thedudemanguyperson

Incorrect.

I'll leave you to find the correct answer with the fact that reflecting $(x,y)$ across $y=-x$ yields $(-y,-x)$. 

$x=1\pmod{3}, x=2\pmod{4}, x=1\pmod{5}$.

If $x=1\pmod{3,5}$, $x=1\pmod{\text{lcm}(3,5)=15}$.

Now $x=15k+1=4j+2$. Taking the equation modulo $4$, $1-k=2\pmod{4}\implies k=3\pmod{4}$. So minimum $x$ is $15*3+1=45+1=\boxed{46}$.

Nice one.

$(3x-12)^2+(ax-4)^2=16\implies 9x^2-72x+144+a^2x^2-8ax+16=16\implies (9+a^2)x^2-(72+8a)x+144=0$.

Discriminant is zero, so $(72+8a)^2=4(144)(9+a^2)\implies a=\boxed{\frac{9}{4}}$.

I will avoid using the Lagrange Interpolation Formula here because it is insanely overkill.

Using the method of undetermined coefficients the answer is $\boxed{1}$.

Another way is to look at finite differences and take the last number.

There is no diagram, so there is no problem!! 

$\angle XYV=\angle WYV\implies \triangle WYV\sim\triangle VYX\implies \frac{WY}{VY}=\frac{YV}{YX}$. First option is correct.

(a) $\angle CBA=\frac{1}{2}(180-36)=72\implies \angle ABD=36$ implying $BD=AD$. Now $\angle BDA=180-72=108\implies \angle BDC=72, \angle DCB=180-38-72=72$, implying the result.

(b) $\frac{BC}{AB}=\frac{CD}{BC}\implies \frac{x}{AB}=\frac{y}{x}\implies x^2=AB\cdot y$. From (a), $AD=BC=x$, and from problem, $AB=BC=x+y$ as well. So $x^2=(x+y)y$. 

(c) $\frac{x^2}{y}=x+y\implies \frac{x}{y}=1+\frac{y}{x}\implies r+1=\frac{1}{r}\implies r^2+r-1=0\implies r=\frac{\sqrt{5}-1}{2}$.

(d) By the law of cosines in $\triangle BCD$, $y^2=2x^2-2x^2\cos(36)\implies y^2=2x^2(1-\cos(36))\implies 1-\cos(36)=\frac{1}{2}\left(\frac{y}{x}\right)^2=\frac{r^2}{2}$. The rest is easy. $\cos(36)=\frac{\sqrt{5}+1}{4}$ and double angle yields $\cos(72)=\frac{\sqrt{5}-1}{4}$.

Done  

The side length is $\frac{6}{\sqrt{3}}=2\sqrt{3}$. And the area is $6\sqrt{3}$ from there. The circumradius is then $\frac{2\sqrt{3}}{\sqrt{3}}=2$. The inradius is equal to the area divided by the semiperimeter, which is $1$. The result is $4\pi-\pi=\boxed{3\pi}$.

The distance between $(1,2)$ and $(7,-4)$ is $\sqrt{6^2+6^2}=6\sqrt{2}$ by the distance formula. This is the length of the diagonal of the square. Recall that the diagonal is equal to $\sqrt{2}\cdot\text{side length}$. Therefore, $\text{side length}=6$, and $\text{Area}=6^2=\boxed{36}$.