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You said your problem had to do with Newton's law of cooling, is that the equation Y(t)=Ys+Aoe^kt?
hey nick are you still there?
it say's I have 2 replys yet I only saw one questioned response from nick am I doing something wrong?

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All right, I think I figured out your problem, it may take a while, but I'll explain it.
c'mon please help? I'll show you what I think I know lol
In that equation Y(t)=Ys+Aoe^kt, the Ys=surrounding temperature at t=0. In this case, let t=0 be the 12:00 AM time. So the Ys in this problem is 65. Ao = Y(0) - Ys, which is 78-65 = 13, the 78 being the temperature at time 0. So, your equation should start out looking like Y(t)=65+13e^kt.
Now, we need to solve for k. To do that, set your Y(t) to the second temperature taken, which was 74 degrees. You also know t, since the second temperature was recorded an hour later or 60 minutes later. So it should be 74=65+13e^(60k).
Move the 65 over, which makes it 9=13e^(60k). To get rid of the e, take the log of both sides, i.e. ln(9/13)=lne^(60k).
That will give you ln(9/13)=60k. Solve for k and you should get something like -.0061...
Now that you have the k value, you can solve for when the body died. Set the equation equal to 98.6. So you should now have 98.6=65+13e^(-.0061...t)
Solve t the same way you found the k value and you will be able to determine when the boy died.
k thank you
No problem, glad I could help.
yes you did I was in panic mode. The equation we were given varies but it's the same idea

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