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

Answer 1

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.