The equation you've provided is \( v = 0.12r^2(10 - r) \), which likely represents a relationship where \( v \) depends on the radius \( r \). Let's explore it.

### Step 1: Expand the Equation
We can expand the given equation to simplify it.

\[
v = 0.12r^2(10 - r)
\]
Distribute \( r^2 \) across the terms inside the parentheses:
\[
v = 0.12 \times r^2 \times 10 - 0.12 \times r^2 \times r
\]
\[
v = 1.2r^2 - 0.12r^3
\]

Now the equation is written as:
\[
v = 1.2r^2 - 0.12r^3
\]

### Step 2: Analyze the Behavior of the Function
This is a cubic function, and its general shape will depend on the values of \( r \). Specifically:
- For small values of \( r \), the term \( 1.2r^2 \) will dominate.
- As \( r \) increases, the cubic term \( -0.12r^3 \) will cause the value of \( v \) to decrease for larger \( r \).

Would you like to explore further by finding critical points or calculating specific values of \( r \)?

Here are 5 follow-up questions:
1. Would you like to find the maximum or minimum value of \( v \) for this equation?
2. Should I calculate the derivative to explore the rate of change of \( v \)?
3. Would you like a graph of the function for different values of \( r \)?
4. Should we solve for \( r \) when \( v = 0 \)?
5. Do you want to analyze the physical interpretation of this equation (e.g., in a real-world context)?

**Tip**: In optimization problems, it's often useful to find the derivative to locate maximum or minimum values by identifying critical points.

Explore the cubic function v = 0.12r^2(10 - r) with step-by-step expansion and analysis of its behavior. Learn how to simplify the equation, find critical points, and understand its real-world applications. Suitable for high school students in Grades 10-12.

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