Answer :
Let's analyze the function given:
[tex]\[ v(r) = \frac{4}{3} \pi r^3 \][/tex]
This is the formula to calculate the volume [tex]\( v \)[/tex] of a sphere (or a rubber ball, in this context) with radius [tex]\( r \)[/tex].
Here, we need to determine what [tex]\( v\left(\frac{5}{7}\right) \)[/tex] represents.
1. Understanding the function [tex]\( v(r) \)[/tex]:
- [tex]\( v(r) \)[/tex] gives the volume when the radius [tex]\( r \)[/tex] is known.
- In particular, [tex]\( v\left(\frac{5}{7}\right) \)[/tex] asks for the volume when the radius [tex]\( r \)[/tex] is [tex]\( \frac{5}{7} \)[/tex] feet.
2. Interpreting [tex]\( v\left(\frac{5}{7}\right) \)[/tex]:
- This signifies that we substitute [tex]\( r = \frac{5}{7} \)[/tex] into the volume formula.
3. Resulting interpretation:
- When we substitute [tex]\( r = \frac{5}{7} \)[/tex] feet into the volume formula, we determine the volume of the rubber ball for that specific radius.
4. Analyzing the options:
- the radius of the rubber ball when the volume equals [tex]\(\frac{5}{7}\)[/tex] cubic feet: This is incorrect because [tex]\( v\left(\frac{5}{7}\right) \)[/tex] gives the volume, not the radius.
- the volume of the rubber ball when the radius equals [tex]\(\frac{5}{7}\)[/tex] feet: This is correct because we are given the radius as [tex]\(\frac{5}{7}\)[/tex] feet and we find the corresponding volume.
- that the volume of the rubber ball is 5 cubic feet when the radius is 7 feet: Incorrect, as it mixes up the values and confuses the units.
- that the volume of the rubber ball is 7 cubic feet when the radius is 5 feet: Incorrect, as it also mixes up values and units.
Therefore, the correct interpretation of [tex]\( v\left(\frac{5}{7}\right) \)[/tex] is:
[tex]\[ \boxed{\text{the volume of the rubber ball when the radius equals } \frac{5}{7} \text{ feet}} \][/tex]
Thus, [tex]\( v\left(\frac{5}{7}\right) \)[/tex] represents the volume of the rubber ball when the radius equals [tex]\(\frac{5}{7}\)[/tex] feet. The evaluated volume is approximately 1.526527042560638 cubic feet.
[tex]\[ v(r) = \frac{4}{3} \pi r^3 \][/tex]
This is the formula to calculate the volume [tex]\( v \)[/tex] of a sphere (or a rubber ball, in this context) with radius [tex]\( r \)[/tex].
Here, we need to determine what [tex]\( v\left(\frac{5}{7}\right) \)[/tex] represents.
1. Understanding the function [tex]\( v(r) \)[/tex]:
- [tex]\( v(r) \)[/tex] gives the volume when the radius [tex]\( r \)[/tex] is known.
- In particular, [tex]\( v\left(\frac{5}{7}\right) \)[/tex] asks for the volume when the radius [tex]\( r \)[/tex] is [tex]\( \frac{5}{7} \)[/tex] feet.
2. Interpreting [tex]\( v\left(\frac{5}{7}\right) \)[/tex]:
- This signifies that we substitute [tex]\( r = \frac{5}{7} \)[/tex] into the volume formula.
3. Resulting interpretation:
- When we substitute [tex]\( r = \frac{5}{7} \)[/tex] feet into the volume formula, we determine the volume of the rubber ball for that specific radius.
4. Analyzing the options:
- the radius of the rubber ball when the volume equals [tex]\(\frac{5}{7}\)[/tex] cubic feet: This is incorrect because [tex]\( v\left(\frac{5}{7}\right) \)[/tex] gives the volume, not the radius.
- the volume of the rubber ball when the radius equals [tex]\(\frac{5}{7}\)[/tex] feet: This is correct because we are given the radius as [tex]\(\frac{5}{7}\)[/tex] feet and we find the corresponding volume.
- that the volume of the rubber ball is 5 cubic feet when the radius is 7 feet: Incorrect, as it mixes up the values and confuses the units.
- that the volume of the rubber ball is 7 cubic feet when the radius is 5 feet: Incorrect, as it also mixes up values and units.
Therefore, the correct interpretation of [tex]\( v\left(\frac{5}{7}\right) \)[/tex] is:
[tex]\[ \boxed{\text{the volume of the rubber ball when the radius equals } \frac{5}{7} \text{ feet}} \][/tex]
Thus, [tex]\( v\left(\frac{5}{7}\right) \)[/tex] represents the volume of the rubber ball when the radius equals [tex]\(\frac{5}{7}\)[/tex] feet. The evaluated volume is approximately 1.526527042560638 cubic feet.