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?

Question

Container A: y equals 16 x plus 104
Container B: y equals negative 2 x squared plus 40x plus 160
A. 360 ounces
B. 328 ounces
C. 232 ounces
D. 136 ounces

Answer 1

To find the time 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

Simplifying and rearranging:

2x^2 - 24x - 56 = 0

Dividing by 2:

x^2 - 12x - 28 = 0

This quadratic equation can be factored as:

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

So the two times when the containers hold the same amount of water are x = 14 and x = -2. Since the containers are filling up with water, we only care about the positive solution x = 14.

To find the amount of water in each container at this time, we substitute x = 14 into either equation:

Container A: y = 16(14) + 104 = 328 ounces

Container B: y = -2(14)^2 + 40(14) + 160 = 328 ounces

So the two containers hold the same amount of water, 328 ounces, at the time when they reach this level.

The answer is B. 328 ounces.

Answer 2

f an object is dropped from a height of 144 ft, the function h of t equals negative 16t squared plus 144 gives the height of the object after t seconds. When will the object hit the ground?
A. 1.5 s
B. 3 s
C. 6 s
D. 9 s

Answer 3

The object will hit the ground when its height is equal to 0. So we solve the equation:

-16t^2 + 144 = 0

Dividing both sides by -16:

t^2 - 9 = 0

This quadratic equation can be factored as:

(t - 3)(t + 3) = 0

So the two times when the object is at a height of 0 are t = 3 and t = -3. Since time cannot be negative, we only care about the positive solution t = 3.

Therefore, the object will hit the ground after 3 seconds.

The answer is B. 3 s