document.write( "Question 636765: Please help. Due tomorrow.
\n" ); document.write( "1.)A box is to be formed by cutting squares measuring 2 cm on a side from a square piece of cardboard and then folding the sided. If the volume of the box is to be 392 cm squared, what is the original size of the cardboard?
\n" ); document.write( "2.)Manuel is constructing a model house. He wants each window to have an area of 315cm squared, and he wants the length of each window to be 6 cm more than the width. What are the dimensions of each window?
\n" ); document.write( "3.)A circle has a radius of 20 cm. By how much must the radius be increased so that the area will increase to 144 pie cm squared?
\n" ); document.write( "4.)An airplane flies 900 miles against a headwind if 25 miles per hour. The plane took 15 minutes longer for this flight than with a tailwind of 25 miles per hour. How fast could the plane fly in still air?
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Algebra.Com's Answer #401256 by KMST(5328)\"\" \"About 
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1.)The box is formed by cutting squares measuring 2 cm on a side from a square piece of cardboard That looks like this, with the cutting lines marked in red:
\n" ); document.write( " The folding lines, shown in blue, show the central square that will be the base of the box.
\n" ); document.write( "The box sides (the folded flaps on all four sides) will 2 cm high, and that will be the height of the box.
\n" ); document.write( "The volume of the box, \"V\", is the surface area of the base, \"A\", times the height, \"h\" of the box.
\n" ); document.write( "\"V=A%2Ah\", so \"A=V%2Fh\"
\n" ); document.write( "If the volume of the box is to be 392 cubic centimeters, then the surface area of the base of the box is
\n" ); document.write( "\"A\"=(392 cubic centimeters)/(2 centimeters)=\"392cm%5E3%2F2cm=196cm%5E2\"=196 square centimeters.
\n" ); document.write( "What is the original size of the cardboard?
\n" ); document.write( "THE FIFTH GRADER SOLUTION:
\n" ); document.write( "The area of the square that is the base of the box is side of the base of the box squared.
\n" ); document.write( "\"196=14%5E2\", so the base of the box is a 14 cm by 14 cm square.
\n" ); document.write( "With 2 cm of cardboard to each side (4 extra cm), the width of the square piece of cardboard, before cutting was
\n" ); document.write( "\"14cm%2B4cm=highlight%2818cm%29\"
\n" ); document.write( "IF YOU KNOW ENOUGH ALGEBRA TO COMPLICATE THE PROBLEM:
\n" ); document.write( "You start by saying, let \"x\" cm be the length of each side of the square piece of cardboard, which looks like this, with the cutting lines marked in red:
\n" ); document.write( " The folding lines, shown in blue, show the central square that will be the base of the box.
\n" ); document.write( "That length of each side of that square base of the box is
\n" ); document.write( "\"x-2-2=x-4\" cm
\n" ); document.write( "That surface area of the base of the box (in square cm) is
\n" ); document.write( "\"A=%28x-4%29%5E2=x%5E2-8x%2B16\"
\n" ); document.write( "The height of the box is 2 cm.
\n" ); document.write( "The volume of the box (area of the base times height) is
\n" ); document.write( "\"%28x%5E2-8x%2B16%29%2A2=392\" square centimeters.
\n" ); document.write( "If we are smart, we divide both sides by 2 now to get a simpler equation:
\n" ); document.write( "\"%28x%5E2-8x%2B16%29%2A2%2F2=392%2F2\" --> \"x%5E2-8x%2B16=196\"
\n" ); document.write( "Then we subtract 196 from both sides, to get a quadratic equation in standard form.
\n" ); document.write( "\"x%5E2-8x%2B16-196=196-196\" --> \"x%5E2-8x-180=0\"
\n" ); document.write( "To solve that quadratic equation, if we are not good at factoring, we have to \"complete the square\", or use the dreaded quadratic formula.
\n" ); document.write( "If we are good at factoring, we realize that \"x%5E2-8x-180=%28x-18%29%28x%2B10%29\"
\n" ); document.write( "and find that the solutions are \"x=-10\" and \"x=18\".
\n" ); document.write( "Since the length of the side of the original square piece of cardboard cannot be a negative number, the only solution is
\n" ); document.write( "The length of the side of the original square piece of cardboard is \"highlight%28x=18%29\" cm.
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\n" ); document.write( "2.)
\n" ); document.write( "THE FIFTH GRADER SOLUTION:
\n" ); document.write( "The fifth grader may factor to find that the prime factorization of \"315\" is
\n" ); document.write( "\"315=3%2A3%2A5%2A7\" , and would rearrange that to find
\n" ); document.write( "\"315=%283%2A5%29%2A%283%2A7%29=15%2A21\" to figure out that the windows are 15 cm by 21 cm.
\n" ); document.write( "That fifth grader is smart and lucky, but we should try to use algebra, because the fifth grader approach would not work for a similar problem where the answers were not integers. (However, problems are often rigged so they have integer answers, and it is good to have a smart fifth grader around, so he/she can check your algebra answers).
\n" ); document.write( "THE ALGEBRA SOLUTION:
\n" ); document.write( "Let the width of each window be \"x\" cm.
\n" ); document.write( "Then the length of each window (in cm) would be \"x%2B6\".
\n" ); document.write( "The area (in square centimeters) of each window, calculated multiplying length times width, would be
\n" ); document.write( "\"%28x%2B6%29x=315\"
\n" ); document.write( "We work with that equation to transform it into a quadratic equation in a form that we like better.
\n" ); document.write( "\"%28x%2B6%29x=315\" --> \"x%5E2%2B6x=315\" --> \"x%5E2%2B6x-315=0\"
\n" ); document.write( "We can solve it by \"completing the square\", or by factoring, of by using the quadratic formula.
\n" ); document.write( "Completing the square:
\n" ); document.write( "\"x%5E2%2B6x=315\" --> \"x%5E2%2B6x%2B9=315%2B9\" --> \"%28x%2B3%29%5E2=324\" --> \"%28x%2B3%29%5E2=%28sqrt%28324%29%29%5E2\" --> \"%28x%2B3%29%5E2=18%5E2\"
\n" ); document.write( "The two solutions would be:
\n" ); document.write( "\"x%2B3=-18\" --> \"x=-21\", which does not make sense, and
\n" ); document.write( "\"x%2B3=18\" --> \"x%2B3-3=18-3\" --> \"highlight%28x=15%29\"
\n" ); document.write( "So the width of the windows is \"highlight%2815%29\" cm and the length is \"15%2B6=highlight%2821%29\" cm.
\n" ); document.write( "Factoring:
\n" ); document.write( "From \"x%5E2%2B6x-315=0\" we realize that
\n" ); document.write( "since \"315=15%2A25\" and \"21-15=6\",
\n" ); document.write( "\"x%5E2%2B6x-315=%28x%2B21%29%28x-15%29\" , so we re-write the equation as
\n" ); document.write( "\"%28x%2B21%29%28x-15%29=0\"
\n" ); document.write( "One of those factor must be zero, and realize that the solution coming from
\n" ); document.write( "\"x-15=0\" --> \"highlight%28x=15%29\" gives us a valid window width of \"highlight%2815%29\" cm, for a length of \"15%2B6=highlight%2821%29\" cm.
\n" ); document.write( "On the other hand, making the other factor zero, leads to an invalid solution:
\n" ); document.write( "\"x%2B21=0\" --> \"x=-21\" , because the width of the windows cannot be a negative number.
\n" ); document.write( "
\n" ); document.write( "3.)The area, \"a\" of a circle of radius \"r\" can be calculated as \"A=pi%2Ar%5E2\"
\n" ); document.write( "The area (in square centimeters) of circle with a radius of 20 cm is \"a=pi%2A20%5E2=400pi\" .
\n" ); document.write( "That is more than \"144pi\", so either
\n" ); document.write( "there is a typo somewhere in the problem, or
\n" ); document.write( "it is a trick question, where they call a decrease in area from
\n" ); document.write( "\"400pi\" \"cm%5E2\" down to \"144pi\" \"cm%5E2\" is called
\n" ); document.write( "an increase of \"144%7D%7D+%7B%7B%7Bcm%5E2\"-\"400pi\" \"cm%5E2\"=\"-256pi\" \"cm%5E2\".
\n" ); document.write( "If the area of a circle with radius \"r\" is
\n" ); document.write( "\"pi%2Ar%5E2=144picm%5E2\" , dividing both sides by \"pi\" , we get
\n" ); document.write( "\"pi%2Ar%5E2%2Fpi=144picm%5E2%2Fpi\" --> \"r%5E2=144cm%5E2\" --> \"r=sqrt%28144cm%5E2%29=12cm\"
\n" ); document.write( "Since \"12cm-20cm=highlight%28-8cm%29\"
\n" ); document.write( "I would say that the radius must be increased \"highlight%28-8cm%29\" so that the area will increase (by \"-256pi\" \"cm%5E2\") to 144 pie cm squared?
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\n" ); document.write( "4.)If the speed of an airplane relative to the air is \"x\" miles per hour,
\n" ); document.write( "the plane will move at \"x\" miles per hour with respect to the ground in still air.
\n" ); document.write( "With a tail wind of \"25\" miles per hour, the same airplane will move at \"x%2B25\" miles per hour with respect to the ground.
\n" ); document.write( "Against a wind of \"25\" miles per hour, the same airplane will move at \"x-25\" miles per hour with respect to the ground. An airplane flies 900 miles against a headwind of 25 miles per hour.
\n" ); document.write( "We know that speed, time and distance are related like this:
\n" ); document.write( "\"speed=distance%2Ftime\" <--> \"speed%2Atime=distance\" <--> \"time=distance%2Fspeed\"
\n" ); document.write( "If we want to figure how long a trip will take, we divide distance by speed.
\n" ); document.write( "\"15+minute=15minute%281hour%2F60minute%29=0.25hour\"
\n" ); document.write( "If the plane took 15 minutes longer to fly the same 900-mile distance against that wind than with the wind, the equation relating the times, in hours is
\n" ); document.write( "\"900%2F%28x-25%29=900%2F%28x%2B25%29%2B0.25\"
\n" ); document.write( "To get rid of denominators, we multiply both sides of the equation time \"%28x-25%29%28x%2B25%29\"
\n" ); document.write( "
\n" ); document.write( "\"900%28x%2B25%29=%28900%2F%28x%2B25%29%29%28x-25%29%28x%2B25%29%2B0.25%28x-25%29%28x%2B25%29\"
\n" ); document.write( "\"900%28x%2B25%29=900%28x-25%29%2B0.25%28x-25%29%28x%2B25%29\"
\n" ); document.write( "\"900%28x%2B25%29=900%28x-25%29%2B0.25%28x%5E2-625%29\"
\n" ); document.write( "\"900x%2B22500=900x-22500%2B0.25x%5E2-156.25%29\"
\n" ); document.write( "Adding \"-900x-22500\" to both sides, or subtracting \"900x%2B22500\" from both sides, we get
\n" ); document.write( "\"900x%2B22500-900x-22500=900x-22500%2B0.25x%5E2-156.25-900x-22500\"
\n" ); document.write( "\"0=-22500%2B0.25x%5E2-156.25-22500\" --> \"0.25x%5E2-45156.25=0\" --> \"4%280.25x%5E2-45156.25%29=4%2A0\" --> \"x%5E2-180625=0\" --> \"x=sqrt%28180625%29\" --> \"highlight%28x=425%29\" miles per hour.
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