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Let's break down and solve each of these problems one by one:

### 1. \( x \) varies with the square of \( y \) such that when \( y = -2 \), \( x = 20 \). What is \( y \) when \( x = 5 \)?
- The relationship can be written as \( x = k y^2 \), where \( k \) is a constant.
- Given that \( x = 20 \) when \( y = -2 \):
  \[
  20 = k (-2)^2
  \]
  \[
  20 = 4k \implies k = 5
  \]
- Now, to find \( y \) when \( x = 5 \):
  \[
  5 = 5 y^2 \implies y^2 = 1 \implies y = \pm 1
  \]
Thus, \( y = 1 \) or \( y = -1 \).

---

### 2. As the ship travels at constant speed, the distance (\( d \)) varies directly with time (\( t \)). If it travels a distance of 48 nautical miles in 0.50 hours, how far is the ship after 5.0 hours?
- The relationship is \( d = k t \), where \( k \) is the constant speed.
- Given \( d = 48 \) nautical miles and \( t = 0.5 \) hours:
  \[
  48 = k \times 0.5 \implies k = 96 \text{ nautical miles per hour}
  \]
- Now, to find the distance when \( t = 5 \) hours:
  \[
  d = 96 \times 5 = 480 \text{ nautical miles}
  \]
Thus, the ship will travel **480 nautical miles** in 5 hours.

---

### 3. A ship travels a particular distance such that the time of travel (\( t \)) varies inversely with the speed (\( v \)). It takes \( 7 \frac{1}{2} \) hours to complete the distance when the speed is 16 knots. What time is required when its speed is 24 knots?
- Since \( t \) varies inversely with \( v \), the relationship is \( t v = k \), where \( k \) is a constant.
- Given \( t = 7.5 \) hours and \( v = 16 \) knots:
  \[
  7.5 \times 16 = k \implies k = 120
  \]
- Now, to find \( t \) when \( v = 24 \) knots:
  \[
  t \times 24 = 120 \implies t = \frac{120}{24} = 5 \text{ hours}
  \]
Thus, the required time is **5 hours**.

---

### 4. The cost of picture production (\( C \)) varies jointly with the product of the length (\( L \)) and width (\( W \)) of the picture. A picture whose length is 35 cm and width is 25 cm costs Php 280.00. What is the cost of reproduction when the length is 70 cm and the width is 15 cm?
- The relationship is \( C = k L W \), where \( k \) is a constant.
- Given \( C = 280 \), \( L = 35 \), and \( W = 25 \):
  \[
  280 = k \times 35 \times 25 \implies k = \frac{280}{875} = 0.32
  \]
- Now, to find the cost when \( L = 70 \) cm and \( W = 15 \) cm:
  \[
  C = 0.32 \times 70 \times 15 = 336 \text{ Php}
  \]
Thus, the cost of reproduction is **Php 336.00**.

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Would you like more details on any of these solutions? Here are some related questions to expand on the concepts:

1. What would happen to the time required if the speed in question 3 were halved?
2. How does changing the proportionality constant \( k \) affect direct variation in question 2?
3. Can you find the value of \( y \) if \( x \) were 10 in question 1?
4. How would joint variation change if the cost in question 4 also depended on the material used?
5. What other forms of variation can be modeled similarly to direct and inverse variations?

**Tip:** Always identify the type of variation in word problems—whether direct, inverse, or joint—to set up the correct relationship between variables.

1. x varies with the square of y such that when y = -2, x = 20. What is y when x is 5? 2. As the ship travels at constant speed, the distance (d), varies directly with time (t). If it travels a distance of 48 nautical miles in 0.50 hours, how far is the ship after 5.0 hours? 3. A ship travels a particular distance such that the time of travel (t) varies inversely with the speed (v). It takes 7 1/2 hours to complete the distance when the speed is 16 knots. What time is required when its speed is 24 knots? 4. The cost of picture production (C) varies jointly with the product of the length (L) and width (W) of the picture. A picture whose length is 35 cm and the width is 25 cm costs Php 280.00. What is the cost of reproduction when the length is 70 cm and the width is 15 cm?

Solve problems involving direct, inverse, and joint variation. Includes real-world applications like distance and time for ships and picture production costs. Suitable for students in Grades 7-10.

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