Answered

" A bowl of soup at 200Á F. is placed in a room of constant temperature of 60Á 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 :

syed514
T(t)=60+140e−0.075t T(12)=60+140e−0.075∗12 T(12)=60+140e−0.9 T(12)=60+140(0.4065696597)
        =116.84
 So the temperature will be approximately 117 degrees

Other Questions