Question In the investigation of a homicide, the time of death is important. The normal body temperature of most healthy people is 98.6°F . Suppose that when a body is discovered at noon, its temperature is 82°F . Two hours later it is 72° F . If the temperature of the surroundings is 65°F , what was the approximate time of death? .

Please like the answer.  If u have any doubt, please comment below TQ Let 'to be temp when body discovered ' t_{2}^{\prime} be temp after two hoursGiven to =82^{\circ} \mathrm{F}\text { s } t_{2}=72^{\circ} \mathrm{F}8 temperatuse of suraindigs \left(t_{5}\right)=65^{\circ} \mathrm{F}W.K.T\begin{aligned}T(t) & =\left(t_{0}-t_{s}\right) e^{k t}+t_{s} \\& =(82-65) e^{k t}+65 \\T(t) & =n e^{k t}+65 \rightarrow \text { (1) }\end{aligned}if t=2\begin{array}{l}\Rightarrow T(2)=17 e^{2 K}+65 \\72=17 e^{2 k}+65 \Rightarrow m e^{2 k}=7 \\\Rightarrow e^{2 x}=\frac{7}{17} \\e^{x}=\sqrt{7 / 17} \\\therefore T(t)=17 \cdot\left(\frac{7}{17}\right)^{t / 2}+65 \rightarrow(2) \\\end{array}giventhat bedy temp generat T(t)=98.6 \mathrm{~F}\begin{aligned}\Rightarrow 98.6 & =\pi\left(\frac{7}{17}\right)^{+12}+65 \\\Rightarrow \ln \left(\frac{87}{17}\right)^{+1 / 2} & =\ln \left(\frac{98.6-65}{17}\right) \\\Rightarrow t & =\frac{\ln 1.976}{\ln (7 / 7)} \\& =-1.535\end{aligned}given body is found atroon (1,Pm)\text { Time of death }=12-1.5\simeq 10: 30 Am (approbtratily) ...