thedudemanguyperson

Clearly we'll have an equation of the form $ax+b=ax+c$. So $y=-4$, so $-4x+4x=1$, impossible.

Ans: $\boxed{-4}$.

a) $\frac{15}{12}=\frac{5}{4}$

b) $12\left(\frac{4}{5}\right)^3=\frac{768}{125}$

c) $12*\left(\frac{5}{4}\right)^{16}\approx 426.32$

$5\ge -3\implies  f(5)=7+4*5=\boxed{27}$

$2x^2-x+11=0$

$r^2+s^2=(r+s)^2-2rs$

By Vieta: $r+s=\frac{1}{2}, rs=\frac{11}{2}$

Answer: $\frac{1}{4}-11=\boxed{\frac{-43}{4}}$.

Firstly, you can always put it on a grid - use coordinates.

My Solution is synthetic, and uses the mass points method:

Suppose $C$ has mass $1$. We get $A$ to have mass $1$ as well. Therefore, $B$ has mass $2$, and $D$ has mass $2$. Since $A$ has mass $1$ and $D$ has mass $2$, $FD=\frac{1}{2}AF\implies FD=\frac{1}{3}AD$, or the height of $\triangle FDE$ is $\frac{1}{3}$ of the height of the big triangle. The base is $\frac{1}{2}$ of the big triangle. Therefore, the area is $\frac{1}{2}*\frac{1}{3}=\frac{1}{6}$ of the big triangle. The big triangle has area $\frac{1}{2}\cdot 14\cdot 18=126$, so the answer is $\boxed{21}$.

It's pretty clear that $\boxed{x=1}$ works.

$(a+c+e)+(b+d+f)i=-3i\implies (a+c-a-c)+(7+d+f)i=-3i\implies (d+f+7)i=-3i\implies d+f+7=-3\impies d+f=-\boxed{-10}$.

Here we go.

a) $\frac{17\pi}{5}$ is in the third quadrant, and the angle is $\frac{7\pi}{5}-\pi=\boxed{\frac{2\pi}{5}}$.

b) $614$ is in the third quadrant, and the angle is $614-360-180=\boxed{74}$

c) $\frac{21\pi}{4}$ is in the first quadrant, so the angle is $\boxed{\frac{\pi}{4}}$

d) $240$ is in the third quadrant, so the angle is $240-180=\boxed{60}$.

$18^2=18*18=324$

$324-140=\boxed{184}$.