Mitchell poured the contents of a completely filled cone into an empty cylinder, and the cylinder became two-thirds full. the cylinder had a radius of 5 cm and a height of 15 cm.
what were possible dimensions of the cone? use 3.14 to approximate pi.
a.
h = 5 cm; r = 15 cm
b.
h = 7.5 cm; r = 10 cm
c.
h = 10 cm; r = 7.5 cm
d.
h = 15 cm; r = 5 cm

First, we define the volume of a cone as V= 1/3x Pi x r²xH and the volume of a cylinder as v= Pi x R² x h. So, according to the question we have V=2/3v=(2/3)xR²xPixh=(2/3) x 3.14x5²x15=785cm^3. V=785cm^3. So it is (1/3) x3.14xr²xh=785, this implies r=10, and h=7.5, because (1/3) x3.14x100x7.5=785. Finally, the dimensions of the cone are r=10, h=7.5 and V=785cm^3

1. 8 or -82. 14 or -143. -11, -3, 0, 7, 10