A bowl of soup at 165° F. is placed in a room of constant temperature of 75° F. The temperature T of the soup t minutes after it is placed in the room is given by T(t) = 75 + 90 e^– 0.075 t
Find the temperature of the soup 20 minutes after it is placed in the room. (Round to the nearest
degree.)

Question

A bowl of soup at 165° F. is placed in a room of constant temperature of 75° F. The temperature T of the soup t minutes after it is placed in the room is given by T(t) = 75 + 90 e^– 0.075 t
Find the temperature of the soup 20 minutes after it is placed in the room. (Round to the nearest
degree.)

anonymous · Answer

plug in t=20 in the given equation

anonymous · Answer

75 + 90e^-.075 x 20
75 + 90e^-1.5
75 + 90 x .223
95.08

after rounding it is 95