The answer is 15 in³.
The volume of the cone is:
[tex]V_1= \pi r_1 ^{2} \frac{h_1}{3} = \frac{ \pi r_1^{2} h_1}{3} [/tex]
where:
V₁ - the volume of the cone
r₁ - the radius of the cone
h₁ - the height of the cone
The volume of the cylinder is:
[tex]V_2= \pi r_2h_2^{2} [/tex]
where:
V₂ - the volume of the cone
r₂ - the radius of the cone
h₂ - the height of the cone
Since the cone fits exactly inside of the cylinder, they have the same radius and the height:
r₁ = r₂
h₁ = h₂
Also:
V₁ = 5
Now, let's write two volume formulas together:
[tex]V_1= \frac{ \pi r^{2} h}{3} [/tex]
[tex]V_2= \pi rh^{2} [/tex]
We can include V₂ into V₁:
[tex]V_1= \frac{V_2}{3} [/tex]
⇒ [tex]V_2=3*V_1[/tex]
[tex]V_2=3*5 in^{3} [/tex]
[tex]V_2=15 in^{3} [/tex]
