thedudemanguyperson

Let $b\sqrt{c}=\sqrt{d}$

So $(a+\sqrt{d})^2=79+24\sqrt{7}\implies a^2+d+2a\sqrt{d}=79+24\sqrt{7}$. Let's try $d=7$, then $a^2+7=79, 2a=24$, which doesn't work. Now trying $\sqrt{d}=3\sqrt{7}$, then $d=63$ and $a^2=16, a=4$, which indeed works. Then $a+\sqrt{d}=4+\sqrt{63}=4+3\sqrt{7}$ and $abc=12*7=\boxed{84}$.

$\frac{75}{108+75+90+54}=\frac{25}{109}$

So it's $\frac{25}{109}*360\approx \boxed{83^{\circ}}$

Since $\mathbb{A}^{-1}=\frac{1}{det\mathbb{A}}adj(\mathbb{A})$, so its invertible whenever the determinant is nonzero.

The sum of the coefficients of $f(x)$ is $f(1)$. So put $x=1$ up there and see what you get.

$\frac{5850-3350}{3350}$ plug it into your calculator.

WLOG $g\equiv 0$. $3+0=\boxed{3}$.

$(x^4-3x+2)^2$

Upon reduction to $(x^4+x+1)^2$ we see $x^3$ is impossible to get so the answer is $\boxed{0}$.

Clearly $-2-x\ne x^3+2x+2$ for most $x$. Fix the problem please.

You can fill in the dashed boxes. Then solve the sudoku. There are sudoku solvers online if you are lazy :P

Substitute $x=0$ and find what you get.