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contestada

Suppose you turn off the engine of your car when the temperature of the engine reaches 191 °F. If the outside temperature is a constant 66 °F, then the temperature T of the engine t minutes after you turn off the engine satisfies the equation below.
ln

T − 66
125

= −0.07t

(a) Solve the equation for T.
T =


(b) Find the temperature of the engine 30 minutes after you turn it off. Give your answer to the nearest tenth of a degree.

°F

Respuesta :

antonsandiego
a) 
Take  Newton’s law of cooling as a basis and make a transformation : 
From : [tex]ln(T-65/125) = -0.07t [/tex] to :[tex]ln[ (T - 65 ) / 125 ] = -0.07t [/tex]
Then solve : [tex]ln[ (T - 65 ) / 125 ] = -0.07t[/tex]
[tex](T - 65 ) / 125 = e^(-0.07t) [/tex]
And the  result :[tex]T = 65 + 125e^(-0.07t) [/tex]

b) The easiest way how  you can find the temperature of the engine 30 minutes after you turn it off : [tex]T = 65 + 125e^(-0.07t) [/tex]
[tex]T = 65 + 125e^(-0.07*30)[/tex]
Here is the solution in the nearest tenth of a degree  : T = 80.3℉ 

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