Question

select the correct answer.

the square of (y) varies directly as the cube of (x). when (x = 4), (y = 2). which equation can be used to find other combinations of (x) and (y)?

(y^2 = \frac{1}{16}x^3)
(y = \frac{1}{2}x)
(xy = 8)
(x^3y^2 = 128)

Explanation:

Set up the direct variation equation

$$ y^2 = k \cdot x^3 $$

Solve for the constant of variation \(k\)

$$
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$$

Write the final equation

$$ y^2 = \frac{1}{16}x^3 $$

Answer:

• (A) \(y^2 = \frac{1}{16}x^3\) (Correct answer)
• (B) \(y = \frac{1}{2}x\)
• (C) \(xy = 8\)
• (D) \(x^3y^2 = 128\)