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7. (0/3 points) DETAILS PREVIOUS ANSWERS SALGTRIG4 4.6.025.MI. This exercise uses Newtons Law of Cooling. A hot bowl of soup
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Answer #1

ans given Tlt) -0.05t 59 + 143 e Initially at to T(D) : 59 + 143 00 59 + 143 202°F t=10 min Thool -0.05X10 59 + 143 e 59 + 14

a) initial temperature = 202°F

b) temperature after 10 min = 145.7°F

c) time for temperature to be 100°F = 25 minutes

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7. (0/3 points) DETAILS PREVIOUS ANSWERS SALGTRIG4 4.6.025.MI. This exercise uses Newton's Law of Cooling. A...
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