Ken predicts that the average temperature will be −5°C in December. If he predicts the temperature will rise 6°C in January and
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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.
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.