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Compute the average rate of change in *D* over the interval from *t* = 3 to *t* = 6.

To compute the average rate of change of \( D \) over the interval from \( t = 3 \) to \( t = 6 \), we need to find the change in \( D \) (distance in yards) over the change in \( t \) (time in minutes) between these two points.

### Step-by-Step Solution

1. **Identify the coordinates at \( t = 3 \) and \( t = 6 \):**

   From the graph:
   - At \( t = 3 \), \( D \approx 0 \) yards.
   - At \( t = 6 \), \( D \approx 600 \) yards.

2. **Calculate the change in \( D \) and \( t \):**
   \[
   \Delta D = D(6) - D(3) = 600 - 0 = 600 \text{ yards}
   \]
   \[
   \Delta t = 6 - 3 = 3 \text{ minutes}
   \]

3. **Compute the average rate of change:**
   \[
   \text{Average rate of change} = \frac{\Delta D}{\Delta t} = \frac{600 \text{ yards}}{3 \text{ minutes}} = 200 \text{ yards per minute}
   \]

### Conclusion

The average rate of change in \( D \) over the interval from \( t = 3 \) to \( t = 6 \) is **200 yards per minute**.

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Would you like more details or have any other questions?

Here are some related questions you might be interested in:

1. What is the average rate of change in \( D \) over the interval from \( t = 6 \) to \( t = 12 \)?
2. How do you interpret the average rate of change in the context of this graph?
3. Can the average rate of change be negative? If so, what would that indicate on this graph?
4. What does the slope of a segment on a distance-time graph represent?
5. How can you determine the instantaneous rate of change from a graph?

**Tip:** The average rate of change between two points on a graph is equivalent to the slope of the straight line connecting those points.

Learn how to calculate the average rate of change in distance over the time interval from t = 3 to t = 6 with a detailed step-by-step solution. This problem involves graphical interpretation and linear approximation, suitable for high school students.

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