Question

A bowl of soup at 200° F. is placed in a room of constant temperature of 60° F. The
temperature T of the soup t minutes after it is placed in the room is given by
T(t) = 60 + 140 e
– 0.075 t
Find the temperature of the soup 12 minutes after it is placed in the room. (Round to the nearest
degree.)

Answer 1

\[\Large T(t) = 60+140e^{-0.075t}\]


\[\Large T(12) = 60+140e^{-0.075*12}\]


\[\Large T(12) = 60+140e^{-0.9}\]


\[\Large T(12) = 60+140(0.4065696597)\]


I'll let you finish.

Answer 2

Thank you, I got 116.92 degrees F

Answer 3

then you round that to the nearest degree to get the final answer of 117 degrees F

Answer 4

Yes, Thank you