thedudemanguyperson

3 out 5 voted for Tommy. This is equal to 9 out of 15 (copy the ratio three times). So 15 voted in total.

I think the link has malware.

Let the side length of the base be $s$. Then $\frac{1}{3}s^2\cdot 30=63960$. Solve for $\sqrt{s^2+30^2}$.

Yep, believe so :)

If $a-b=1$ there are $10, 21, 32, ..., 98=9$ integers.

If $a-b=2$ then we have $20, 42, 64, 86=4$ integers.

If $a-b=3$, then we have $30, 63, 96=3$ integers.

If $a-b=4$, then we have $40, 84=2$ integers.

If $a-b=5$, then we have $50=1$ integer.

If $a-b=6$, then we have $60$.

If $a-b=7$, we have $70$, if $a-b=8$ we have $80$, if $a-b=9$ we have $90$.

So it's $9+4+3+2+1+1+1+1+1=\boxed{23}$. 

We have:

$c=\frac12 e$, $8(c-36)=e$. Now solve for $e+c$.

The point is something like $(x,\frac{1}{2}x^2-9)$ for some constant value of $x$. The distance between this and the origin is $a=\sqrt{x^2+(\frac{1}{2}x^2-9)^2}$, and so $a^2=x^2+(\frac{1}{2}x^2-9)^2=x^2+\frac{1}{4}x^4-9x^2+81=\frac{1}{4}x^4-8x^2+81=\frac14(x^4-32x^2+324)$ and factor the inside as $(x^2+a)^2+b$ which has minimum value of $b$. Then multiply by $\frac14$ and finish. :)

It's equal to $(3^{33}+7^{54})(8^{326}+(-1)^{2016})\pmod{10}$.

$3,7,8$ all cycle modulo $10$ and $(-1)^{2016}$ is trivial to deal with.

Hint: The $y$ coordinate turns into $4-(y-4)=8-y$. Now put $y\mapsto 8-y$ in your equation and expand :)

Consider the circle centered at $(2,2)$ with radius $1$. P must lie in this circle, inside the square. This is a quarter of the circle, which you can find the area of. Then find the area of the square and use geometric probability principle.