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:

$w=\frac{-b\pm \sqrt{b^2-4ac}}{2a}
\\
\\=\frac{--72\pm \sqrt{(-72)^2-4(6)(-798)}}{2(6)}
\\
\\=\frac{72\pm \sqrt{5184--19152}}{12}
\\
\\=\frac{72\pm \sqrt{5184+19152}}{12}
\\
\\=\frac{72\pm \sqrt{24336}}{12}
\\
\\=\frac{72\pm 156}{12}
\\
\\=\frac{72+156}{12} \text{ or } \frac{72-156}{12}
\\
\\=\frac{228}{12} \text{ or } \frac{-84}{12}=19 \text{ or } -7$

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.

4 0

3 years ago

Help in a hurry PLSSSSSSSSS

Mila [183]

I don’t think functions can have repeating x values

3 0

2 years ago

Read 2 more answers