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= 16x+104
Container B: y= -2x²+40x+160
A. 360 oz.
B. 328 oz.
C. 232 oz.
D. 136 oz.
well when is the amount in A equal to the amount in B ??
16 x + 104 = -2 x^2 + 40 x + 160
well first divide everything by 2 to make it easy
8 x + 52 = - x^2 +20 x + 80
now put everything on the left
x^2 + 8 x - 20 x + 52 - 80 = 0
or
x^2 - 12 x - 28 = 0
go here
x = -2 or x = 14
use x = 14
y = 16(14) +104 = 328 ounces
or
y = -2 * 14^2 + 40 * 14 + 160 = -392 + 560 + 160 = 328 remarkable
The answer is B 328
thanks:)
https://www.mathsisfun.com/quadratic-equation-solver.html
sorry, forgot quad eqn link
thanks i used 3 calculators and could not get the answer
To find the amount of water the containers hold at the time when they hold the same amount, we need to find the x-value where the equations of the two containers are equal.
Let's set the equations equal to each other and solve for x:
16x + 104 = -2x² + 40x + 160
To solve this quadratic equation, we need to rearrange it into standard form (ax² + bx + c = 0):
2x² + 24x + 56 = 0
Now, we can solve this quadratic equation by factoring, completing the square, or using the quadratic formula. Let's use the quadratic formula:
x = (-b ± √(b² - 4ac))/(2a)
In this equation, a = 2, b = 24, and c = 56. Substituting these values into the formula:
x = (-24 ± √(24² - 4(2)(56)))/(2(2))
Simplifying the expression under the square root:
x = (-24 ± √(576 - 448))/(4)
x = (-24 ± √128)/(4)
x = (-24 ± 8√2)/(4)
x = -6 ± 2√2
Now, we have two possible values for x:
x₁ = -6 + 2√2
x₂ = -6 - 2√2
Since time cannot be negative, we discard the negative value and consider only the positive one:
x = -6 + 2√2
Now, we can substitute this value of x into either equation to find the amount of water the containers hold at that time. Let's use the equation for Container A:
y = 16x + 104
y = 16(-6 + 2√2) + 104
y = -96 + 32√2 + 104
y = 8 + 32√2
So, at the time when the containers hold the same amount of water, they both hold 8 + 32√2 ounces.
Therefore, the correct answer is not provided in the given options.
D. 136 oz.
Because we want to find the time when both containers hold the same amount of water, we need to set the two equations equal to each other and solve for x:
16x + 104 = -2x² + 40x + 160
Rearrange the equation to set it equal to zero:
2x² + 24x - 56 = 0
Now we can use the quadratic formula to solve for x:
x = (-24 ± √(24² - 4(2)(-56))) / 2(2)
x = (-24 ± √(576 + 448)) / 4
x = (-24 ± √1024) / 4
x = (-24 ± 32) / 4
Now we have two possible solutions for x: (-24 + 32) / 4 = 2 and (-24 - 32) / 4 = -14.
Since time can't be negative, we take x = 2. Plug this value back into either equation to find y:
y = 16(2) + 104
y = 32 + 104
y = 136
So, at the time when the containers hold the same amount of water, they both hold 136 oz.