## Prealgebra (7th Edition)

$50^oF=10^oC$ $59^oF=15^oC$ $68^oF=20^oC$ $77^oF=25^oC$
$\frac{5(F-32)}{9}$ substitute each given value for F and solve $\frac{5(50-32)}{9}=\frac{5(18)}{9}=\frac{(5)(2)(/\!\!9)}{/\!\!9}=10$ $\frac{5(59-32)}{9}=\frac{5(27)}{9}=\frac{(5)(3)(/\!\!9)}{/\!\!9}=15$ $\frac{5(68-32)}{9}=\frac{5(36)}{9}=\frac{(5)(4)(/\!\!9)}{/\!\!9}=20$ $\frac{5(77-32)}{9}=\frac{5(45)}{9}=\frac{(5)(5)(/\!\!9)}{/\!\!9}=25$