Water is added to two containers for 16 minutes. The equations below model the ounces of water, y, in each container after x minutes. At the time after the start when the containers hold the same amount of water, how much water do they hold?
Container A: y=16x+104
Container b: y=-2x^2+40x+160
To find when the containers hold the same amount of water, we need to set the two equations equal to each other:
16x + 104 = -2x^2 + 40x + 160
Rearranging the equation, we get:
2x^2 + 24x + 56 = 0
Dividing the entire equation by 2:
x^2 + 12x + 28 = 0
Now we need to factor the quadratic equation:
(x + 2)(x + 14) = 0
Therefore, x = -2 or x = -14. Since time cannot be negative, the containers will hold the same amount of water after 14 minutes from the start.
Now, substituting x = 14 into either equation, we can find how much water they hold at that time:
For Container A: y = 16(14) + 104 = 224 + 104 = 328 ounces of water
For Container B: y = -2(14)^2 + 40(14) + 160 = -2(196) + 560 + 160 = -392 + 560 + 160 = 328 ounces of water
Therefore, both containers hold 328 ounces of water when 14 minutes have passed.