Answer:
[tex]T(t)=-50e^{-0.4t}+42[/tex]
Step-by-step explanation:
We are given that
[tex]T(0)=-8^{\circ} C[/tex]
[tex]T'(t)=20e^{-0.4t}[/tex]
Integrating on both sides
[tex]T(t)=20\int e^{-0.4 t}[/tex]
[tex]T(t)=\frac{20}{-0.4}e^{-0.4t}+C[/tex]
Using the formula
[tex]\int e^{ax}dx=\frac{e^{ax}}{a}+C[/tex]
Substitute t=0 and T(0)=-8
[tex]-8=-50+C[/tex]
[tex]C=-8+50=42[/tex]
Substitute the value of C
[tex]T(t)=-50e^{-0.4t}+42[/tex]
