Question 628074
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If the sides of the cutout squares are 6", then the depth of the box when the sides are folded up must be 6 inches.  Divide the volume by 6 to get the area of the bottom of the box, 99 square inches.


The original piece of metal had a width of *[tex \LARGE w] and therefore had a length of *[tex \LARGE w\ +\ 30]


Since you took 6 X 6 inches out of each corner, subtract 12 inches from each of the width and the length to get expressions for the width and length of the bottom of the box.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (w\ +\ 18)(w\ -\ 12)\ =\ 99]


Expand the binomials, put the quadratic into standard form, and then solve for *[tex \LARGE w].  Then calculate *[tex \LARGE w\ +\ 30]


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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