### Instruction: Two cones have the same volume. If one has a base with radius 3 times as large as the other's and a height of 24 inches, how many inches tall is the other?

Note: The volume of a cone is $\frac{1}{3} \pi r^2 h,$ where $r$ is the radius and $h$ is the height.### Response: I know that the volumes of the cones are equal, so I can set their formulas equal to each other and simplify.
Let $r_1$ and $h_1$ be the radius and height of the first cone, and $r_2$ and $h_2$ be the radius and height of the second cone.
Then I have $\frac{1}{3} \pi r_1^2 h_1 = \frac{1}{3} \pi r_2^2 h_2$.
I can cancel out the $\frac{1}{3} \pi$ terms on both sides, and get $r_1^2 h_1 = r_2^2 h_2$.
Now I can plug in the values that I know.
The problem says that $r_1 = 3r_2$ and $h_1 = 24$.
So I have $(3r_2)^2 \cdot 24 = r_2^2 \cdot h_2$.
Expanding the square, I get $9r_2^2 \cdot 24 = r_2^2 \cdot h_2$.
Dividing both sides by $r_2^2$, I get $9 \cdot 24 = h_2$.
Multiplying, I get $h_2 = 216$.
So the height of the other cone is 216 inches.