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= -2x2 + 40x + 160

Answer 1

To find out when the two 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 and simplifying this equation gives us:

2x^2 - 24x - 56 = 0

Now we need to solve this quadratic equation for x. Factoring out a 2 from the equation gives:

2(x^2 - 12x - 28) = 0

(x - 14)(x + 2) = 0

So, we have two possible solutions for x: x = 14 or x = -2. Since time cannot be negative, we take x = 14 minutes.

To find out how much water the containers hold at this time, we substitute x = 14 into either equation. Let's use Container A's equation:

y = 16(14) + 104
y = 224 + 104
y = 328

Therefore, at the time when the containers hold the same amount of water (after 14 minutes), they both hold 328 ounces of water.