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
A. 360 ounces
B. 328 ounces
C. 232 ounces
D. 136 ounces
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.
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
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
To find the amount of water that both containers hold at the same time, we need to set the equations for Container A and Container B equal to each other and solve for the value of y.
The equation for Container A is y = 16x + 104.
The equation for Container B is y = -2x^2 + 40x + 160.
Setting the two equations equal to each other, we have:
16x + 104 = -2x^2 + 40x + 160
Rearranging the equation, we get:
-2x^2 + 24x + 56 = 0
To solve this quadratic equation, we can use either factoring, completing the square, or the quadratic formula. In this case, factoring would be the most efficient method.
The equation can be factored as follows:
-2(x^2 - 12x - 28) = 0
By factoring the quadratic within the parentheses, we get:
-2(x - 14)(x + 2) = 0
This equation is true when either:
x - 14 = 0 or x + 2 = 0
Solving for x in each case, we find:
x = 14 or x = -2
Since we are interested in the time after the start when the containers hold the same amount of water, we only consider positive values of x. Therefore, the solution is x = 14.
To find the amount of water both containers hold at this time, we substitute x = 14 into either equation.
For Container A:
y = 16x + 104
y = 16(14) + 104
y = 224 + 104
y = 328
Therefore, at the time when the containers hold the same amount of water, they both hold 328 ounces. So, the correct answer is B. 328 ounces.