The two containers hold 328 ounces at the they hold same amount of water.
<u>Step-by-step explanation:</u>
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, the two equations must be equal.
⇒
The first step is to divide everything by 2 to make it simplified.
⇒ 
Now put everything on the left
.

Add the like terms together to further reduce the equation

Factorizing the equation to find the roots of the equation.
Here, b = -12 and c = -28
where,
- b is the sum of the roots ⇒ -14 + 2 = -12
- c is the product of the roots ⇒ -14 × 2 = -28
- Therefore, (x-14) (x+2) = 0
- The solution is x = -2 or x = 14
Take x = 14 and substitute in any of the given two equations,
⇒ 
⇒ 
⇒ 328 ounces
∴ The two containers hold 328 ounces at the they hold same amount of water.
Answer:
0.79
Step-by-step explanation:
19 / 24 is 0.79166667
So it's 0.79
Answer:
y = ½x - 14
Step-by-step explanation:
Given the linear equation, y = 3x - 4, where the <u>slope</u>, m = 3, and the <u>y-intercept</u> is (0, -4):
The slope of a linear equation represents the steepness of the line's graph. The higher the value of the slope, the steeper the line. Hence, the slope of the other line must be less than three, but is greater than zero: 0 < <em>m</em> < 3. (a negative slope will show a <em>declining</em> line).
Next, the vertical translation of the line involves changing the value of the parent graph's y-intercept. Since the prompt states that the equation must represent a downward vertical shift of 10 units, then the y-intercept of the other line must be (0, -14).
The linear equation that I have chosen that meets the requirements of the given prompt is: y = ½x - 14. <em>You're more than welcome to choose a different slope</em>, as long as it is less than 3, but is greater than 0 (must be a positive slope).
Attached is a graph of both equations, to demonstrate that the other equation represents a graph with a steeper slope than the original graph.