algebra word problem?
917544
This type of problem is solved by finding the fraction of the job each, then both, can do in an hour.  If t = the hours for the faster person, she can do 1/t of the job in an hour, the slower can do 1/(t+2) of the job, and together they do 1/5 of the job in an hour.\n\nSo 1/t + 1/(t+2) = 1/5
836446
Remember the order of operations "Please Excuse My Dear Aunt Sally"?\nParentheses\nExponents\nMultiply\nDivide\nAdd \nSubtract\n\nif n = 2 the plug in the variable  (4 * 2)^2\n\nwork the problem inside the parentheses and square the answer
786009
-2(x-3) + 4(x+7)\n-2x+6   +4x+28\n2x+34
1051508
Your equation would look like x^2-ax+b.  Which means when you solve the equation you would two answers one of which will contain a plus sign other will contain a minus sign since the middle term is negative.  Thus the answer is C.
189233
The problem with this type of algebra problem is that it's hard to know the syntax (grammar) of the problem if it's just written out in words. For example, is it 2x - 9, all over x, or is x the denominator only for 9? It's confusing!So in the future, to make it clear, use brackets like this:\n\n****IF I am reading your problem CORRECTLY****, I would write it as:\n(((2x-(9/x))^2)-4) - (((x-(7/x))^2)-4) = ?\n\nBut put aside the whole brackets thing for a minute. \n\nNow, normally, it helps to break down the math probelem into parts, but it looks like you're learning to do exponents. If this is the case, you must have learned that exponents allow you to work certain problems quicker without having to break down the problem into small parts and solving them seperately. To remember what those exponent rules are, try both the links at the bottom of this post. You'll use exponent rules all through your life, so know them!
1169010
Find d?\n\nWhere are a, b, and c?\n\nwhy don't you write out the entire problem if you want help?
975217
Line AC = 6, line BC= 2x+2, line AB=x+1\nSo 2x + 2 = 6\n2x = 4\nx = 2 <-- option C
893957
well darling ;\nimagine this like that\n\ny^2 + ( a+b)y + ab \nyou need to find " a+b " & " ab" \nin our function \na+b = 16 & ab = 64 \n( y + a)( y + b) \n( y + 8 ) ( y + 8)\n{ ' 8 + 8 = 16 '   &  ' 8 * 8 = 64 ' }\nso\n( y + 8)^2\n\nGood Luck sweetheart.
1022192
i have  a book about algebra..... and some website...see if they could help u
5554
Solving the problem stepwise -\n(x - 8 ) ( x - 2 )\n= x(x-2) - 8(x-2)\n=x^2-2x-8x+2\n=x^2-10x+2\n\nThese two sites will help you for the concepts.
745882
C is correct but not complete... only the 3 is factored out... the real answer would be:\n\n3(x-1)(x+4)
470404
((a^2 - b^2)/8a) X ((4a^3)/(a + b)^2))\n(4a^3(a^2 - b^2))/(8a(a + b)^2)\n((4a^3)/(8a)) * ((a^2 - b^2)/(a + b)^2)\n((a^(3 - 1))/2) * (((a - b)(a + b))/((a + b)(a + b)))\n((a^2)/2) * ((a - b)/(a + b))\n\nANS : ((a^2)(a - b))/(2(a + b))\n\nOn your problem, here is where you went wrong\n\nTo make it easier, you should had kept your problem in\n\n(4a^3(a - b))/(8a(a + b)) form, then had done this\n\n((4a^3)/(8a)) * ((a - b)/(a + b))\n\n(4/8)(a^(3 - 1))((a - b)/(a + b))\n(1/2)(a^2)((a - b)/(a + b))\n((a^2)(a - b))/(a + b)
727014
start with the O... they are always -2\nsince there are 2, and the compound must balance...\nPb must be +4\n\nso you get +4 -2
680155
i'd love to help but i'm having trouble with it myself\n\nuse ur math textbook it will have it unless ur math book say algebra and is red ( that my alg. book and it doesn't help) \n\nSRY good luck
156773
when x=3\na) 9\nb) 38\nc) 23\nd) 9\ne) 4 and 2/3\n\nwhen x=2\na) 7\nb) 30\nc) 17\nd) 7\ne) 6\n\nwhen x=1\na) 5\nb) 22\nc) 11\nd) 5\ne) 10\n\nwhen x=0\n\na) 3\nb) 14\nc) 5\nd) 3\ne) ERROR, can not divide by zero
191567
Please use parentheses to explain the problem more effectively. Assuming the problem as,\n\n(x-1)/2x + x/(x+3)\n\nThe solution can be,\n\n(3x^2+2x-3)/(2x^2+6)
863644
take out 4x first\n4x(x2-6x+8)\n4x(x-4)(x-2)