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
360 ounces
328 ounces
232 ounces
136 ounces
To find the amount of water in each container at the time when they hold the same amount of water, we need to set the two equations equal to each other and solve for x:
16x + 104 = -2x^2 + 40x + 160
0 = -2x^2 + 24x + 56
0 = -x^2 + 12x + 28
0 = x^2 - 12x - 28
Now we need to factor this equation:
0 = (x - 14)(x + 2)
Setting each factor equal to zero, we get:
x - 14 = 0
x = 14
x + 2 = 0
x = -2
Since x = -2 is not a valid solution in the context of the problem, we will use x = 14.
Now, we can substitute x = 14 back into either equation to find the amount of water in each container:
Container A: y = 16(14) + 104
y = 360 ounces
Container B: y = -2(14)^2 + 40(14) + 160
y = 360 ounces
Therefore, at the time after the start when the containers hold the same amount of water, they both hold 360 ounces.