https://file.aiquickdraw.com/mathbot/file/1729048716780ngj03i2g.jpg

Let's analyze the mathematical problems shown in the image you've uploaded.

### Problem 16:
Oscar participates in a charity walk. The graph shows his distance (in miles) from the water stop as a function of time. The question asks how many miles Oscar walked and to explain the answer.

#### Analysis:
1. The graph has three key points:
   - Starting at the **water stop** (0 miles) at time \( t = 0 \).
   - Reaching a **water stop** at \( t = 30 \) minutes.
   - Ending the walk at \( t = 60 \) minutes, which is also at 0 miles (he returns to the water stop).

Thus, Oscar walked:
- From \( t = 0 \) to \( t = 30 \), he walked **2 miles** away from the water stop.
- From \( t = 30 \) to \( t = 60 \), he walked **2 miles** back to the water stop.

Therefore, Oscar walked a total of \( 2 \text{ miles} + 2 \text{ miles} = 4 \text{ miles} \).

#### Explanation:
Oscar's walk consisted of two phases: going 2 miles away from the water stop and then returning 2 miles back, so he covered a total of 4 miles.

---

### Problem 17:
You are asked to find the rate of change over the interval \( 5 \leq t \leq 6 \).

#### Analysis:
Looking at the graph, between \( t = 5 \) and \( t = 6 \), the graph is **horizontal** at a height of 10 ft. A horizontal line represents no change in height.

The rate of change is calculated by the formula:
\[
\text{Rate of change} = \frac{\Delta y}{\Delta x}
\]
where \( \Delta y \) is the change in height and \( \Delta x \) is the change in time.

Between \( t = 5 \) and \( t = 6 \), the height does not change (\( \Delta y = 0 \)), and the time interval is 1 second (\( \Delta x = 1 \)).

So, the rate of change is:
\[
\frac{0}{1} = 0
\]

#### Conclusion:
The rate of change over the interval \( 5 \leq t \leq 6 \) is **0**.

---

### Problem 18:
This problem asks for the rate of change over the interval \( 8 \leq t \leq 10 \).

#### Analysis:
Between \( t = 8 \) and \( t = 10 \), the graph shows a **decrease in height** from 10 ft to 0 ft.

The rate of change is:
\[
\frac{\Delta y}{\Delta x} = \frac{0 - 10}{10 - 8} = \frac{-10}{2} = -5
\]

#### Conclusion:
The rate of change over the interval \( 8 \leq t \leq 10 \) is **-5 ft/s**, indicating a decrease in height.

---

Would you like further details on any of these problems or have any additional questions?

Here are some related questions to explore:
1. How do you interpret the slope of a line in a distance vs. time graph?
2. What does a zero rate of change imply in real-world scenarios?
3. How can you calculate the total distance traveled from a velocity vs. time graph?
4. What is the significance of a negative slope in a real-world situation?
5. How can piecewise linear graphs represent real-life events?

**Tip:** The rate of change in a graph is equivalent to the slope, which helps measure how fast or slow the quantity is increasing or decreasing over time.

How many miles did Oscar walk? What is the rate of change for the given time intervals?

Learn how to calculate the total distance walked and the rate of change from a distance-time graph. This problem involves analyzing a graph, finding the slope, and interpreting piecewise linear functions.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation