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The image you uploaded presents a solution to a related rates problem. Here's a breakdown of the steps shown:

### Problem Breakdown:
1. **Given Relation**:
   \[
   \frac{dr}{dt} = \frac{1}{20} (2h) \cdot \frac{dh}{dt}
   \]
   This equation likely comes from a physical situation where \(r\) and \(h\) are related variables (such as radius and height), and the equation represents how the rate of change of one variable relates to the rate of change of the other.

2. **Simplifying the Equation**:
   - The equation is then simplified:
     \[
     \frac{dr}{dt} = \frac{2h}{20} \cdot \frac{dh}{dt} = \frac{h}{10} \cdot \frac{dh}{dt}
     \]

   - This step shows that the equation simplifies the term \(\frac{2h}{20}\) to \(\frac{h}{10}\).

3. **Given Data Substitution**:
   - The next equation uses specific values:
     \[
     \frac{dr}{dt} = -\frac{1}{5}
     \]
     \[
     \frac{h}{10} \cdot \frac{dh}{dt} = -\frac{1}{5}
     \]
     This step substitutes the value \(\frac{dr}{dt} = -\frac{1}{5}\) into the simplified equation.

4. **Solving for \(\frac{dh}{dt}\)**:
   - Solving for \(\frac{dh}{dt}\):
     \[
     \frac{dh}{dt} = \frac{-\frac{1}{5}}{\frac{h}{10}} = -\frac{1}{5} \cdot \frac{10}{h} = -\frac{2}{3} \, \text{in/sec}
     \]

### Final Answer:
\(\frac{dh}{dt} = -\frac{2}{3}\) in/sec.

This indicates the rate at which height \(h\) is changing with respect to time is \(-\frac{2}{3}\) inches per second.

Would you like more details on any part of this solution, or do you have any questions?

Here are 5 related questions:
1. What physical situation could correspond to this related rates problem?
2. How would the solution change if \(\frac{dr}{dt}\) was positive instead of negative?
3. What assumptions might be made about the relationship between \(r\) and \(h\)?
4. How do you determine the signs of \(\frac{dh}{dt}\) and \(\frac{dr}{dt}\)?
5. What units are used for \(r\) and \(h\), and how do they affect the interpretation of \(\frac{dh}{dt}\)?

**Tip:** When solving related rates problems, always pay attention to the signs of the rates, as they indicate whether the quantities are increasing or decreasing.

Given the equation \(\frac{dr}{dt} = \frac{1}{20} (2h) \cdot \frac{dh}{dt}\), solve for \(\frac{dh}{dt}\) when \(\frac{dr}{dt} = -\frac{1}{5}\).

Learn how to solve a related rates problem involving the equation \(\frac{dr}{dt} = \frac{1}{20} (2h) \cdot \frac{dh}{dt}\) and find \(\frac{dh}{dt}\) when \(\frac{dr}{dt} = -\frac{1}{5}\). This solution covers key concepts in differentiation and algebra, suitable for students in Grades 11-12.

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