Question

Help pleaseeeeee with 9 and 10 ill try to give 100 points

Answer 1

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
`\Delta{A}` is 0 - 16,000 ft or -16,000 ft. The change in time
`\Delta{t}` is 5 hr - 4.5 hr or 0.5 hr. So the rate of change of the altitude is

`\frac{\Delta{A}}{\Delta{t}} = \frac{-16,000\:\text{ft}}{0.5\:\text{hr}} = -32,000\:\text{ft/hr}`

10. The initial temperature is 58°F and the final temperature is 76°F, therefore
`\Delta{T} = 76°F - 58°F = 18°F.` The time elapsed is
`\Delta{t} = 8\:\text{hrs}.` The rate of change of the temperature is

`\frac{\Delta{T}}{\Delta{t}} = \frac{18\text{°F}}{8\:\text{hrs}} = 2.25\text{°F/hr}`