thedudemanguyperson

$AP=BQ\implies AP=PQ\sqrt{2}$ is desired. Follows trivially from LoC and coordinates as $\angle ACP=\angle BCP=45^{\circ}$. Can't think of anything synthetic...

$\overline{AB}=4-0=4$

$\overline{BC}=x-0=x$

Area = $\overline{AB}\cdot\overline{BC}=4x=28\implies \boxed{x=7}$.

Since $dist((-2,2), (2,-2))=\sqrt{4^2+4^2}=4\sqrt{2}$, the area is $(4\sqrt2)^2=\boxed{32}$.

Equivalent to https://web2.0calc.com/questions/finding-the-semi-major-axis-on-an-ellipse

Inversely proportional, i.e. $\boxed{\frac{3}{2}}$. (reciprocal of two thirds)

The sides are being multiplied by $5$, so $3x+8=4*5=20\implies 3x=12\implies\boxed{x=4}$.

There are $\binom{12}{5}$ ways to choose all the balls.

There are $\binom{3}{2}$ ways to catch the red ball, and $\binom{3}{1}$ ways to catch each of the other ones.

The answer is $\frac{\binom{3}{2}\left(\binom{3}{1}\right)^3}{\binom{12}{5}}=\frac{81}{792}=\boxed{\frac{9}{88}}$.

It's obviously tangent at $(k/2, k/2)$. So $\frac{k^2}{2}=2k+1\implies k^2=4k+2\implies k^2-4k-2=0\implies k=2\pm\sqrt{6}$. We take positive root, so $\boxed{k=2+\sqrt{6}}$.

At $3$, it must be continuous. So $2*3+a=3+2\implies \boxed{a=-1}$.

Let's look at the right hand side: $\frac{2x(x-1)-6(x+1)}{x^2-1}=\frac{2x^2-2x-6x-6}{x^2-1}=\frac{2x^2-8x-6}{x^2-1}$. So $2x^2-8x-6=-9x\implies 2x^2+x-6=0$, so the sum of the solutions is $\boxed{-\frac12}$ by Vieta's formulae.