Alright, let's address the problem step-by-step.
### Part A: Solve the formula for [tex]\( r \)[/tex]
Given the formula for the volume [tex]\( V \)[/tex] of a cone:
[tex]\[ V = \frac{1}{3} \pi r^2 h \][/tex]
We need to solve for [tex]\( r \)[/tex].
1. First, isolate [tex]\( r^2 \)[/tex] by multiplying both sides of the equation by 3:
[tex]\[ 3V = \pi r^2 h \][/tex]
2. Next, divide both sides by [tex]\( \pi h \)[/tex] to solve for [tex]\( r^2 \)[/tex]:
[tex]\[ r^2 = \frac{3V}{\pi h} \][/tex]
3. Finally, take the square root of both sides to solve for [tex]\( r \)[/tex]:
[tex]\[ r = \sqrt{\frac{3V}{\pi h}} \][/tex]
Thus, the radius [tex]\( r \)[/tex] is:
[tex]\[ r = \sqrt{\frac{3V}{\pi h}} \][/tex]
### Part B: Determine the radius of the cone
We are given:
[tex]\[ V = 125 \, \text{m}^3 \][/tex]
[tex]\[ h = 12 \, \text{cm} \][/tex]
Note: The units for [tex]\( V \)[/tex] and [tex]\( h \)[/tex] are not consistent (one is in cubic meters and the other in centimeters). Assuming there's a typo and since volumes are usually in cubic units, let's correct the height to meters for consistency:
[tex]\[ h = 0.12 \, \text{m} \][/tex]
Using the formula derived in Part A, we substitute the given values:
[tex]\[ r = \sqrt{\frac{3 \times 125}{\pi \times 0.12}} \][/tex]
Performing the calculation will give us the radius. The calculated radius is approximately:
[tex]\[ r \approx 3.15 \, \text{m} \][/tex]
Therefore, the radius of the cone, rounded to the nearest hundredth, is [tex]\( 3.15 \, \text{m} \)[/tex].
