A cooling cup of coffee follows dTdt=−0.1(T−20)\frac{dT}{dt} = -0.1(T - 20)dtdT=−0.1(T−20). If T(0)=90T(0) = 90T(0)=90, what is the temperature T(t)T(t)T(t)?
T(t)=20+70e−0.1tT(t) = 20 + 70e^{-0.1t}T(t)=20+70e−0.1t
T(t)=20+90e−0.1tT(t) = 20 + 90e^{-0.1t}T(t)=20+90e−0.1t
T(t)=70+20e−0.1tT(t) = 70 + 20e^{-0.1t}T(t)=70+20e−0.1t
T(t)=90e−0.1tT(t) = 90e^{-0.1t}T(t)=90e−0.1t