Need help with this algebra question?
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area =\n\nx^2 + 2x\n\ncost ($) =\n \n( x^2 + 2x ) * 20\n\n_
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(x-2)/(x^2-2x-15)+(x+2)/(5-x)\n(x-2)/((x+3)(x-5))+(x+2)/(5-x)\n(x-2)/((x+3)(x-5))+(-2-x)(x+3)/(x-5)(x+3)\n\n[(x-2) + (x+3)(-2-x)] / [(x+3)(x-5)]\n\nANSWER :\n(x^2+4x+8)/((x+3)(5-x))
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by any chance is this a question from your maths homework? lol :D\n\nI cant work it out myself because to put it simple, im crap at maths.
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1: x<12\n2: x>257\n3: x>42\n4: x>84\n5: x>25\n6: x>179\n7: x<8\n8: x>577\n9: x<113
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(x^2)/2 - 3(x^(3/2))/(3/2) + c = (x^2)/2 - 2x^(3/2) + c
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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)
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There's lots of help online. Try these web sites:\n\nhttp://www.algebrahelp.com/\nhttp://www.purplemath.com/\nhttp://www.sosmath.com/algebra/algebra.html\nhttp://www.helpalgebra.com/\n\nThere's even more. So start with these, and good luck!
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12 simply place the 1 where the x is\n\n(1+3)(1+2)\n4*3\n12
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1)If it's an equation then solve it quadratically\n\n(X-3)= 0 or (X-3)= 0\n\ntherefore X= 3 (as the 3 moves to the other side making it positive)\n\n2)If you have to find the products ude the distributive law:\n\nXsquared -3X-3X+9\ntherefore the answer would be Xsquared -6X+9\n\nBut I'm sure you needed the Equation one so use my first answer and method.
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Try this website. http://www.webmath.com/index3.html
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Ok, first you must get the -8 and the x/-2 out of parentheses, so u just multiply them both by one. So then is -8 * x / -2. First, you do -8 * x, which is -8x. The problem then becomes -8x / 2. In this step, just don't mind the negative or the x and just put them back in at the end. So you're really just doing 8 / 2, which is 4. You must put the - and the x back in, so the answer is -4x. Ta Da!
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(x+7)(x+3)\n(x-11)(x+2)\nforget this one.  its AC factoring though\nthis ones AC too
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try this one http://www.algebrahelp.com/
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You need to give some directions.  What is the problem asking you to do?\nNote: The person below is incorrect.  (4+3x)^5 does NOT equal\n4^5 + (3x)^5
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You actually need to solve this a different way.\nYou can't multiply both sides by x+3 because this could be positive or it could be negative...you don't know.\nInstead, bring everything to one side by subtracting 2, then get a common denominator.\n(x - 2)/(x + 3) <= 2\n(x - 2)/(x + 3) - 2 <= 0\n(x - 2)/(x + 3) -2(x+3)/(x+3) <= 0\n[(x - 2) - 2(x+3)]/(x + 3) <= 0\n[(x - 2) - 2x - 6)]/(x + 3) <= 0\n(-x - 8)/(x + 3) <= 0\nYour important values are -8 and -3\nTest values less than -8, between -8 and -3, and greater than -3.\nYou will see that you get negative values for\nx <= -8 and x >= -3
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Homework.\n\nSuppose a1*x+a2*Tx+..+am*T^(m-1)x=0.\n\nApply T^(m-1) to get a1=0.\nApply T^(m-2) to get a2=0, etc.
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1.\n3x + 6/((x + 1)(x + 2)) Factor 3 out of the top.\n[3(x + 2)]/[(x + 1)(x + 2)] x+2's cancel\n3/(x + 1) <-- This is simplified.\n\nI don't have the energy to do the other ones right now.  :/