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 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

Answer 1

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.