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A bottle of white wine at room temperature 68F68^\circ\text{F} is placed in a refrigerator at 44 p.m. Its temperature after tt hr is changing at the rate of 18e0.6t-18e^{-0.6t} F/hour^\circ\text{F}/\text{hour}. By how many degrees will the temperature of the wine have dropped by 77 p.m.? What will the temperature of the wine be at 77 p.m.?

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Q. A bottle of white wine at room temperature 68F68^\circ\text{F} is placed in a refrigerator at 44 p.m. Its temperature after tt hr is changing at the rate of 18e0.6t-18e^{-0.6t} F/hour^\circ\text{F}/\text{hour}. By how many degrees will the temperature of the wine have dropped by 77 p.m.? What will the temperature of the wine be at 77 p.m.?
  1. Calculate Time Difference: The wine is placed in the fridge at 44 p.m. and we want to know the temperature at 77 p.m., so we calculate the time difference.\newlineTime difference = 77 p.m. - 44 p.m. = 33 hours.
  2. Find Temperature Change Rate: The temperature change rate is given by the function 18e0.6t-18e^{-0.6t}. We need to plug t=3t = 3 into the function to find the temperature drop after 33 hours. Temperature drop = 18e(0.63)-18e^{(-0.6*3)}.
  3. Calculate Exponent Part: Now we calculate the exponent part: 0.6×3=1.8-0.6 \times 3 = -1.8.
  4. Calculate e1.8e^{-1.8}: Next, we calculate e1.8e^{-1.8}. Using a calculator, e1.80.1653e^{-1.8} \approx 0.1653.
  5. Find Temperature Drop: Now we multiply this by 18-18 to find the temperature drop.\newlineTemperature drop = 18×0.16532.9754-18 \times 0.1653 \approx -2.9754.\newlineOops, we made a mistake here. The temperature drop should be a positive value since we're talking about how much it has decreased, not the direction of change.

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