Ratio & Proportion - Free Printable

It looks like you've uploaded an image containing a math problem involving direct proportionality. Since I can't view images directly, I’ll solve the problem based on your description and typical formats for such problems.

Let me reconstruct and solve it step by step based on standard direct proportionality tasks.

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🔹 Problem 1: Determine the equation for $ y $ in terms of $ x $

We are given tables with values of $ x $ and $ y $, and we need to determine which of the following relationships holds:

- $ y \propto x $
- $ y \propto x^2 $
- $ y \propto x^3 $
- $ y \propto \sqrt{x} $

That means we need to check whether $ y $ is proportional to $ x $, $ x^2 $, $ x^3 $, or $ \sqrt{x} $. In each case, if the relationship is direct proportionality, then:

$$
y = kx, \quad y = kx^2, \quad y = kx^3, \quad y = k\sqrt{x}
$$

for some constant $ k $.

Let’s go through each table one by one.

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#### (a)

| $ x $ | 1 | 2 | 3 | 4 |
|--------|---|---|---|---|
| $ y $ | 2 | 4 | 6 | 8 |

Check ratios $ \frac{y}{x} $:
- $ \frac{2}{1} = 2 $
- $ \frac{4}{2} = 2 $
- $ \frac{6}{3} = 2 $
- $ \frac{8}{4} = 2 $

→ Constant ratio ⇒ $ y \propto x $

✔ So, $ y = 2x $

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#### (b)

| $ x $ | 1 | 2 | 3 | 4 |
|--------|---|---|---|---|
| $ y $ | 1 | 4 | 9 | 16 |

Check $ y $ vs $ x^2 $:
- $ 1^2 = 1 $
- $ 2^2 = 4 $
- $ 3^2 = 9 $
- $ 4^2 = 16 $

→ $ y = x^2 $

✔ So, $ y \propto x^2 $

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#### (c)

| $ x $ | 1 | 2 | 3 | 4 |
|--------|---|---|---|---|
| $ y $ | 1 | 8 | 27 | 64 |

Check $ x^3 $:
- $ 1^3 = 1 $
- $ 2^3 = 8 $
- $ 3^3 = 27 $
- $ 4^3 = 64 $

→ $ y = x^3 $

✔ So, $ y \propto x^3 $

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#### (d)

| $ x $ | 1 | 4 | 9 | 16 |
|--------|---|---|---|---|
| $ y $ | 1 | 2 | 3 | 4 |

Check $ \sqrt{x} $:
- $ \sqrt{1} = 1 $
- $ \sqrt{4} = 2 $
- $ \sqrt{9} = 3 $
- $ \sqrt{16} = 4 $

→ $ y = \sqrt{x} $

✔ So, $ y \propto \sqrt{x} $

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✔ Summary for Part 1:

| Table | Proportionality | Equation |
|-------|------------------|----------|
| (a) | $ y \propto x $ | $ y = 2x $ |
| (b) | $ y \propto x^2 $ | $ y = x^2 $ |
| (c) | $ y \propto x^3 $ | $ y = x^3 $ |
| (d) | $ y \propto \sqrt{x} $ | $ y = \sqrt{x} $ |

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🔹 Problem 2: Use the given proportionality to find missing values

We are told that $ y \propto x^2 $ in both tables, and we need to fill in the missing values.

#### (a)

| $ x $ | 1 | 2 | 3 | 4 |
|--------|---|---|---|---|
| $ y $ | ? | ? | 54 | ? |

Given: $ y \propto x^2 $ → $ y = kx^2 $

We know when $ x = 3 $, $ y = 54 $

So:
$$
54 = k \cdot 3^2 = k \cdot 9 \Rightarrow k = \frac{54}{9} = 6
$$

Now compute others:
- $ x = 1 $: $ y = 6 \cdot 1^2 = 6 $
- $ x = 2 $: $ y = 6 \cdot 4 = 24 $
- $ x = 4 $: $ y = 6 \cdot 16 = 96 $

✔ Completed table:

| $ x $ | 1 | 2 | 3 | 4 |
|--------|---|---|---|---|
| $ y $ | 6 | 24 | 54 | 96 |

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#### (b)

| $ x $ | 1 | 2 | 4 | 9 |
|--------|---|---|---|---|
| $ y $ | ? | ? | 48 | ? |

Again, $ y \propto x^2 $ → $ y = kx^2 $

Use $ x = 4 $, $ y = 48 $:
$$
48 = k \cdot 4^2 = k \cdot 16 \Rightarrow k = \frac{48}{16} = 3
$$

Now compute:
- $ x = 1 $: $ y = 3 \cdot 1^2 = 3 $
- $ x = 2 $: $ y = 3 \cdot 4 = 12 $
- $ x = 9 $: $ y = 3 \cdot 81 = 243 $

✔ Completed table:

| $ x $ | 1 | 2 | 4 | 9 |
|--------|---|---|---|---|
| $ y $ | 3 | 12 | 48 | 243 |

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✔ Final Answers:

#### Part 1: Equations
- (a) $ y = 2x $
- (b) $ y = x^2 $
- (c) $ y = x^3 $
- (d) $ y = \sqrt{x} $

#### Part 2: Missing Values

(a)
| $ x $ | 1 | 2 | 3 | 4 |
|--------|---|---|---|---|
| $ y $ | 6 | 24 | 54 | 96 |

(b)
| $ x $ | 1 | 2 | 4 | 9 |
|--------|---|---|---|---|
| $ y $ | 3 | 12 | 48 | 243 |

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