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  1. PART A The formula for the volume of
    a cone is V = r²h, where r is the
    radius of the circular base and h is the
    height. Solve the formula for h.
    h=
    PART B To the nearest tenth, what is the
    height, in meters, of a cone with a base
    radius of 4 m and a volume of 40 m³?
    m
    L

Verified Answer :

GinnyAnswer

Certainly! Let's tackle each part of the problem step-by-step.

### Part A
The given formula for the volume of a cone is:

\(\[ V = \frac{1}{3}\pi r^2 h \]\)

We need to solve this formula for \(\( h \)\). Follow these steps:

1. Multiply both sides by 3 to clear the fraction:

\(\[ 3V = \pi r^2 h \]\)

2. Divide both sides by \(\(\pi r^2\)\) to isolate \(\( h \)\):

\(\[ h = \frac{3V}{\pi r^2} \]\)

So, the formula for \(\( h \)\) is:

\(\[ h = \frac{3V}{\pi r^2} \]\)

### Part B
Given:
- Volume, \(\( V = 40 \, \text{m}^3 \)\)
- Radius, \(\( r = 4 \, \text{m} \)\)

We need to find the height \(\( h \)\) using the formula derived in Part A. Plug the given values into the formula:

\(\[ h = \frac{3 \cdot 40}{\pi \cdot 4^2} \]\)

Now, simplify the formula step-by-step:

1. Compute the denominator:

\(\[ \pi \cdot 4^2 = \pi \cdot 16 \]\)

2. Compute the numerator:

\(\[ 3 \cdot 40 = 120 \]\)

3. Combine the results:

\(\[ h = \frac{120}{16\pi} \]\)

4. Simplify the fraction:

\(\[ h = \frac{120}{16\pi} = \frac{15}{2\pi} \approx \frac{15}{6.2832} \approx 2.3874 \]\)

Finally, round to the nearest tenth:

\(\[ h \approx 2.4 \, \text{m} \]\)

### Conclusion

- Part A: The formula for \(\( h \)\) is:
\(\[ h = \frac{3V}{\pi r^2} \]\)

- Part B: The height of the cone, to the nearest tenth, is:
\(\[ h \approx 2.4 \, \text{m} \]\)

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