Precalculus (6th Edition)

$y=\frac{1}{3}$
$y$ varies inversely as $x$ therefore $y=\frac{k}{x}$ $y=20$ when $x=\frac{1}{4}$. Substitute these values into $y=\frac{k}{x}$ to obtain: $\require{cancel} \\y=\frac{k}{x} \\20=\frac{k}{\frac{1}{4}} \\20(\frac{1}{4})=\frac{k}{\frac{1}{4}}\cdot \frac{1}{4} \\20(\frac{1}{4})=\frac{k}{\cancel{\frac{1}{4}}}\cdot \cancel{\frac{1}{4}} \\5=k$ Thus, the equation for $y$ is $y=\frac{5}{x}$. To find the value of $y$ when $x=15$, substitute 15 to x in the equation above to obtain: $\require{cancel} \\y=\frac{5}{15} \\y=\frac{1}{3}$