Problema Solution
A bowl of soup at 195F, is placed in a room of constant temperature of 65F. The temperature T of the soup t minutes after it is placed in the room is given by T(t)=65+130e^- 0.075t
Find the temperature of the soup 6 minutes after it is placed in the room (round to the nearest degree)
Answer provided by our tutors
We assume that "e^- 0.075t" raises the enter expression "e^(- 0.075t)". Since 't' is in seconds, T(t) provides:
64+130e^(-0.075(6))
click here to simplify the expression
146.89
So the temperature of the soup would be approximately 147 degrees.