A thermometer is taken from a room where the temperature is [tex]23^{o}C[/tex] to the outdoors, where the temperature is [tex]-11^{o}C[/tex]. After one minute the thermometer reads [tex]14^{o}C[/tex].(adsbygoogle = window.adsbygoogle || []).push({});

A.) What will the reading on the thermometer be after 3 more minutes?

my work:

[tex] T=23^{o}C[/tex]

[tex] T_s = -11^{o}C[/tex]

[tex] 23 - (-11) = 34e^{kt}[/tex] <-- initial temp right?

after 1 min...

[tex]t(1) = 34e^{k1} = 14 - (-11)[/tex]

[tex]t(1) = 34e^{k1} = 25[/tex]

solved for k and got -0.307484

my equation:

[tex] y(0) = 34e^{-0.307484*t} - 11[/tex]

so i just subbed in 3 for t and solved and got 2.516, but it's incorrect. anyone know where i went wrong?

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# Homework Help: Temperatures - math

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