[tex]\bf \textit{volume of a cone}\\\\
V=\cfrac{\pi r^2 h}{3}\qquad
\begin{cases}
r=radius=\frac{diameter}{2}\\
h=height\\
----------\\
diameter=3\\
r=\frac{3}{2}\\
V=12
\end{cases}\implies 12=\cfrac{\pi \left( \frac{3}{2} \right)^2 h}{3}[/tex]
solve for "h"
