Question 1116731
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Body temperature at time zero is 90F, so


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ T(0)\ =\ 90\ =\ 70\ +\ ce^0]


So


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 90\ =\ 70\ +\ c]


Hence


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ c\ =\ 20]


Then


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ T(1)\ =\ 88.5\ =\ 70\ +\ 20e^k]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{28.5}{20}\ =\ e^k]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ k\ =\ \ln(0.925)]


Then


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 98.6\ =\ 70\ +\ 20e^{\ln(0.925)t}]


Solve for *[tex \Large t] which will be a negative number of hours so that you can count back from 11:00 PM to approximate time of death.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
<img src="http://c0rk.blogs.com/gr0undzer0/darwin-fish.jpg">
*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  


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