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If \(y \propto \frac{1}{x^2}\) and x = 3 when y = 4, find y when x = 2.
index-0
1
index-1
3
index-2
9
index-3
18
Correct Answer: C

\(y \propto \frac{1}{x^2}\)

\(y = \frac{k}{x^2}\)

\(4 = \frac{k}{3^2}\)

\(k = 4 \times 3^2 = 36\)

\(y = \frac{36}{x^2}\)

When x = 2,

\(y = \frac{36}{2^2} = 9\)

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