Answer
$0.012$ foot-candles
Work Step by Step
If the intensity $I$ varies inversely with $d^{2}$, then
there is a nonzero constant $k$ such that
$I=\displaystyle \frac{k}{d^{2}}$
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If $ I=0.075$ when $d=2$,
substituting, we find k:
$0.075=\displaystyle \frac{k}{2^{2}}\quad/\times 2^{2}$
$k=0.3$
Thus, we write: $\displaystyle \quad I=\frac{0.3}{d^{2}}.$
If $d=5$,
$I=\displaystyle \frac{0.3}{5^{2}}=0.012$ foot-candles
