The temperature $T(t)$ of a body at time $t=0$ is $160^{\circ} \mathrm{F}$ and it decreases continuously as per the differential equation $\frac{d T}{d t}=-K(T-80)$, where $K$ is a positive constant. If $T(15)=120^{\circ} \mathrm{F}$, then $T(45)$ is equal to

<p>$$\begin{aligned} & \frac{\mathrm{dT}}{\mathrm{dt}}=-\mathrm{k}(\mathrm{T}-80) \\ & \int_\limits{160}^{\mathrm{T}} \frac{\mathrm{dT}}{(\mathrm{T}-80)}=\int_\limits0^{\mathrm{t}}-\mathrm{Kdt} \\ & {[\ln |\mathrm{T}-80|]_{160}^{\mathrm{T}}=-\mathrm{kt}} \\ & \ln |\mathrm{T}-80|-\ln 80=-\mathrm{kt} \\ & \ln \left|\frac{\mathrm{T}-80}{80}\right|=-\mathrm{kt} \\ & \mathrm{T}=80+80 \mathrm{e}^{-\mathrm{kt}} \\ & 120=80+80 \mathrm{e}^{-\mathrm{k} .15} \\ & \frac{40}{80}=\mathrm{e}^{-\mathrm{k} 15}=\frac{1}{2} \\ & \therefore \mathrm{T}(45)=80+80 \mathrm{e}^{-\mathrm{k} .45} \\ & =80+80\left(\mathrm{e}^{-\mathrm{k} .15}\right)^3 \\ & =80+80 \times \frac{1}{8} \\ & =90 \end{aligned}$$</p>