https://file.aiquickdraw.com/mathbot/file/1729187627549xnbh1jub.jpg

The problem provides the formula for the volume of a sphere:

\[
V = \frac{4}{3} \pi r^3
\]

We are told that the volume of the bouncy ball is \( 5100 \pi \, \text{mm}^3 \). To find the radius \( r \), we'll use this information.

### Step 1: Set up the equation

Given that \( V = 5100 \pi \), substitute into the volume formula:

\[
5100 \pi = \frac{4}{3} \pi r^3
\]

### Step 2: Cancel \( \pi \) from both sides

\[
5100 = \frac{4}{3} r^3
\]

### Step 3: Solve for \( r^3 \)

Multiply both sides by 3 to get rid of the fraction:

\[
15300 = 4r^3
\]

Now, divide by 4:

\[
r^3 = \frac{15300}{4} = 3825
\]

### Step 4: Find the cube root of 3825

\[
r = \sqrt[3]{3825}
\]

Using a calculator:

\[
r \approx 15.68 \, \text{mm}
\]

Thus, the radius of the bouncy ball is approximately **15.68 mm** (rounded to 2 decimal places).

Would you like further details or clarifications?

Here are 5 related questions to expand on this:
1. How would the formula change if the volume were given in cubic centimeters instead of millimeters?
2. What would the radius be if the volume were doubled?
3. How do you calculate the surface area of a sphere using the radius?
4. What is the relationship between the radius and the diameter of the sphere?
5. Can you derive the volume formula for a sphere from its geometry?

**Tip:** When solving for a variable, always perform operations step-by-step and check if units are consistent throughout the calculation.

The bouncy ball below is a sphere with a volume of 5100π mm³. Calculate its radius, r. If your answer is a decimal, give it to 2 decimal places.

Learn how to calculate the radius of a bouncy ball modeled as a sphere with a volume of 5100π mm³. Step-by-step solution provided, including use of the volume formula and algebraic manipulation. Suitable for students in Grades 8-10.

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