## Algebra 2 (1st Edition)

$r=5,h=9$
The volume of the cone is:$\frac{1}{3}\pi r^2h$, $h=r+4$ Thus our equation is: $\frac{1}{3}\pi r^2(r+4)=75\pi\\r^2(r+4)=225\\(r-5)(r^2+9r+45)=0$ The second term has a negative determinant and thus never equals 0. Thus, the only valid zero for the function is $r=5$. Hence the dimensions are $r=5,h=9$.