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Algebra 10 Online
OpenStudy (anonymous):

The temperature of Andrea's coffee is given by the equation y = 108e-0.32t + 72 in degrees Fahrenheit t minutes after she pours it. Andrea prefers her coffee at 115 oF. How long should she wait before drinking her coffee? Round your answer to the nearest tenth of a minute.

OpenStudy (caozeyuan):

Two thing I need clearification: 1, y is the temperature in degree F. 2, the correct form of the equation is\[y=108e ^{-0.32t}+72\]

OpenStudy (anonymous):

yes thats the correct form of the equation. and thats the exact question, so im not sure if y is the temperature in degree, but i assume it is

OpenStudy (caozeyuan):

Ok, so y=115, right?

OpenStudy (caozeyuan):

It is stated in the question

OpenStudy (anonymous):

oh yeah nvm

OpenStudy (caozeyuan):

assuming that y is indeed the temp, otherwise we are screwed

OpenStudy (anonymous):

so now what?

OpenStudy (caozeyuan):

so 108*e(t)=115-72=43, where e(t) denotes the exponential part of the equation

OpenStudy (caozeyuan):

\[e(t)=\frac{ 43 }{ 108 }\]

OpenStudy (caozeyuan):

\[\ln (e(t))=\ln(\frac{ 43 }{ 108 })\]

OpenStudy (caozeyuan):

\[\ln(e(t))=-0.32t\]

OpenStudy (caozeyuan):

\[-0.32t=ln(\frac{ 108 }{ 43 })\]

OpenStudy (caozeyuan):

\[t=\frac{ \ln(\frac{ 108 }{ 43 }) }{ -0.32 }\]

OpenStudy (caozeyuan):

put the RHS into calculator and that's your answer

OpenStudy (caozeyuan):

I am soooo sorry! you should have noticed that I have switched places!

OpenStudy (caozeyuan):

explanation\[t=\frac{ \ln(\frac{ 43 }{ 108 }) }{ -0.32 }=\frac{- \ln(\frac{ 43 }{ 108 }) }{ 0.32 }=\frac{ \ln(\frac{ 108 }{ 43 }) }{ 0.32 }\]

OpenStudy (caozeyuan):

that should be you REALLY CORRECT answer

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