Question

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Answer 1

Answer:

Step-by-step explanation:

Given

Height of cone $h=10\ in.$

radius of cone $r=3\ in.$

Volume of Kyle cone

$V_k=\frac{1}{3}\pi r^2h$

$V_k=\frac{1}{3}\pi (3)^2(10)=30\ pi \ in.^3$

For cylinder A

$h=7\ in.$

$r=2\ \in.$

Volume of cylinder $V_A=\pi r^2h$

$V_A=\pi (2)^2(7)=28\pi \ in.^3$

So, $V_a$

Volume of cylinder $V_B=\pi r^2h$

$h=30\ in \ r=1\ in.$

$V_B=\pi (1)^2(30)=30\pi \ in.^3$

So, $V_B=V_k$

Volume of cone $V_C=\frac{\pi}{3} r^2h$

$h=6\ in.\ r=4\ in.$

$V_A=\frac{1}{3}\times \pi (4)^2(6)=32\pi \ in.^3$

So, $V_C>V_k$

For sphere D

$r=3\ in.$

Volume $V_D=\frac{4}{3}\pi r^3$

$V_D=\frac{4}{3}\times \pi (3)^3=36\ \pi \ in.^3$

So, $V_D>V_k$

For Sphere E

$r=2\ in.$

$V_D=\frac{4}{3}\times \pi (2)^3=10.66\ \pi \ in.^3$

$V_E$