Question

Number 39 not a test, exam, homework or anything gradedPrecalc college level

Answer 1

we have the formula

`T(t)=T_s+D_0e^(-kt)`

step 1

Find out the value of k

we have

T=150 degrees

Ts=70

To=200

Do=200-70=130

t=10 min

substitute

`\begin{gathered} 150=70+130e^(-10k) \\ 150-70=130e^(-10k) \\ 80=130e^(-10k) \\ (80)/(130)=e^(-10k) \end{gathered}`

Apply ln on both sides

`\ln ((80)/(130))=\ln e^(-10k)`
`\begin{gathered} \ln ((80)/(130))=-10k\cdot\ln e^{} \\ \ln ((80)/(130))=-10k \\ k=0.0486 \end{gathered}`

substitute the value of k in the given formula

`T(t)=T_s+D_0e^(-0.0486t)`

Now

For t=20 min

Ts=70

To=200

Do=200-70=130

substitute

`\begin{gathered} T(t)=70+130\cdot e^(-0.0486\cdot10) \\ T(t)=150 \end{gathered}`

The answer is option A