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Sagot :
Let's address the problem step-by-step using the formula for the volume of a cylinder, [tex]\( V = \pi r^2 h \)[/tex], where [tex]\( V \)[/tex] is the volume, [tex]\( r \)[/tex] is the radius, and [tex]\( h \)[/tex] is the height.
We'll solve for the missing heights first and then derive a general expression for the volume [tex]\( V \)[/tex] given a radius [tex]\( r \)[/tex] and height [tex]\( h \)[/tex].
### Step 1: Find the missing height for each cylinder
To solve for the height [tex]\( h \)[/tex], we rearrange the volume formula:
[tex]\[ h = \frac{V}{\pi r^2} \][/tex]
#### First Cylinder
Given:
- Volume, [tex]\( V = 96\pi \)[/tex] cubic inches
- Radius, [tex]\( r = 4 \)[/tex] inches
Substitute the given values into the rearranged formula:
[tex]\[ h = \frac{96\pi}{\pi \times 4^2} \][/tex]
[tex]\[ h = \frac{96\pi}{\pi \times 16} \][/tex]
[tex]\[ h = \frac{96\pi}{16\pi} \][/tex]
[tex]\[ h = \frac{96}{16} \][/tex]
[tex]\[ h = 6 \text{ inches} \][/tex]
So, the missing height for the first cylinder is [tex]\( 6 \)[/tex] inches.
#### Second Cylinder
Given:
- Volume, [tex]\( V = 31.25\pi \)[/tex] cubic inches
- Radius, [tex]\( r = 2.5 \)[/tex] inches
Substitute the given values into the rearranged formula:
[tex]\[ h = \frac{31.25\pi}{\pi \times 2.5^2} \][/tex]
[tex]\[ h = \frac{31.25\pi}{\pi \times 6.25} \][/tex]
[tex]\[ h = \frac{31.25\pi}{6.25\pi} \][/tex]
[tex]\[ h = \frac{31.25}{6.25} \][/tex]
[tex]\[ h = 5 \text{ inches} \][/tex]
So, the missing height for the second cylinder is [tex]\( 5 \)[/tex] inches.
### Step 2: General Expression for Volume [tex]\( V \)[/tex]
For a general cylinder with radius [tex]\( r \)[/tex] and height [tex]\( h \)[/tex], the volume [tex]\( V \)[/tex] is given by:
[tex]\[ V = \pi r^2 h \][/tex]
This formula remains the same as the original volume formula for a cylinder.
### Summary
[tex]\[ \begin{array}{|c|c|c|} \hline \text{Volume (cubic inches)} & \text{Radius (inches)} & \text{Height (inches)} \\ \hline 96\pi & 4 & 6 \\ \hline 31.25\pi & 2.5 & 5 \\ \hline V & r & \frac{V}{\pi r^2} \\ \hline \end{array} \][/tex]
So:
1. The height for the first cylinder is [tex]\( 6 \)[/tex] inches.
2. The height for the second cylinder is [tex]\( 5 \)[/tex] inches.
3. The volume [tex]\( V \)[/tex] for a radius [tex]\( r \)[/tex] and height [tex]\( h \)[/tex] is given by [tex]\( \ V = \pi r^2 h \)[/tex].
We'll solve for the missing heights first and then derive a general expression for the volume [tex]\( V \)[/tex] given a radius [tex]\( r \)[/tex] and height [tex]\( h \)[/tex].
### Step 1: Find the missing height for each cylinder
To solve for the height [tex]\( h \)[/tex], we rearrange the volume formula:
[tex]\[ h = \frac{V}{\pi r^2} \][/tex]
#### First Cylinder
Given:
- Volume, [tex]\( V = 96\pi \)[/tex] cubic inches
- Radius, [tex]\( r = 4 \)[/tex] inches
Substitute the given values into the rearranged formula:
[tex]\[ h = \frac{96\pi}{\pi \times 4^2} \][/tex]
[tex]\[ h = \frac{96\pi}{\pi \times 16} \][/tex]
[tex]\[ h = \frac{96\pi}{16\pi} \][/tex]
[tex]\[ h = \frac{96}{16} \][/tex]
[tex]\[ h = 6 \text{ inches} \][/tex]
So, the missing height for the first cylinder is [tex]\( 6 \)[/tex] inches.
#### Second Cylinder
Given:
- Volume, [tex]\( V = 31.25\pi \)[/tex] cubic inches
- Radius, [tex]\( r = 2.5 \)[/tex] inches
Substitute the given values into the rearranged formula:
[tex]\[ h = \frac{31.25\pi}{\pi \times 2.5^2} \][/tex]
[tex]\[ h = \frac{31.25\pi}{\pi \times 6.25} \][/tex]
[tex]\[ h = \frac{31.25\pi}{6.25\pi} \][/tex]
[tex]\[ h = \frac{31.25}{6.25} \][/tex]
[tex]\[ h = 5 \text{ inches} \][/tex]
So, the missing height for the second cylinder is [tex]\( 5 \)[/tex] inches.
### Step 2: General Expression for Volume [tex]\( V \)[/tex]
For a general cylinder with radius [tex]\( r \)[/tex] and height [tex]\( h \)[/tex], the volume [tex]\( V \)[/tex] is given by:
[tex]\[ V = \pi r^2 h \][/tex]
This formula remains the same as the original volume formula for a cylinder.
### Summary
[tex]\[ \begin{array}{|c|c|c|} \hline \text{Volume (cubic inches)} & \text{Radius (inches)} & \text{Height (inches)} \\ \hline 96\pi & 4 & 6 \\ \hline 31.25\pi & 2.5 & 5 \\ \hline V & r & \frac{V}{\pi r^2} \\ \hline \end{array} \][/tex]
So:
1. The height for the first cylinder is [tex]\( 6 \)[/tex] inches.
2. The height for the second cylinder is [tex]\( 5 \)[/tex] inches.
3. The volume [tex]\( V \)[/tex] for a radius [tex]\( r \)[/tex] and height [tex]\( h \)[/tex] is given by [tex]\( \ V = \pi r^2 h \)[/tex].
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