Step-by-step explanation:
9. The height at the 4.5-hr mark is 16,000 ft. At the 5.0-hr mark, the plane has landed and thus, it's altitude is zero. The change in altitude [tex]\Delta{A}[/tex] is 0 - 16,000 ft or -16,000 ft. The change in time [tex]\Delta{t}[/tex] is 5 hr - 4.5 hr or 0.5 hr. So the rate of change of the altitude is
[tex]\dfrac{\Delta{A}}{\Delta{t}} = \dfrac{-16,000\:\text{ft}}{0.5\:\text{hr}} = -32,000\:\text{ft/hr}[/tex]
10. The initial temperature is 58°F and the final temperature is 76°F, therefore [tex]\Delta{T} = 76°F - 58°F = 18°F.[/tex] The time elapsed is [tex]\Delta{t} = 8\:\text{hrs}.[/tex] The rate of change of the temperature is
[tex]\dfrac{\Delta{T}}{\Delta{t}} = \dfrac{18\text{°F}}{8\:\text{hrs}} = 2.25\text{°F/hr}[/tex]
