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## Re: Differential Equation

Thermometer was taken outside

$T = -10 + (70 + 10)e^{-kt}$

$T = -10 + 80e^{-kt}$

At 1:02 pm, t = 2 and T = 26°F

$26 = -10 + 80e^{-2k}$

$26 = -10 + 80e^{-2k}$

$\frac{36}{80} = e^{-2k}$

$e^{-k} = \left( \frac{9}{20} \right)^{1/2}$

Hence,

$T = -10 + 80\left( \frac{9}{20} \right)^{t/2}$

At 1:05pm, t = 5

$T = -10 + 80\left( \frac{9}{20} \right)^{5/2}$

$T = 0.8673^\circ F$ ← thermometer reading at 1:05 pm

Thermometer was brought back to the room

$T = 70 + (0.8673 - 70)\left( \frac{9}{20} \right)^{t/2}$

$T = 70 + (0.8673 - 70)\left( \frac{9}{20} \right)^{t/2}$

$T = 70 - 69.1327\left( \frac{9}{20} \right)^{t/2}$

At 1:09 pm, t = 4

$T = 70 - 69.1327\left( \frac{9}{20} \right)^{4/2}$

$T = 56^\circ F$

answer