@ElectricPavlov Nice solution! However, I believe the asker requested the answer in "simplest radical form".
Moving everything inside the outer radical into the inner radical, we get $\sqrt[3]{\sqrt[3]{x^4}}$. Simplifying square roots, you get $\boxed{\sqrt[9]{x^4}}$
Thank you! Your answer was actually correct!
Maybe this is a bug, but the LaTeX renders just fine for me...
$3(8x^2-5x+1)=8x^2(3)-5x(3)+1(3)=\boxed{24x^2-15x+3}$
$2r(4r-8+3)=2r(4r-5)=4r(2r)-5(2r)=\boxed{8r^2-10r}$
$\frac{1in}{0.51mi}=\frac{x}{2.8mi}$
$0.51x=2.8(1)$
$x=\frac{2.8}{0.51}$
$x=$about$\boxed{5.5}$
It must be rotated $\frac{360}{5}=\boxed{72^{\circ}}$
If it is perpendicular to $y=x+8$, the slope must be $-1$, so the equation for the line is $y=-x+c$ for some c. Substituting $(4,-10)$, we get $-10=-(4)+c$, so $c=-6$. The equation of the line is $\boxed{y=-x-6}$
