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At the moment a hot cake is put in a cooler, the difference between the cake's and the cooler's temperatures is 50^(@) Celsius. This causes the cake to cool and the temperature difference loses (1)/(5) of its value every minute. Write a function that gives the temperature difference in degrees Celsius, D(t),t minutes after the cake was put in the cooler. D(t)= ◻

At the moment a hot cake is put in a cooler, the difference between the cake's and the cooler's temperatures is 50∘ 50^{\circ} Celsius. This causes the cake to cool and the temperature difference loses 15 \frac{1}{5} of its value every minute.\newlineWrite a function that gives the temperature difference in degrees Celsius, D(t),t D(t), t minutes after the cake was put in the cooler.\newlineD(t)= D(t)= □ \square

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Q. At the moment a hot cake is put in a cooler, the difference between the cake's and the cooler's temperatures is 50∘ 50^{\circ} Celsius. This causes the cake to cool and the temperature difference loses 15 \frac{1}{5} of its value every minute.\newlineWrite a function that gives the temperature difference in degrees Celsius, D(t),t D(t), t minutes after the cake was put in the cooler.\newlineD(t)= D(t)= □ \square
  1. Initial Temperature Difference: The initial temperature difference is 5050 degrees Celsius, so D(0)=50D(0) = 50.
  2. Decay Rate Calculation: The temperature difference decreases by (1)/(5)(1)/(5) of its current value each minute, so the decay factor is 1−(1)/(5)=(4)/(5)1 - (1)/(5) = (4)/(5) per minute.
  3. Exponential Decay Function: The function for the temperature difference is an exponential decay function, D(t)=D(0)⋅(decay factor)tD(t) = D(0) \cdot (\text{decay factor})^{t}.
  4. Substitution into Equation: Substitute D(0)=50D(0) = 50 and decay factor =45= \frac{4}{5} into the equation: D(t)=50×(45)tD(t) = 50 \times \left(\frac{4}{5}\right)^{t}.

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