We simply choose $2$ players from a group of 10, so the answer is $\binom{10}{2}=\boxed{45}$.

$x=-7$, so $49a-7b+c=0$.