A bowl of soup at 175°F is placed in a room of constant temperature of 70°F. The temperature T of the soup t minutes after it is placed in the room is given by T(t) = 70 + 105e^(-0.075t). Find the temperature of the soup 18 minutes after it is placed in the room (to the nearest degree). Suppose $6500 is invested in an account at an annual interest rate of 7.7% compounded continuously. How long (to the nearest tenth of a year) will it take the investment to double in size?
Added by Alex S.
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075t, where t is the time in minutes since the soup was placed in the room. We want to find the temperature of the soup 18 minutes after it was placed in the room. So we substitute t = 18 into the formula: T(18) = 70+105e-0.075(18) T(18) = 70+105e-1.35 T(18) ≈ Show more…
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