3 years ago

" A bowl of soup at 200Á F. is placed in a room of constant temperature of 60Á F. The

1 answer:

7 0

T(t)=60+140e−0.075t T(12)=60+140e−0.075∗12 T(12)=60+140e−0.9 T(12)=60+140(0.4065696597)
=116.84
So the temperature will be approximately 117 degrees

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