Ryan places a bottle of water inside a cooler. As the water cools, its temperature \( C(t) \) in degrees Celsius is given by the following function, where \( t \) is the number of minutes since the bottle was placed in the cooler. \[ C(t)=4+20 e^{-0.045 t} \] Ryan wants to drink the water when it reaches a temperature of 17 degrees Celsius. How many minutes should he leave it in the coolen? Round your answer to the nearest tenth, and do not round any interimediate computations.
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To find out how many minutes \( t \) Ryan should leave the bottle in the cooler for it to reach 17 degrees Celsius, we need to set the equation \( C(t) = 17 \) and solve for \( t \) in the equation: \[ 17 = 4 + 20 e^{-0.045 t} \] First, we can simplify the equation: \[ 17 - 4 = 20 e^{-0.045 t} \implies 13 = 20 e^{-0.045 t} \] Now, divide both sides by 20: \[ e^{-0.045 t} = \frac{13}{20} \] Next, take the natural logarithm of both sides: \[ -0.045 t = \ln\left(\frac{13}{20}\right) \] Solving for \( t \): \[ t = \frac{\ln\left(\frac{13}{20}\right)}{-0.045} \] Calculating \( \ln\left(\frac{13}{20}\right) \): \[ \ln\left(\frac{13}{20}\right) \approx -0.1744 \] Now substituting back into the equation for \( t \): \[ t \approx \frac{-0.1744}{-0.045} \approx 3.87 \] Rounding to the nearest tenth: \[ t \approx 3.9 \] Ryan should leave the bottle in the cooler for approximately 3.9 minutes.
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