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Apply: Radical Functions Module Assessment

A cylindrical pipe is 9 ft long and has a volume of 100 ft³. Find its approximate diameter to the nearest hundredth of a foot.

Sagot :

Sure! Let's solve the problem step-by-step.

We are given:
- The length \( L \) of the cylindrical pipe is 9 feet.
- The volume \( V \) of the cylindrical pipe is 100 cubic feet.

We need to find the approximate diameter of the pipe to the nearest hundredth of a foot.

To solve this, we use the formula for the volume of a cylinder:

[tex]\[ V = \pi r^2 L \][/tex]

where:
- \( V \) is the volume,
- \( r \) is the radius,
- \( L \) is the length,
- and \( \pi \) is a constant approximately equal to 3.14159.

First, let's rearrange this formula to solve for the radius \( r \):

[tex]\[ r^2 = \frac{V}{\pi L} \][/tex]

Given \( V = 100 \) cubic feet and \( L = 9 \) feet, we substitute these values into the formula:

[tex]\[ r^2 = \frac{100}{\pi \times 9} \][/tex]

Next, we compute the division inside the square root:

[tex]\[ r^2 = \frac{100}{28.27433} \approx 3.535 \][/tex]

Now, take the square root of both sides to solve for \( r \):

[tex]\[ r \approx \sqrt{3.535} \approx 1.881 \][/tex]

So the radius \( r \) is approximately 1.881 feet.

To find the diameter \( d \), we use the relationship between the radius and the diameter:

[tex]\[ d = 2r \][/tex]

Substituting the value we found for \( r \):

[tex]\[ d \approx 2 \times 1.881 \approx 3.761 \][/tex]

Finally, rounding this to the nearest hundredth of a foot, we get:

[tex]\[ d \approx 3.76 \][/tex]

Therefore, the approximate diameter of the cylindrical pipe is 3.76 feet.