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Which of the following functions gives the radius, [tex]\( r(v) \)[/tex], of a conical artifact that is 20 inches tall as a function of its volume, [tex]\( v \)[/tex], in cubic inches?

A. [tex]\( r(v)=\sqrt{\frac{3 \pi}{3 v}} \)[/tex]

B. [tex]\( r(v)=\sqrt{\frac{3 \pi}{20 \pi}} \)[/tex]

C. [tex]\( r(v)=\left(\frac{3 v}{20 \pi}\right)^2 \)[/tex]

D. [tex]\( r(v)=\sqrt[3]{\frac{3 \pi}{20 \pi}} \)[/tex]


Sagot :

To determine which function gives the radius [tex]\( r(v) \)[/tex] of a conical artifact that is 20 inches tall as a function of its volume [tex]\( v \)[/tex] in cubic inches, let’s start by recalling the volume formula for a cone:
[tex]\[ V = \frac{1}{3} \pi r^2 h \][/tex]

Given the height [tex]\( h = 20 \)[/tex] inches, we can rewrite the formula:
[tex]\[ V = \frac{1}{3} \pi r^2 \times 20 \][/tex]

Simplify the expression:
[tex]\[ V = \frac{20 \pi r^2}{3} \][/tex]

Now we solve for [tex]\( r \)[/tex]:

1. Isolate [tex]\( r^2 \)[/tex] by multiplying both sides of the equation by [tex]\(\frac{3}{20 \pi}\)[/tex]:
[tex]\[ r^2 = \frac{3V}{20 \pi} \][/tex]

2. Take the square root of both sides to solve for [tex]\( r \)[/tex]:
[tex]\[ r = \sqrt{\frac{3V}{20 \pi}} \][/tex]

Examining the provided options:

A. [tex]\( r(v) = \sqrt{\frac{3 \pi}{3 v}} \)[/tex]

B. [tex]\( r(v) = \sqrt{\frac{3 \pi}{20 \pi}} \)[/tex]

C. [tex]\( r(v) = \left(\frac{3 v}{20 \pi}\right)^2 \)[/tex]

D. [tex]\( r(v) = \sqrt[3]{\frac{3 \pi}{20 \pi}} \)[/tex]

Clearly, none of these options provide the correct formula. The correct formula should be:

[tex]\[ r(v) = \sqrt{ \frac{3V}{20 \pi}} \][/tex]
Nonetheless, based on the calculation and choices given, the answer doesn’t exactly match any of the provided choices. Let's look more closely at all provided functions but none of them match the correct expression for [tex]\( r(v) \)[/tex]. Therefore, the correct function in line with our calculations is not among the provided options.

Hence the correct answer should be:
[tex]\[ r(v) = \sqrt{\frac{3V}{20 \pi}} \][/tex]
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