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)
A. 360 ounces
B. 328 ounces
C. 232 ounces
D. 136 ounces
To find the amount of water each container holds when they have the same amount, we need to set the equations equal to each other and solve for x:
16x + 104 = -2x^2 + 40x + 160
Rearranging the equation and combining like terms:
2x^2 - 24x - 56 = 0
Dividing both sides of the equation by 2:
x^2 -12x - 28 = 0
Now we can solve this quadratic equation by factoring, completing the square, or using the quadratic formula. Unfortunately, this equation does not factor easily, so let's use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
In this equation, a = 1, b = -12, and c = -28. Substituting these values into the quadratic formula, we get:
x = (-(-12) ± √((-12)^2 - 4(1)(-28))) / (2(1))
Simplifying:
x = (12 ± √(144 + 112)) / 2
x = (12 ± √256) / 2
Now we have two possible values for x:
x = (12 + 16) / 2 = 14
x = (12 - 16) / 2 = -2
Since time cannot be negative, we ignore the value x = -2.
Now we substitute x = 14 into either equation to find the amount of water in each container:
For Container A: y = 16x + 104
y = 16(14) + 104
y = 224 + 104
y = 328 ounces
Therefore, when the containers hold the same amount of water, they both hold 328 ounces.
The answer is B. 328 ounces