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.
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
1 answer