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Peter walks 3 minutes at a constant rate and travels 330 meters.
If we graph this relationship with time along the x-axis and distance along the y-axis, the slope of the line representing this relationship is  ?
.

If Peter walks for an hour, the point on the line that represents the distance he walks with respect to time is  ?
.

To find the slope of the line representing the relationship between time and distance for Peter's walk, we use the fact that he travels 330 meters in 3 minutes. The slope \( m \) is the rate of travel in meters per minute.

\[ m = \frac{\text{change in distance}}{\text{change in time}} = \frac{330 \text{ meters}}{3 \text{ minutes}} = 110 \text{ meters per minute} \]

So, the slope \( m \) is \( 110 \).

Next, we need to find the distance Peter walks in one hour (60 minutes). Using the slope, we can calculate this as follows:

\[ \text{Distance after 60 minutes} = 60 \times 110 = 6600 \text{ meters} \]

So, the point on the line after 60 minutes is \( (60, 6600) \).

- The slope of the line is \( 110 \).
- A point on this line after 60 minutes is \( (60, 6600) \).

Would you like more details or have any questions?

Here are some questions you might find interesting:
1. How do you convert minutes into hours in rate calculations?
2. What is the difference between speed and velocity?
3. How can you use the slope to predict values for different times?
4. What other real-world scenarios can be modeled with a linear relationship?
5. How do you find the y-intercept in a distance-time graph?

**Tip:** Always double-check your unit conversions, especially when dealing with time and distance, to ensure your calculations are accurate.

Learn how to interpret linear relationships using slope and distance-time graphs. This problem explores the concept of slope as the rate of change and its application in calculating distances over time. Suitable for high school students.

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