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A chef uses a warming oven to maintain the temperature of food until all items are ready for plating. Potato croquettes, whose temperature is 200 C, are placed in the warming oven. The relationship between the core temperature of the potato in degrees Celsius, and the cooking time, t in minutes, is modelled by the equation 200 – T=180(0.94)t. If the chef needs to plate the entrée before the potato croquettes have a core temperature of 1400 C, how much time does he have?

A Chef Uses A Warming Oven To Maintain The Temperature Of Food Until All Items Are Ready For Plating Potato Croquettes Whose Temperature Is 200 C Are Placed In class=

Sagot :

Given

[tex]200-T=180(0.94)^t[/tex]

Solution

When T=140

[tex]\begin{gathered} 200-T=180(0.94)^{t} \\ \\ 200-140=180(0.94)^t \\ 60=180(0.94)^t \\ \\ 60=180\cdot\:0.94^t \\ Divide\text{ both side 180} \\ \frac{180\cdot \:0.94^t}{180}=\frac{60}{180} \\ \\ simplify \\ 0.94^t=\frac{1}{3} \end{gathered}[/tex]

Apply exponent rule

[tex]\begin{gathered} t\ln \left(0.94\right)=\ln \left(\frac{1}{3}\right) \\ \\ t=-\frac{\ln \left(3\right)}{\ln \left(0.94\right)} \\ \\ \\ t=17.7552 \end{gathered}[/tex]

The final answer

t =1 7.7552years

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