The original width would be 19 and the original length would be 32.
Let w be the width. Then 2w-6 would be the length. However, after cutting a 3-inch square from each corner, both the width and length left over to fold into a box would be 6 inches smaller; thus the dimensions would be w-6 and 2w-6-6 or 2w-12.
Since the section cut out is 3 inches long, 3 will be the height of the box.
Volume is found by multiplying the length, width and height of the box; thus we have:
1014=(w-6)(2w-12)(3)
We multiply the binomials and have:
1014 = [w*2w-12*w-6*2w-6(-12)](3)
1014 = (2w²-12w-12w+72)(3)
1014 = (2w²-24w+72)(3)
1014 = 6w² - 72w + 216
When solving a quadratic equation, we want it set equal to 0. Subtract 1014 from each side:
1014-1014 = 6w² - 72w + 216 - 1014
0 = 6w² - 72w - 798
We will use the quadratic formula to solve this:

Since we cannot have a negative number for a measurement, 19 has to be the width; then 2(19)-6 = 32 would be the length.
Answer:
D.Associative Property
Step-by-step explanation:
Answer:
sorry don't know what is ans
Answer:
We are given the parent function,
.
So, we have,
1. Reflection over x-axis. 2. Reflection over y-axis.
This changes f(x) to -f(x). This changes f(x) to f(-x)
New function is,
New function is 
3. Translated down by 2 units 4. Translated left by 1 unit
This changes f(x) to f(x)-2 This changes f(x) to f(x+1)
New function is,
New function is, 
5. Translated right by 2 units 6. Translated up by 1 unit
This changes f(x) to f(x-2) This changes f(x) to f(x)+1
New function is
New function is, 
<span>-7x + 2y = 0
6x + 6y = 0
</span><span>ordered pair: (−2,2)
the way to figure out if that ordered pair is a solution is pretty simple, you just plug the ordered pair into both equations, and check if the outcome is true.
</span>-7(-2) + 2(2) = 0
14 + 4 = 0
18 = 0 is NOT true, so this isn't a solution. still, check your second equation as well.
6x + 6y = 0
6(-2) + 6(2) = 0
-12 + 12 = 0
for this equation, it is a solution. however, because the solution must apply to the system as a whole, and not just one of the equations, the ordered pair still is not a solution.