If p varies inversely as the square of q and p = 8 when q = 4, find q when p = 32
p ∝ \(\frac{1}{q}\)
p = \(\frac{k}{q}\)
K = q\(^2\)p
= 4\(^2\)(8)
∴ p = \(\frac{128}{\text{q}}\)
32 = \(\frac{128}{\text{q}}\)
q\(^2\) = \(\frac{128}{32}\)
q\(^2\) = 4
q = √4 = \(\pm\) 2
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