- 38.79
- 3053.63
- 904.78
- 2544.69
- 226.19
- 402.12
- 1072.33
- 1526.81
- 28.73
- 113.1
- 3801.33
- 268.08
- 2094.4
- 75.4
- 94.25
- 37.7
- 1884.96
- 2065.24
- 19861.7
- 1385.44
- 287.98
- 4.19
- 3619.11
- 113.1
- 50.27
I did this really quick so I hope all the answers are right, and double check them if you have time just in case
48 fluid ounces. There's 8 oz in one cup:)
I think the answer is c sorry if I’m wrong
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.