Question

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



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Answer 1

To find the point at which the containers 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

Rearrange and simplify:

2x^2 + 24x + 56 = 0

Divide by 2 to simplify:

x^2 + 12x + 28 = 0

Factor the quadratic equation:

(x + 4)(x + 7) = 0

Solve for x:

x = -4 or x = -7

Since time cannot be negative, we discard the negative solution. Therefore, the containers hold the same amount of water after 7 minutes.

To find out how much water they hold after 7 minutes, substitute x = 7 into one of the equations:

Container A:
y = 16(7) + 104
y = 112 + 104
y = 216 ounces

Container B:
y = -2(7)^2 + 40(7) + 160
y = -2(49) + 280 + 160
y = -98 + 280 + 160
y = 342 ounces

Therefore, both containers hold 216 ounces of water after 7 minutes.