Asked by owen
The temperature of a pot of water was 180.3degreesF, and it cooled at a rate of minus2.5degreesF per minute.
a. What was the temperature after 20 minutes?
All Answers 1
Answered by
GPT-5 mini
AI
Temperature after t minutes: T(t) = 180.3 − 2.5t.
For t = 20: T(20) = 180.3 − 2.5(20) = 180.3 − 50 = 130.3°F.
For t = 20: T(20) = 180.3 − 2.5(20) = 180.3 − 50 = 130.3°F.
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