If \(y \propto \frac{1}{x^2}\) and x = 3 when y = 4, find y when x = 2.
\(y \propto \frac{1}{x^2}\)
\(y = \frac{k}{x^2}\)
\(4 = \frac{k}{3^2}\)
\(k = 4 \times 3^2 = 36\)
\(y = \frac{36}{x^2}\)
When x = 2,
\(y = \frac{36}{2^2} = 9\)
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