Hi I need help with my Homework.. I already did these.. just making sure their correct, thank you:\n\n09) Perform the indicated operation:  7/x+3 - 3/x^2-9\n\n10) Perform the indicated operation: (1/x+2)+9\n\n11) Perform the indicated operation: (17/x^2+3x-28)-(2/x-4)
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Assuming I've read the problems right (writing math problems on a single line is tough without using a lot of brackets, so I assume for 9 that the space indicates two expressions, for 10 that 1/x+2 is 1/(x+2) and not (1/x)+2, and that the polynomial in 11 is all under 17 instead of (17/x^2)+3x... )\n\n09) Given: 7/(x+3) - 3/(x^2-9)\nFactor Denominator: 7/(x+3) - 3/(x+3)(x-3)\nRewrite with LCD: 7(x-3)/(x+3)(x-3) - 3/(x+3)(x-3)\nMultiply and Rewrite: 7x-21/(x^2-9) - 3/(x^2-9)\nSubtract: (7x-24)/(x^2-9)\n\n10) Given: (1/x+2)+9\nRewrite with LCD: (1/(x+2))+[(9x+18)/(x+2)]\nAdd fraction w/same denom.: (1+9x+18)/(x+2)\nSimplify: (9x+19)/(x+2)\n\n11) Given: [17/(x^2+3x-28)] - [2/(x-4)]\nFactor first denom: [17/(x+7)(x-4)] - [2/(x-4)]\nRewrite with LCD: [17/(x+7)(x-4)] - [2(x+7)/(x+7)(x-4)]\nSimplify numerator: [17/(x+7)(x-4)] - [(2x+14)/(x+7)(x-4)]\nSubtract fraction w/like denom: [17-(2x+14)]/[(x+7)(x-4)]\nSubtraction rule for polynomial: [17-2x-14]/[(x+7)(x-4)]\nSimplify: (3-2x)/(x^2+3x-28)
32840
There are a couple of ways to work this out, I will take the easiest to explain.\n\n(x-3)(x^3+3x^2+9x+27) Can be rewritten as:\n\n(x)(x^3+3x^2+9x+27)+(-3)(x^3+3x^2+9x+27) Now, multiply those out.\n\nx^4+3x^3+9x^2+27x-3x^3-9x^2-27x-81 Now, sort by like powers of x\n\nx^4+3x^3-3x^3+9x^2-9x^2+27x-27x-81 Simplify\n\nx^4-81
484842
Well, you got me stumped!\n\n-------------------------------------\n\nI started working it out and got pretty far.\n\nThe above given information might be incorrect.  Maybe?  I don't know.\n\nI tried it several times.  I even double and triple checked my work!\n\n---------------------------------------------\n\nThis is as far as I got and then I got stumped:\n(D-E) (B+C) = (10J + AFGH) / A\n\n------------------------------------------------------\n\nGreat question!\n\nI look forward to the continuing discussion.\n\nI love the way you worded your question.  Right on!!!\n\nKeep those challenge questions coming!  ;o}
524185
[(7x^3)(-x^7)]/14x^4\n= (-7x^10)/14x^4\n= -x^6/2
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I got   3(x² - 5x + 4) or 3( x - 4 )( x - 1 )\n\nLet me show you my work...\n\n3( x - 2 )² - 3(x - 2) - 6\n\nSimplify\n\n3( x -2 )( x - 2 ) - 3( x- 2 ) - 6\n\nDistribute\n\n3( x² - 4x + 4) - 3x + 6 - 6\n\nDistribute and Simplify\n\n3x² - 12x + 12 - 3x\n\nSimplify\n\n3x² - 15x + 12\n\nYou can now simplify by factoring out 3\n\n3(x² - 5x + 4)\nor\n3( x - 4 )( x - 1 )
599378
divided by (X-3)\n\n Let's work with (2x^2-9x+9) first\n\n2X^2-9X+9\n= (x^2-6x+9)+(x^2-3x)\n= (x-3)^2 + x(x-3)\n= (x-3) [(x-3) + x)]\n=(x-3) (2x-3)\n\nnow divide that for (x-3)\n\n(x-3) (2x-3)\n---------------\n   (x-3)\n\n= 2x-3   (this is your answer)\n\nGood luck!
443562
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
435310
I prefer "long" multiplication but it doesn't line up as well...\n\n3x^2 + 7x - 6\n* _____x+12\n--------------------  Multiplying term by term\n_____36x^2 + 84X - 72\n3x^3 + 7x^2 - 6x\n--------------------------------  Add term by term\n3x^3 + 43x^2 + 78x - 72   <--  final answer
216609
Remember Order of Operations - in this case, Parenthesis, then Add/Subtract.\n\n1.) First, we do the parenthesis:\n3x-5(3x-7)=2(x+9)+45\n3x-(15x-35)=(2x+18)+45\n\n2.) Next, we add/subtract from the parenthesis:\n3x-(15x-35)=(2x+18)+45\n-12x + 35 = 2x + 63\n\n3.) Then, we solve by subtraction and division:\n-12x + 35 = 2x + 63\n-12x - 2x = 63 - 35\n-14x = 28\nx = -2\n\nCheck:\n3(-2) - 5(3(-2)-7) = 2(-2+9)+45\n-6 - 5(-6-7) = 2(7) + 45\n-6 + 65 = 14 + 45\n59 = 59
226564
if u meant {4/(q^2+2q+1)}- {q/7(q+1)}\nthe answer is\n\n(28-q^2-q)/{7*(1+q)^2}
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.. |10 - 4x|\n---------------- > 3\n|x² - 7x +10| \n\nSo \n\nNow |10 - 4x| = 10 - 4x for 10 - 4x ≥ 0 ie x ≤ 2.5\nand = -(10 - 4x) for 10 - 4x < 0 ie x > 2.5\n\nAlso 3|x² - 7x +10| = 3(x² - 7x +10)  for x² - 7x +10 ≥ 0 ie x ≤ 2 and x ≥ 5\nand = -3(x² - 7x +10) for x² - 7x +10 < 0 ie 2 < x < 5\n\nFind points of intertsection:\n\nWhen x < 2\n10 - 4x = 3(x² - 7x +10)\n3x² - 17x +20 = 0\n(x - 4)(3x - 5) = 0\nOnly solution in domain is x = 5/3\n\nWhen 2 < x ≤ 2.5\n10 - 4x = -3(x² - 7x +10)\nSo 3x² - 25x + 40 = 0\nx = (25 ± √(625 - 480)/6\nOnly solution in domain is x = (25 - √(145))/6 (≈ 2.160)\n\nWhen 2.5 ≤ x < 5\n4x - 10 = -3(x² - 7x +10)\nSo 3x² - 17x +20 = 0\n(x - 4)(3x - 5) = 0\nOnly solution in the domain is x = 4\n\nWhen x > 5\n4x - 10 = 3(x² - 7x +10)\nSo3x² - 25x + 40 = 0\nx = (25 ± √(625 - 480)/6\nOnly solution in domain is x = (25 + √(145))/6 (≈ 6.174)\n\nSo \n.. |10 - 4x|\n---------------- > 3 occurs when |10 - 4x| > 3|x² - 7x +10|\n|x² - 7x +10|\n\nand this is when 5/3 < x < (25 - √(145))/6 \nand when 4 < x < (25 + √(145))/6\n\nWHHHOOOOOOOOPPPSSS YESSSSSSSSSSSSS asymptotes!! at 2 and 5!! so x cannot be 2 or 5\n\n5/3 < x < 2, 2 < x < (25 - √(145))/6, 4 < x < 5, 5 < x < (25 + √(145))/6 (although you can always argue that any number > -∞ and ignore the discontinuities)
458162
1)1,8,27,64,125\n  x^3 = 1^3, 2^3, 3^3, 4^3, 5^3\n\n2)9, 13, 15, 19, 21, 25\n a(n)+ n-2 = {a(n-1) +a(n+1)}/2\n\n3)00001, 00011, 00101, 00111, 01001\n  1,3,5,7,9 in binary\n\n4)17,23,5,18,20,25,21,9,15,16\nok, your teacher is an ass for this one..  heres the breakdown\nq is the 17th letter, w is the 23 letter, e, r, t, y, u, i, o, p\nthe top row of a standard U.S. keyboard. i think i earned 10 points for that one alone\n\n\n5)\nFOUR\nF I V E\nN I N E\n\nR=0\nF=2\nI=3\nN=5\nE=6\nU=7\nV=8\nO=9\n\n2970 + 2386 = 5356\nDONE.  Hope this works for you
487528
Order of operations: Parentheses Exponent Multiplication Division Addition Subtraction.\n\n2 - 2[4-9] ÷ (-5)\n2 - 2[-5] ÷ (-5)\n2 - [-10] ÷ (-5)  ~~ same as 2 + 10 ÷ (-5)\n2 + [-2]  ~~ same as 2 -2\n= 0\n\nI'd double check that if I were you; it's a bit early for me. :*)
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[2x/(x+3)(x-3)]*[(x-3)/4(x)]\ncancelling 2 and x-3 we will get\n=x/(x+3)2(x)\nsimplify we will get\n=x/(2x^2+6x)
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(3 a^-7 b^7) / (9 a^7 b^-5) = ((3)(1/a^7)(b^7)) / ((9)(a^7)(1/b^5)) =\n\n((3 b^7) / a^7) / ((9 a^7) / b^5) = ((3 b^7) / a^7) times (b^5 / (9 a^7))\n\n= (b^12) / (3 a^14)    OR    3^-1 a^-14 b^12
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(x+3)(x^2+2x+4)\n\nStart by taking the first term in the first parenthesis and multiplying it by each term in the second parenthesis:\n\nx*x^2 + x*2x + x*4 = x^3 + 2x^2 + 4x\n\nNow take the second term and multiply it by each term in the second parenthesis:\n\n3*x^2 + 3*2x + 3*4 = 3x^2 + 6x +12\n\nNow add all of that together\n\nx^3 + 2x^2 + 4x + 3x^2 + 6x +12\n\nand combine terms\n\nx^3 + 5x^2 + 10x +12
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1) 1/4, 3/8, 2/5\n2) 5/9, 2/3, 7/8\n3) 9/10, 11/12, 15/16\n4) 5/9, 7/12, 2/3\n\n\nExplanation:\n1/4:\n   ____\n4 )1.0\n\n\n\n\n      .2\n   ____\n4 )1.0\n    -  .8\n_______\n        .2\n\n\n\n\n\n\n\n      .25\n   ____\n4 )1.00\n    -  .80\n_______\n        .20\n    -   .20\n_______\n         0\n\n\n\nI doubt the explanation will work.