Respuesta :
Answer:
Step-by-step explanation:
Given
Height of cone [tex]h=10\ in.[/tex]
radius of cone [tex]r=3\ in.[/tex]
Volume of Kyle cone
[tex]V_k=\frac{1}{3}\pi r^2h[/tex]
[tex]V_k=\frac{1}{3}\pi (3)^2(10)=30\ pi \ in.^3[/tex]
For cylinder A
[tex]h=7\ in.[/tex]
[tex]r=2\ \in.[/tex]
Volume of cylinder [tex]V_A=\pi r^2h[/tex]
[tex]V_A=\pi (2)^2(7)=28\pi \ in.^3[/tex]
So, [tex]V_a<V_k[/tex]
Volume of cylinder [tex]V_B=\pi r^2h[/tex]
[tex]h=30\ in \ r=1\ in.[/tex]
[tex]V_B=\pi (1)^2(30)=30\pi \ in.^3[/tex]
So, [tex]V_B=V_k[/tex]
Volume of cone [tex]V_C=\frac{\pi}{3} r^2h[/tex]
[tex]h=6\ in.\ r=4\ in.[/tex]
[tex]V_A=\frac{1}{3}\times \pi (4)^2(6)=32\pi \ in.^3[/tex]
So, [tex]V_C>V_k[/tex]
For sphere D
[tex]r=3\ in.[/tex]
Volume [tex]V_D=\frac{4}{3}\pi r^3[/tex]
[tex]V_D=\frac{4}{3}\times \pi (3)^3=36\ \pi \ in.^3[/tex]
So, [tex]V_D>V_k[/tex]
For Sphere E
[tex]r=2\ in.[/tex]
[tex]V_D=\frac{4}{3}\times \pi (2)^3=10.66\ \pi \ in.^3[/tex]
[tex]V_E<V_k[/tex]
