## Precalculus (6th Edition) Blitzer

The value of $\frac{x}{y}$ is $-\frac{\sqrt{3}}{3}$ .
Consider the provided values of $x$ and $y$: $x=-\frac{1}{2}$ and $y=\frac{\sqrt{3}}{2}$ Now, calculate $\frac{x}{y}$ as: \begin{align} & \frac{x}{y}=\frac{-\frac{1}{2}}{\frac{\sqrt{3}}{2}} \\ & =\frac{-1}{2}\cdot \frac{2}{\sqrt{3}} \\ & =-\frac{1}{\sqrt{3}} \end{align} Now, rationalize the denominator by multiplying the numerator and the denominator by $\sqrt{3}$ : \begin{align} & \frac{x}{y}=-\frac{1}{\sqrt{3}}\cdot \frac{\sqrt{3}}{\sqrt{3}} \\ & =-\frac{\sqrt{3}}{3} \end{align} Therefore, the value of $\frac{x}{y}$ is $-\frac{\sqrt{3}}{3}$ .