Let's analyze the problem from the uploaded image.

We are tasked with determining the relationship between $x$ and $y$ based on their variation and finding the constant $k$. The given relationship options are:

1. $y = kx$ (direct variation)
2. $y = \frac{k}{x}$ (inverse variation)

From the table:

• $x = 3, 4, 12, 20$
• $y = \frac{4}{3}, 1, \frac{1}{3}, \frac{1}{5}$

Step 1: Verify the Variation

For $y = \frac{k}{x}$, $k$ can be calculated as: $k = y \cdot x$

Using the values in the table:

1. For $x = 3, y = \frac{4}{3}$: $k = 3 \cdot \frac{4}{3} = 4$

2. For $x = 4, y = 1$: $k = 4 \cdot 1 = 4$

3. For $x = 12, y = \frac{1}{3}$: $k = 12 \cdot \frac{1}{3} = 4$

4. For $x = 20, y = \frac{1}{5}$: $k = 20 \cdot \frac{1}{5} = 4$

Since $k$ is constant across all values, the relationship is confirmed to be $y = \frac{k}{x}$, with $k = 4$.

Final Answer:

• The relationship is $y = \frac{k}{x}$.
• The value of $k$ is $\mathbf{4}$.

Would you like further clarification or details? Here are some follow-up questions you might explore:

1. How does direct variation differ from inverse variation in terms of $x$ and $y$?
2. What happens to $y$ as $x$ increases in an inverse variation?
3. How can this type of problem be used in real-world applications?
4. What other forms of variation (e.g., joint or combined) exist in mathematics?
5. How would the equation change if $k$ were negative?

Tip: Always substitute values into the equation to confirm relationships in variation problems.