Question 1111612
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You have to solve the equation for *[tex \Large t] and then subtract that amount of time from 10:25 AM.  The thing that makes your problem unsolvable as stated is that you did not specify any units for the constant *[tex \Large k].  However, presuming that *[tex \Large k\ =\ 0.1336] degrees per hour, proceed as follows:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ T(t)\ =\ T_e\ +\ (T_o\ -\ T_e)e^{-kt}]


Substitute given data


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 80\ =\ 68\ +\ (98.6\ -\ 68)e^{-0.1336t}]


Which simplifies to


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ e^{-0.1336t}\ =\ \frac{12}{30.6}]


Take the natural log of both sides


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -0.1336t\ =\ \ln{\frac{12}{30.6}}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t\ =\ \frac{\ln{\frac{12}{30.6}}}{-0.1336}]


Your calculator should then give you a number of hours to subtract from 10:25 AM presuming my assumption regarding the units on the constant *[tex \Large k] was correct.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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