A pie comes out of an oven at $335^{\circ} \mathrm{F}$ and is placed to cool in a $70^{\circ} \mathrm{F}$ kitchen. The temperature of the pie $T$ (in ${ }^{\circ} \mathrm{F}$ ) after $t$ minutes is given by $T=70+265 e^{-0.017 t}$. The pie is cool enough to cut when the temperature reaches $110^{\circ} \mathrm{F}$. Determine the time this will take. Round to the nearest minute. The pie is cool enough to cut after approximately $\square$ minutes.

Step 1 ：We are given a pie that comes out of an oven at $335^{\circ} \mathrm{F}$ and is placed to cool in a $70^{\circ} \mathrm{F}$ kitchen. The temperature of the pie $T$ (in ${ }^{\circ} \mathrm{F}$ ) after $t$ minutes is given by $T=70+265 e^{-0.017 t}$. The pie is cool enough to cut when the temperature reaches $110^{\circ} \mathrm{F}$. We need to determine the time this will take.

Step 2 ：We need to solve the equation $70+265 e^{-0.017 t} = 110$ for $t$.

Step 3 ：First, we subtract 70 from both sides of the equation to get $265 e^{-0.017 t} = 40$.

Step 4 ：Next, we divide both sides of the equation by 265 to get $e^{-0.017 t} = \frac{40}{265}$.

Step 5 ：Then, we take the natural logarithm of both sides of the equation to get $-0.017 t = \ln\left(\frac{40}{265}\right)$.

Step 6 ：Finally, we divide both sides of the equation by -0.017 to get $t = \frac{\ln\left(\frac{40}{265}\right)}{-0.017}$.

Step 7 ：Using a calculator, we find that $t \approx 112$.

Step 8 ：Final Answer: The pie is cool enough to cut after approximately $\boxed{112}$ minutes.