## Precalculus (6th Edition) Blitzer

Range: $59^{\mathrm{o}}F$ to $95^{\mathrm{o}}F$, inclusive, or $[59^{\mathrm{o}}F,95^{\mathrm{o}}F].$
Given: $15\leq C\leq 35\qquad$ Replace C with the expression for F $15\displaystyle \leq\frac{5}{9}(F-32)\leq 35\qquad$ Solve for F. Begin by multiplying all parts with $\displaystyle \frac{9}{5}$ $15\displaystyle \cdot\frac{9}{5}\leq\frac{9}{5}\cdot\frac{5}{9}(F-32)\leq\frac{9}{5}\cdot 35$ $27\leq F-32\leq 63\qquad$ ... add 32 $59\leq F\leq 95$ Range: $\ \ 59^{\mathrm{o}}F$ to $95^{\mathrm{o}}F$, inclusive, or $[59^{\mathrm{o}}F,95^{\mathrm{o}}F].$