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(D) Given that $x$ varies inversely as the square of $y$,we can write the relationship as:$x \propto \frac{1}{y^2}$ or $x = \frac{k}{y^2}$,where $k$ i

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$\frac{1}{9}$

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(D) Given that $x$ varies inversely as the square of $y$,we can write the relationship as:$x \propto \frac{1}{y^2}$ or $x = \frac{k}{y^2}$,where $k$ is a constant.Given $y=2$ when $x=1$,substitute these values into the equation:$1 = \frac{k}{2^2} \implies 1 = \frac{k}{4} \implies k = 4$.Now,we need to find $x$ when $y=6$ using the constant $k=4$:$x = \frac{4}{6^2} = \frac{4}{36} = \frac{1}{9}$.Therefore,the value of $x$ is $\frac{1}{9}$.

$x$ varies inversely as the square of $y$. Given that $y=2$ when $x=1$. The value of $x$ when $y=6$ will be equal to:

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