College Algebra, Chapter 1, 1.6, Section 1.6, Problem 86

As dry air moves upward, it expands and cools at a rate of about $1^\circ C$ for each 100-meter rise, up to about 12km.
a.) Write a formula for the temperature at height $h$ suppose that the ground temperature is $20^\circ C$
We can use the general equation of the line to represent $h$ since the temperature is changing at a constant rate. So,
$y = mx + b$, where $y$ is temperature $T$ and $x$ is height $h$

Since the temperature is cooling, its magnitude is negative.

$\begin{equation} \begin{aligned} T &= \frac{-1^{\circ}C}{100m} (h) + b && \text{Recall that at ground } h = 0\\ \\ 20 &= \frac{-1}{100} (0) + b\\ \\ b &= 20 \end{aligned} \end{equation}$

Thus, $\displaystyle T = \frac{-1}{100}h + 20$
b.) What range of temperatures can be expected if a place takes off and reaches a maximum height of 5km?

$\begin{equation} \begin{aligned} \text{If } h &= 5 \text{km} \left(\frac{1000m}{1\text{km}} \right) = 5000 m \text{ then the max temperature will be,}\\ \\ T &= \frac{-1}{100} (5000) + 20\\ \\ T &= -30 ^\circ C \end{aligned} \end{equation}$


Thus the range of the temperature is, $20 ^\circ C \leq T \leq - 30^\circ C$