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 equals 16 x plus 104
Container B: y equals negative 2 x squared plus 40x plus 160
(1 point)
Responses

360 ounces
360 ounces

328 ounces
328 ounces

232 ounces
232 ounces

136 ounces
136 ounces
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1 answer

To find the point when the two 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

Rearranging terms:

2x^2 + 24x + 56 = 0

Dividing by 2:

x^2 + 12x + 28 = 0

Factoring:

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

Setting each factor to zero:

x = -4 or x = -7

Since time cannot be negative, the containers will hold the same amount of water after 7 minutes.

Plugging in x = 7 to either equation:

Container A: y = 16(7) + 104 = 232 ounces
Container B: y = -2(7)^2 + 40(7) + 160 = 232 ounces

Therefore, the containers will hold 232 ounces of water when they hold the same amount.