To analyze the equation \( v = 0.12r^2(10 - r) \), we can break it down step by step.

### Key components:
- \( v \): Represents a variable, possibly volume or velocity.
- \( r \): Represents another variable, possibly a radius or distance.

The equation can be expanded for further clarity:

\[
v = 0.12r^2(10 - r)
\]

Expanding this:

\[
v = 0.12r^2 \cdot 10 - 0.12r^2 \cdot r
\]

Simplifying:

\[
v = 1.2r^2 - 0.12r^3
\]

### Result:

The equation becomes a cubic function:

\[
v = 1.2r^2 - 0.12r^3
\]

This represents a cubic relationship between \(v\) and \(r\). You can now use this equation to calculate \(v\) for any value of \(r\), or you can further analyze its behavior (e.g., finding critical points, maximum/minimum, etc.).

Would you like to explore any specific aspect of this equation in more detail?

### Related Questions:
1. What are the critical points of the function \(v = 1.2r^2 - 0.12r^3\)?
2. How do you find the maximum or minimum of this function?
3. How does the volume \(v\) behave as \(r\) increases or decreases?
4. How can you interpret this equation in a real-world context (e.g., volume of a container)?
5. What is the significance of the cubic term \( -0.12r^3 \)?

### Tip:
When dealing with cubic functions, analyzing their first and second derivatives can help identify turning points and understand the shape of the graph.

Explore the cubic equation v = 0.12r^2(10 - r) and learn how to simplify and analyze its behavior. This detailed guide covers critical points, maxima and minima using derivative analysis, suitable for Grades 10-12 students.

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