SOLUTION: A can of soda at 82°F is placed in a refrigerator that maintains a constant temperature of 39°F. The temperature T of the soda t minutes after it is placed in the refrigerator is g

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: A can of soda at 82°F is placed in a refrigerator that maintains a constant temperature of 39°F. The temperature T of the soda t minutes after it is placed in the refrigerator is g      Log On


   



Question 1006202: A can of soda at 82°F is placed in a refrigerator that maintains a constant temperature of 39°F. The temperature T of the soda t minutes after it is placed in the refrigerator is given by
T(t) = 39 + 43e^−0.058t.
(a) Find the temperature, to the nearest degree, of the soda 10 minutes after it is placed in the refrigerator.
=°F
(b) When, to the nearest minute, will the temperature of the soda be 43°F?
=min

Answer by Boreal(15235) About Me  (Show Source):
You can put this solution on YOUR website!
T(t) = 39 + 43e^−0.058t.
after 10 minutes
39+43*(e^-0.58)
63 degrees.
==========
43=39+43 e^(-0.058)
(4/43)=e^(-0.058)t
ln of both sides and lx e^x=x
-2.375=-0.058t
t=41 minutes