Word problem!!! AHH!!?
969211
just clarifying first answer... \n\nlet   a = 1st leg\nand b = 2nd leg\n       c = hypotenuse\n\na= 2 + 2b -->> one leg(a) is two inches(2+) more than twice the other leg(2b)\n\nsubstitute a\n[(2+2b)^2] + [b^2] = [13^2]\n[4+4b+(4b^2)] + [b^2} = 169\n5b^2 + 8b + 4-169 = 0\n5b^2 + 8b - 165 = 0\n  A         B     C\nusing quadratic formula which is -B plus or minus the square root of B^2 - 4AC all over 2A\n\n-8 plus/minus squareroot of 64 - [4(5)(-165)] all over 2x5\n\n-8 plus/minus squareroot of 58 all over 10\n\n[-8 + 58] / 10 = 5\n\nor \n\n[-8-58] /10 = -6.6\n\nnegative sides do not exist so one side (b) is 5\n\ngoing back to a= 2 + 2b \na = 2+10\na = 12\n\ntherefore\na = 12,\nb = 5\nc = 13
460264
clever...I like word play ;)
625162
Try dictionary.com, m-w.com, or one of the thousand other good dictionary websites on the web, as for problems, try search on http://mathforum.org/
1007654
Is this your homework???  The answer is B, anyway.
165476
How about (a+c)/(b+d)  ? more interesting!!\n\nbtw this only work when a/b != c/d
698680
one word, five letters, m-o-n-e-y!
347066
ok, i think one will be where you have to solve a word problem dealing with alegebra. One with number sentences that have a letter on one side of the problem and letters on both sides of the problem. i could be wrong, but i'm trying to help.
752305
Find the error. It's impossible!\nCapital "F" and capital "I" and a "." or an"!".\nThe alphabet is just "a red herring"
222592
1. p'q\n\n2. p'\n\n3. pq' + p'\n\nI think i got the correct ones !
872092
It is always understandable when a student feels overwhelmed by "word problems" because they are often presented in an intimidating manner.  If only they could be FUN again ... if YOU wrote the problem, I believe you might find it fun!  :-)\n\nBut let's allow ourselves some joy once we uncover all the unknowns in this problem.  Perhaps we'll get results from converting "written words" into "mathematical expressions."\n\nWhat's most important is that you understand the shape involved (think "shoe box") and how the "openings" are placed (think "windows" on three faces of the shoe box).  All the openings are understood to be identical.\n\nUsing only the letter X -- which stands for the length of an opening -- we can indeed come up with expressions for every other dimension of the container, as well as the openings.\n\nX = Opening LENGTH (given)\nX-2 = Opening HEIGHT ("length is 2 more than height")\n\n2X+8 = Container LENGTH ("8 more than twice X")\n3 = Container HEIGHT (given)\nX+2 = Container WIDTH ("2 more than length of opening")\n\nThe surface area of the container's sides (minus the three openings) is given as 300 square meters.  The equation is also given as 2 (sides) + 2 (front/back) + 2 (top/bottom) - 3 (openings), so let's just set these equal:\n\n300 = 2 (sides) + 2 (front/back) + 2 (top/bottom) - 3 (openings)\n\nAll we need to do is to write, and then substitute, expressions for the areas in parentheses.\n\nSide area = length * height = 3 (2X+8) = 6X + 24\nFront or back area = width * height = 3 * (X+2) = 3X + 6\nTop or bottom area = length * width = (2X+8) * (X+2)\n____________________________ = 2X^2 + 12X + 16\nOpening area = length * height = X (X-2) = X^2 - 2X\n\nTime to substitute:\n\n300 = 2 (6X+24) + 2 (3X+6) + 2 (2X^2+12X+16) - 3 (X^2-2X)\n\nNow that this (long) equation has been written, it is necessary to solve for X, since we need that value in order to calculate what the problem asks for:  The area of each element.\n\nCollect like terms:\n\n300 = 2 (6X+24) + 2 (3X+6) + 2 (2X^2+12X+16) - 3 (X^2-2X)\n300 = 12X + 48 + 6X + 12 + 4X^2 + 24X + 32 - 3X^2 + 6X\n300 = X^2 + 48X + 92\nX^2 + 48X - 208 = 0\n\nWhat we hope is that the trinomial on the right-hand side of this equation is able to be factored easily!  Does it?\nMy approach is to factor '-208' and see whether I can 'build' the linear term's coefficient ('48') from the factors.\n\n-208 = -2 * 104 = -2 * 2 * 52 = -2 * 2 * 2 * 26 = -2 * 2 * 2 * 2 * 13\n\nCan we group these factors into (one positive, one negative) numbers that produce 48 when added together?  Yes!\n\n(-2*2) + (2*2*13) = -4 + 52 = 48\n\nWe got lucky ... the '-4' and '52' provide the perfect pair of factors needed to create a 48 (through addition), so we can now factor.\n\n0 = (X-4)(X+52)\n\nThings are getting better still, because only one of the zeros (some call them 'roots') from the right-hand side will be positive.  The zeros are 4 and -52, and we will only choose the 4, since it represents a physical measurement of length (of an opening).\n\nWe can go back to the expressions above and substitute X=4.\n\nSide area = 6X + 24 = 6(4)+24 = 48 sq. m\nFront/back area = 3X + 6 = 3(4)+6 = 18 sq. m\nTop/bottom area = 2X^2+12X+16 = 2(4^2)+12(4)+16 = 96 sq. m\nOpening area = X^2 - 2X = 4^2-2(4) = 8 sq. m\n\nI hope this helps!
575536
Here is a great article with sample problems.\n\nHope this helps!
781165
If this is a clue from a cryptic crossword puzzle, do you have any letters?\n\nAlso, does it have to be ONE word or could it be several?
614837
I already posted this solution (or how to get it).  Do a search of yahoo answers....
954601
addition\nequation\nknowing too much math\ncorrect answer!
1008520
do your own homework....you have a computer.....should be easy
33805
James should say 8!\nguard  s-i-x   = 3 letters\nguard t-w-e-l-v-e  = 6 letters\nso f-o-u-r-t-e-e-n = 8 letters
611639
do the math moron!!!!!!