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

Help pleaseeeeee with 9 and 10 ill try to give 100 points-example-1
User YROjha
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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}

User Rjmcb
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