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### Question 1:

#### Part (a):
The problem states that "y varies inversely as the square of \( x \)." This means:

\[
y = \frac{k}{x^2}
\]

where \( k \) is the constant of variation.

#### Part (b):
We are given a table of values for \( x \) and \( y \):

\[
\begin{array}{|c|c|c|c|}
\hline
x & 3 & 1.8 & f \\
\hline
y & 2 & r & 8 \\
\hline
\end{array}
\]

We need to calculate:
1. \( k \), the constant of variation.
2. \( r \).
3. \( f \).

We can use the equation \( y = \frac{k}{x^2} \) and the given values to find these.

---

**Step 1: Calculate \( k \) using the first pair \( x = 3 \) and \( y = 2 \):**

\[
y = \frac{k}{x^2}
\]
\[
2 = \frac{k}{3^2} \quad \Rightarrow \quad 2 = \frac{k}{9} \quad \Rightarrow \quad k = 18
\]

So, \( k = 18 \).

---

**Step 2: Calculate \( r \) when \( x = 1.8 \):**

Using \( y = \frac{k}{x^2} \), with \( k = 18 \) and \( x = 1.8 \),

\[
r = \frac{18}{1.8^2} = \frac{18}{3.24} \approx 5.56
\]

---

**Step 3: Calculate \( f \) when \( y = 8 \):**

Using \( y = \frac{k}{x^2} \), with \( y = 8 \) and \( k = 18 \),

\[
8 = \frac{18}{f^2} \quad \Rightarrow \quad f^2 = \frac{18}{8} = 2.25 \quad \Rightarrow \quad f = \sqrt{2.25} = 1.5
\]

So, \( f = 1.5 \).

---

### Question 2:

The problem says \( y \) varies directly as \( x \), which means:

\[
y = kx
\]

We are given the following values:

\[
\begin{array}{|c|c|c|c|}
\hline
x & 2 & 5 & b \\
\hline
y & 12 & a & 48 \\
\hline
\end{array}
\]

#### Step 1: Find the constant \( k \) using \( x = 2 \) and \( y = 12 \):

\[
y = kx \quad \Rightarrow \quad 12 = 2k \quad \Rightarrow \quad k = 6
\]

#### Step 2: Calculate \( a \) when \( x = 5 \):

\[
a = 6 \times 5 = 30
\]

#### Step 3: Calculate \( b \) when \( y = 48 \):

\[
48 = 6 \times b \quad \Rightarrow \quad b = \frac{48}{6} = 8
\]

---

### Final Answers:
1. **For Question 1:**
   - \( k = 18 \)
   - \( r \approx 5.56 \)
   - \( f = 1.5 \)

2. **For Question 2:**
   - \( a = 30 \)
   - \( b = 8 \)

Would you like more details on any of these steps?

---

Here are 5 related questions you might consider:
1. How would the equation change if \( y \) varied directly with the square of \( x \)?
2. Can you solve for \( x \) in terms of \( y \) for an inverse square variation?
3. What happens to \( y \) if \( x \) is halved in the direct variation equation?
4. How would you interpret an inverse cube relationship between \( y \) and \( x \)?
5. How can we determine the units of \( k \) based on real-world problems?

**Tip:** In direct variation problems, solving for the constant \( k \) first simplifies solving for unknown values of \( x \) or \( y \).

Two variables x and y are related such that y varies inversely as the square of x. (a) Write an equation in x, y, and k to describe the inverse variation, where k is the constant of variation. (b) Using the information in the table, calculate the value of (i) k, the constant of variation (ii) r (iii) f.

Also, the table below shows corresponding values of variables x and y, where y varies directly as x. Calculate the values of a and b.

This problem focuses on solving inverse variation (y = k / x^2) and direct variation (y = kx). Learn how to calculate the constant of variation, k, along with unknown variables r, f, a, and b using provided tables. Suitable for Grade 10 students.

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