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 $t=0$ ?
$100^{\circ} \mathrm{F}$
$212^{\circ} \mathrm{F}$
$142^{\circ} \mathrm{F}$
$70^{\circ} \mathrm{F}$
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Answer
Final Answer: \( \boxed{212^{\circ} \mathrm{F}} \)
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: \( \boxed{212^{\circ} \mathrm{F}} \)
