Answer:
[tex]T(t)=-20e^{-0.5t}+13[/tex]
Step-by-step explanation:
The object is taken from a freezer at [tex]-7^0C[/tex]
[tex]T'(t)=10e^{-0.5t}[/tex]
To determine the temperature T(t), at any time t, we take the integral of the average increase in temperature.
[tex]\int T'(t) dt=\int (10e^{-0.5t})dt\\T(t)=-20e^{-0.5t}+k, $where k is a constant of integration$\\At \:t=0, T(t)=-7^0C\\-7=-20e^{-0.5(0)}+k\\k=-7+20=13[/tex]
The temperature of the object at time t is given as:
[tex]T(t)=-20e^{-0.5t}+13[/tex]
