SOLUTION: A cup of coffee is heated to 180°F and placed in a room that maintains a temperature of 75°F. The temperature of the coffee after t minutes is given by T(t) = 75 + 115e^−0.

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: A cup of coffee is heated to 180°F and placed in a room that maintains a temperature of 75°F. The temperature of the coffee after t minutes is given by T(t) = 75 + 115e^−0.      Log On


   

Question 1004659: A cup of coffee is heated to 180°F and placed in a room that maintains a temperature of 75°F. The temperature of the coffee after t minutes is given by
T(t) = 75 + 115e^−0.042t.
(a) Find the temperature, to the nearest degree, of the coffee 30 minutes after it is placed in the room.
=°F
(b) Determine when, to the nearest tenth of a minute, the temperature of the coffee will reach 130°F.
=min
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
+T%28t%29+=+75+%2B+115%2Ae%5E%28-.042t%29+
(a)
+t+=+30+
+T%2830%29+=+75+%2B+115%2Ae%5E%28-.042%2A30%29+
+T%2830%29+=+75+%2B+115%2Ae%5E%28-1.26%29+
+T%2830%29+=+75+%2B+115%2A.28365+
+T%2830%29+=+75+%2B+32.6202+
+T%2830%29+=+107.6202+
+T%2830%29+=+108+ degrees F
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(b)
+T%28t%29+=+75+%2B+115%2Ae%5E%28-.042t%29+
+130+=+75+%2B+115%2Ae%5E%28-.042t%29+
+55+=+115%2Ae%5E%28-.042t%29+
+e%5E%28-.042t%29+=+.47826+
+ln%28+.47826+%29+=+-.042t+
+-.7376+=+-.042t+
+t+=+17.562+
+t+=+17.6+ min
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