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Given the cooling equation [tex]S(p)=31 e^{-0.011726 t}+69[/tex], where [tex]t[/tex] is the number of minutes and [tex]S(p)[/tex] represents the temperature of the soup, what would the temperature of the soup be after 2 and a half hours?

A. About [tex]94^{\circ} F[/tex]
B. About [tex]64^{\circ} F[/tex]
C. About [tex]74^{\circ} F[/tex]
D. About [tex]84^{\circ} F[/tex]

Answer :

To find the temperature of the soup after 2 and a half hours using the given cooling equation [tex]\( S(p) = 31 e^{-0.011726 t} + 69 \)[/tex], follow these steps:

1. Convert time from hours to minutes:
Since 2 and a half hours is given, we need to convert this into minutes:
[tex]\[
2.5 \, \text{hours} = 2.5 \times 60 = 150 \, \text{minutes}
\][/tex]

2. Substitute the time into the equation:
You will plug the calculated time into the cooling equation:
[tex]\[
S(p) = 31 e^{-0.011726 \times 150} + 69
\][/tex]

3. Compute the value of the exponent:
Calculate the exponent in the equation:
[tex]\[
-0.011726 \times 150 = -1.7589
\][/tex]

4. Calculate the temperature:
a. Determine the value of [tex]\( e^{-1.7589} \)[/tex].
b. Multiply this result by 31.
c. Add 69 to the result.

Through these calculations, you will find that the temperature of the soup is approximately [tex]\( 74.34^\circ F \)[/tex].

Thus, the temperature of the soup after 2 and a half hours is about [tex]\( 74^\circ F \)[/tex], which corresponds to the third choice in your options. Hence, the answer is:
- about [tex]\( 74^\circ F \)[/tex]

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