AP Calculus ABeasymcq1 pt

The temperature TT of a cup of coffee, in degrees Fahrenheit, is modeled by T(t)=70+80e0.2tT(t) = 70 + 80e^{-0.2t}, where tt is the time in minutes since it was brewed. What is the average rate of change of the temperature between t=2t=2 and t=5t=5?

A.80e180e0.43\frac{80e^{-1} - 80e^{-0.4}}{3}
B.80e0.480e13\frac{80e^{-0.4} - 80e^{-1}}{3}
C.16e0.7-16e^{-0.7}
D.80e180e0.45\frac{80e^{-1} - 80e^{-0.4}}{5}

Explanation

Core Concept

The average rate of change over an interval is found by evaluating T(5)T(2)52\frac{T(5) - T(2)}{5 - 2}. Substituting the values yields 70+80e1(70+80e0.4)3\frac{70 + 80e^{-1} - (70 + 80e^{-0.4})}{3}, which correctly simplifies to this expression.

Correct Answer

A80e180e0.43\frac{80e^{-1} - 80e^{-0.4}}{3}

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