## Algebra 2 (1st Edition)

$C=\dfrac{5}{9}(F-32)$ and $10^{\circ}$
As we are given that, $F-32=\dfrac{9}{5} C$ The formula for $C$ can be defined as: $C=\dfrac{5}{9}(F-32)$ Plug $50^{\circ}$ for $F$ Then, we have $C=\dfrac{5}{9}(50-32)=10^{\circ}$ Hence, $C=\dfrac{5}{9}(F-32)$ and $10^{\circ}$.