Solve.
-$$\mathrm { F } = \frac { 9 } { 5 } \mathrm { C } + 32 \text { for } \mathrm { C }$$
A) $C = \frac { 9 } { 5 } ( F - 32 )$
B) $C = \frac { F - 32 } { 9 }$
C) $C = \frac { 5 } { F - 32 }$
D) $C = \frac { 5 } { 9 } ( F - 32 )$

Question

Quizplus · Accepted Answer

To solve for $C$, you need to isolate $C$ on one side of the equation. Starting with $F = \frac{9}{5}C + 32$, subtract 32 from both sides to get $F - 32 = \frac{9}{5}C$. Then, multiply both sides by $\frac{5}{9}$ to isolate $C$, resulting in $C = \frac{5}{9}(F - 32)$.

Solve - F=95C+32 for C\mathrm { F } = \frac { 9 } { 5 } \mathrm { C } + 32 \text { for } \mathrm { C }F=59​C+32 for C

Solve
- $\mathrm { F } = \frac { 9 } { 5 } \mathrm { C } + 32 \text { for } \mathrm { C }$