document.write( "Question 143854This question is from textbook College Algebra
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document.write( ": PROBLEM: VOLUME OF A BOX:
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document.write( "A RECTANGULAR PIECE OF METAL IS 10 IN. LONGER THAN IT IS WIDE. SQUARES WITH SIDES 2 IN LONG ARE CUT FROM THE FOUR CORNERS, AND THE FLAPS ARE FOLDED UPWARD TO FORM AN OPEN BOX. IF THE VOLUME OF THE BOX IS 832 IN.^3, WHAT WERE THE ORIGINAL DIMENSIONS OF THE PIECE OF METAL? \n" );
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Algebra.Com's Answer #104678 by Earlsdon(6294)![]() ![]() ![]() You can put this solution on YOUR website! Start by letting the width of the metal sheet be x inches, then the length is x+10 inches. \n" ); document.write( "The volume of the box (832 cu.in.) will be calculated by the length (less 4 inches) times width (less 4 inches) times the height of the box (2 inches). \n" ); document.write( "We can write the equation: \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "The original dimensions of the metal sheet are: \n" ); document.write( "Width = 20 inches and the length = 30 inches. \n" ); document.write( "Check: \n" ); document.write( "The volumes is: \n" ); document.write( "(x-4)(x+6)(2)(2) = 832 Substitute x = 20. \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " |