Problem

Question 21 We made a pot of tea and are waiting for it to cool. The temperature of the tea after $t$ minutes is given by $y(t)=142 e^{-.04 t}+70$. How hot was the tea at time $t=0$ ? $100^{\circ} \mathrm{F}$ $212^{\circ} \mathrm{F}$ $142^{\circ} \mathrm{F}$ $70^{\circ} \mathrm{F}$ Submit Question

Solution

Step 1 :To find the temperature of the tea at time \( t=0 \), we need to substitute \( t=0 \) into the given function \( y(t)=142 e^{-0.04 t}+70 \) and calculate the value.

Step 2 :Substituting \( t=0 \) into the function, we get \( y(0)=142 e^{-0.04 \cdot 0}+70 \).

Step 3 :Simplifying the expression, we have \( y(0)=142 e^{0}+70 \).

Step 4 :Since \( e^{0} \) is equal to 1, the expression becomes \( y(0)=142 \cdot 1 + 70 \).

Step 5 :Adding the values, we get \( y(0)=142 + 70 \).

Step 6 :Calculating the sum, we find that \( y(0)=212 \).

Step 7 :Final Answer: \( \boxed{212^{\circ} \mathrm{F}} \)

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