## Calculus, 10th Edition (Anton)

$T = 15^\circ F$
We need to find the air temperature to the nearest degree. Denote it $T$. We are given that $WCT = -10^\circ F$, $v = 48 \, mi/h$. Recall the definition of the WCT: $$WCT = \left\{\begin{array}{lr} T, & 0\leq v\leq 3\\ 35.74 + 0.6215T − 35.75v^{0.16} + 0.4275T v^{0.16}, & v>3\\ \end{array}\right\}$$ We have that $v > 3$ and therefore we can write: $$WCT = 35.74 + 0.6215T − 35.75v^{0.16} + 0.4275T v^{0.16}$$ Substitute for the value of the $WCT$ to get: $$35.74 + 0.6215T − 35.75v^{0.16} + 0.4275T v^{0.16} = -10$$ Substitude for the value of $v$ to get: $$35.74 + 0.6215T − 35.75 \cdot (48^{0.16}) + 0.4275T \cdot (48^{0.16}) = -10$$ Rewrite the equation as follows: $$T \cdot \left(0.6215 + 0.4275 \cdot \left(48^{0.16}\right)\right) = -10 - 35.74 + 35.75 \cdot (48^{0.16})$$ $$T = \frac{-45.74 + 35.75 \cdot (48^{0.16})}{\left(0.6215 + 0.4275 \cdot \left(48^{0.16}\right)\right)}$$ And plugging it in a calculator yields $T \approx 14.6^\circ F$. We are requested to find the air temperature to the nearest degree; thus the answer is $T = 15^\circ F$.