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 container 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 point)
360 ounces
360 ounces
328 ounces
328 ounces
232 ounces
232 ounces
136 ounces
136 ounces
To find the time after the start when the containers hold the same amount of water, we need to set the equations equal to each other and solve for x:
16x + 104 = -2x^2 + 40x + 160
Rearranging the equation:
2x^2 + 24x - 56 = 0
Factoring:
2(x^2 + 12x - 28) = 0
(x + 14)(x - 2) = 0
Setting each factor equal to 0:
x + 14 = 0 -> x = -14
x - 2 = 0 -> x = 2
Since we cannot have negative time, we discard the solution x = -14.
Therefore, the time after the start when the containers hold the same amount of water is 2 minutes.
Substituting x = 2 into either equation, we can find the amount of water:
Container A: y = 16(2) + 104 = 32 + 104 = 136 ounces
Container B: y = -2(2)^2 + 40(2) + 160 = -2(4) + 80 + 160 = 328 ounces
Therefore, at the time after the start when the containers hold the same amount of water, they hold 136 ounces of water.