MathematicsMedium38×since 2002Q3564The temperature T(t)T(t)T(t) of a body at time t=0t=0t=0 is 160∘F160^{\circ} \mathrm{F}160∘F and it decreases continuously as per the differential equation dTdt=−K(T−80)\frac{d T}{d t}=-K(T-80)dtdT=−K(T−80), where KKK is a positive constant. If T(15)=120∘FT(15)=120^{\circ} \mathrm{F}T(15)=120∘F, then T(45)T(45)T(45) is equal toA90∘^\circ∘ FB85∘^\circ∘ FC80∘^\circ∘ FD95∘^\circ∘ FCheck answerSkip