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To solve this problem, we need to calculate three things:

1. **The total distance walked.** This is the sum of all the individual distances.

2. **The total displacement vector.** We’ll calculate the net movement in the \(x\)- and \(y\)-directions separately by resolving each segment based on its direction, then combine these to find the overall displacement vector.

3. **The straight-line distance back home.** This is the magnitude of the total displacement vector.

Let’s work through each part step-by-step.

---

### 1. Total Distance Walked
The person walks the following distances:
- 4 miles East
- 4 miles Southeast
- 5 miles South
- 3 miles Southwest
- 3 miles East

Adding these up:
\[
4 + 4 + 5 + 3 + 3 = 19 \text{ miles}
\]

So, the total distance walked is **19 miles**.

---

### 2. Total Displacement Vector

To find the displacement vector, we resolve each leg of the journey into \(x\) (East-West) and \(y\) (North-South) components.

#### Step-by-Step Component Analysis
1. **4 miles East**  
   - \(x\)-component: \(4\)
   - \(y\)-component: \(0\)

2. **4 miles Southeast (45° angle)**  
   - \(x\)-component: \(4 \cos(45^\circ) = 4 \cdot \frac{\sqrt{2}}{2} = 2\sqrt{2} \approx 2.83\)
   - \(y\)-component: \(-4 \sin(45^\circ) = -4 \cdot \frac{\sqrt{2}}{2} = -2\sqrt{2} \approx -2.83\)

3. **5 miles South**  
   - \(x\)-component: \(0\)
   - \(y\)-component: \(-5\)

4. **3 miles Southwest (135° angle)**  
   - \(x\)-component: \(3 \cos(135^\circ) = 3 \cdot -\frac{\sqrt{2}}{2} = -1.5\sqrt{2} \approx -2.12\)
   - \(y\)-component: \(-3 \sin(135^\circ) = 3 \cdot \frac{\sqrt{2}}{2} = 1.5\sqrt{2} \approx 2.12\)

5. **3 miles East**  
   - \(x\)-component: \(3\)
   - \(y\)-component: \(0\)

#### Adding Up Components
Now, we sum up all the \(x\)- and \(y\)-components:

- Total \(x\)-component:
  \[
  4 + 2.83 + 0 - 2.12 + 3 = 7.71
  \]

- Total \(y\)-component:
  \[
  0 - 2.83 - 5 + 2.12 + 0 = -5.71
  \]

Thus, the total displacement vector is approximately:
\[
\vec{d} = (7.71, -5.71)
\]

---

### 3. Straight-Line Distance Back Home
The straight-line distance is the magnitude of the displacement vector, calculated as:
\[
\text{Distance} = \sqrt{(7.71)^2 + (-5.71)^2}
\]
\[
= \sqrt{59.4 + 32.6} = \sqrt{92} \approx 9.6 \text{ miles}
\]

---

### Summary of Answers
1. **Total distance walked:** 19 miles
2. **Total displacement vector:** Approximately \((7.71, -5.71)\)
3. **Straight-line distance back home:** Approximately 9.6 miles

Would you like more details on any part of the solution?

---

### Additional Questions
1. How would the total displacement change if one of the directions was altered?
2. Can you find the angle of the displacement vector relative to the starting point?
3. How would the total walked distance be affected if the journey was extended?
4. What if the person took the same route but in reverse—how would the displacement be affected?
5. How does understanding vector components help in navigation?

### Tip
When working with vectors in directions other than the primary axes (North, South, East, West), always resolve them into \(x\) and \(y\) components. This makes calculations and interpretations much clearer.

A person starts walking from home and walks: 4 miles East, 4 miles Southeast, 5 miles South, 3 miles Southwest, 3 miles East. Find the total distance walked, the total displacement vector, and the straight-line distance back home.

This problem explores vector addition and displacement as a person walks in various directions. Learn how to calculate the total distance, find the displacement vector, and determine the straight-line distance back home using trigonometric and vector principles. Suitable for Grades 9-12.

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