## Algebra: A Combined Approach (4th Edition)

$15\pi$ cubic inches
Let $V$ be the volume of a cone. Let $h$ be the height of a cone. Let $r$ be the radius of a cone. $V=khr^2$ $V=32\pi; h=6; r=4$ $32\pi =k(6)(4)^2$ $k=\frac{\pi }{3}$ $V=\frac{\pi }{3}hr^2$ $h=5; r=3$ $V=\frac{\pi }{3}(5)(3)^2$ $V=15\pi$ cubic inches