Add a term to the expression so that it becomes a perfect square trinomial. y2 – 19y +
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Given that the volume of water remaining in the tank after t minutes is given by the function
where V is in gallons, 0 ≤ t ≤ 20 is in minutes, and t = 0 represents the instant the tank starts draining.
The rate at which water is draining four and a half minutes after it begins is given by
![\left. \frac{dV}{dt} \right|_{t=4 \frac{1}{2} = \frac{9}{2} }=\left[40,000\left(1- \frac{t}{20} \right)\left(- \frac{1}{20} \right)\right]_{t= \frac{9}{2} } \\ \\ =\left[-2,000\left(1- \frac{t}{20} \right)\right]_{t= \frac{9}{2} }=-2,000\left(1- \frac{4.5}{20} \right) \\ \\ =-2,000(1-0.225)=-2,000(0.775)=-1,550\, gallons\ per\ minute](https://tex.z-dn.net/?f=%5Cleft.%0A%20%5Cfrac%7BdV%7D%7Bdt%7D%20%5Cright%7C_%7Bt%3D4%20%5Cfrac%7B1%7D%7B2%7D%20%3D%20%5Cfrac%7B9%7D%7B2%7D%20%0A%7D%3D%5Cleft%5B40%2C000%5Cleft%281-%20%5Cfrac%7Bt%7D%7B20%7D%20%5Cright%29%5Cleft%28-%20%5Cfrac%7B1%7D%7B20%7D%20%0A%5Cright%29%5Cright%5D_%7Bt%3D%20%5Cfrac%7B9%7D%7B2%7D%20%7D%20%5C%5C%20%20%5C%5C%20%3D%5Cleft%5B-2%2C000%5Cleft%281-%20%0A%5Cfrac%7Bt%7D%7B20%7D%20%5Cright%29%5Cright%5D_%7Bt%3D%20%5Cfrac%7B9%7D%7B2%7D%20%7D%3D-2%2C000%5Cleft%281-%20%0A%5Cfrac%7B4.5%7D%7B20%7D%20%5Cright%29%20%5C%5C%20%20%5C%5C%20%3D-2%2C000%281-0.225%29%3D-2%2C000%280.775%29%3D-1%2C550%5C%2C%20%0Agallons%5C%20per%5C%20minute)
Therefore, the water is draining at a rate of 1,550 gallons per minute four ans a half minutes after it begins.
Answer option E is the correct answer.
where V is in gallons, 0 ≤ t ≤ 20 is in minutes, and t = 0 represents the instant the tank starts draining.
The rate at which water is draining four and a half minutes after it begins is given by
Therefore, the water is draining at a rate of 1,550 gallons per minute four ans a half minutes after it begins.
Answer option E is the correct answer.