the function c(t)=108e^{-0.08t}+75 calculates the temperature, in degrees fahrenheit, of a cup of coffee…
the function c(t)=108e^{-0.08t}+75 calculates the temperature, in degrees fahrenheit, of a cup of coffee that was handed out a drive - thru window t minutes ago.\na. what is the temperature of the coffee in the instant that it is handed out the window?\nb. after how many minutes is the coffee in the cup 98 degrees fahrenheit?\na. the coffee is degrees fahrenheit the instant it is handed out the window.
Answer
Answer:
a. 183 b. $\frac{\ln(\frac{23}{108})}{- 0.08}\approx19.94$
Explanation:
Step1: Find initial - temperature (a)
When $t = 0$, substitute $t=0$ into $c(t)=108e^{-0.08t}+75$. $c(0)=108e^{-0.08\times0}+75$. Since $e^{0}=1$, then $c(0)=108\times1 + 75=183$.
Step2: Solve for $t$ (b)
Set $c(t)=98$, so $98 = 108e^{-0.08t}+75$. First, subtract 75 from both sides: $98 - 75=108e^{-0.08t}$, which gives $23 = 108e^{-0.08t}$. Then, divide both sides by 108: $\frac{23}{108}=e^{-0.08t}$. Take the natural - logarithm of both sides: $\ln(\frac{23}{108})=\ln(e^{-0.08t})$. Since $\ln(e^{x}) = x$, we have $\ln(\frac{23}{108})=-0.08t$. Finally, solve for $t$: $t=\frac{\ln(\frac{23}{108})}{- 0.08}\approx19.94$.