SOLUTION: A soda can that has a temperature of
80 degrees
F is put into a cooler containing ice (
32degrees
F). The temperature of the soda can
after t minutes can be modeled by f(t)=32
Algebra.Com
Question 1079078: A soda can that has a temperature of
80 degrees
F is put into a cooler containing ice (
32degrees
F). The temperature of the soda can
after t minutes can be modeled by f(t)=32+48(0.9)^t
How long will it take to cool the can of soda down to 50 degrees?
Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
50=32+48*(0.9)^t
18=48*(0.9)^t
(3/8)=0.9^t
ln(3/8)=t ln (0.9)
divide and round at the end
t=9.31 hours
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