i am going to fail this class i just don't understan (x/x-2)-(x+1/x)=(8/x^2-2x) i just don't get it...
537832
(x/x-2)-(x+1/x)=8/(x^2-2x)\n[x(x)-(x-2)(x+1)]/[x(x-2)]=8/[x(x-2)]\nx^2-(x^2-2x+x-2)=8\nx^2-x^2+x+2=8\nx=8-2=6
736337
2/x + 3 = 2\n2/x = -1\n-x = 2\nx = -2
491455
(x^2+4x+1)/(x-2) = x+2+3/(x-2)\n\nx  : x^2+4x+1-x^2-2x =2x+1\n2  : 2x+1 -2x-4 =3
1161184
nb2718 is almost correct except for one thing:\n\nd/dx (-x^2 + 2) = -2x  Where did that minus sign go??  That'd cost you 5 points on a test, or it would have with my calculus teacher!\n\nf(x + h) = -(x +h)^2 + 2 = -[x^2 + 2xh + h^2]+2\n\nf(x) = (-x^2 + 2)\n\nlim h-->0 {[-x^2 - 2xh -h^2] + 2 - [(-x^2 + 2)]}/h =\n\nlim h-->0 {-x^2 - 2xh -h^2 + 2 + x^2 - 2}/h =\n\nlim h-->0 {h(-2x + h)/h}  now divide by h before it goes to zero...\n\nlim h--> {-2x + h} = -2x\n\nQED
529729
(x2)-3 (x-2) over (x2) -4?\nIs this what you mean?\n(x^2)^(-3)*x^(-2)/(x^2)^(-4)\n= x^(-6)*x^(-2)/x^(-8)\n= x^(-6-2-(-8))\n= x^(-8+8)\n= x^0 \n= 1
705864
Alright, don't listen to the above post...\n\nGrade 12 here, answering questions all day for mere fun.\n\n12)\ny - 3 = -1/3 (x + 4)\n\nEXPLAINED\n\nadd 3 to both sides\ny = -1/3 (x + 4) + 3\n\nexpand 1/3(x + 4) \n= x * 1/3 + 4 * -1/3\n= x/3 + 4/3\ny = -1/3x - 4/3 + 3\n\nTo add -4/3 and 3, use 3 = 9/3\ny = -1/3x - 4/3 + 9/3\ny = -1/3x + 5/3\n\nTrue, my bad. I originally added 4/3 instread of -4/3\n\n\n16)\n(-2, -1) (4, -4)\n\nm = Dy / Dx\nm = ((-1) - (-4)) / ((-2) - (4))\nm = 3 / -6\nm = -1/2\n\ny = mx + b\n(-1) = -1/2(-2) + b\n-1 = 1 + b\n-1 - 1 = b\n-2 = b\n\ny = -1/2x - 2\n\nCHECK\ny = -1/2(4) - 2\ny = -2 - 2\ny = -4\nTherefore it works.\n\n18)\n(0, 0) (2, 3)\n\nm = Dy / Dx\nm = ((3) - (0)) / ((2) - (0))\nm = 3/2\n\nb = 0\n\ny = mx + b\ny = (3/2)x + 0\ny = (3/2)x\n\nNote, the lower post copied my answers
455994
The answer is X = -2\n\nYou were right the first time.  You just made a calculation mistake when you tried to check it.\n\nHere's the check I got (I entered periods as spacers):\n\n_1_...+..._2_....=....0\n2+3x......2-3x  \n\n___1__...+...___2__....=....0\n2+3(-2).........2-3(-2)\n\n_1_...+..._2_....=....0\n2-6........2+6\n\n_1_...+..._2_...=...0\n-4...........+8\n\n_1_...+.._1_...=...0\n-4..........+4\n\n0...=...0\n\nTo get to the x = -2, we do what you did with (2 + 3x)(2 - 3x) as the common denominator.  When we multiply it out (multipying the common denominator to both sides of the equation), we get:\n\n2 -3x + 4 + 6x = 0\n6 + 3x = 0\n3x = -6\nx = -2\n\nYou asked a question above about the result of multiplying the common denominator to the right half of the equation.  Of course, multiplying anything times "0" always equals "0".  Right?\n\nGood luck!
106861
How about:\nx ^2/(x^2-4)+1/(x+2) = 0\nx^2/(x^2-4) = -1/(x+2)\nx^2/(x+2)(x-2) = -1/(x+2)\nx^2/(x-2) = -1\nx^2 = -x + 2\nx^2 + x – 2 = 0\n(x+2)(x-1) = 0\ncan you take it from here?
129694
-6 <= 2x-4 < 8 |:2\n-3 <= x-2 < 4   |+2\n-1 <= x < 6 \nso, x is between -1 (included) and 6 (not included)
1008674
You got it right so far. But next...to multiply by 1/4...is the same thing as dividing by 4. So take 4x^2/4=x^2 and then -4/4=-1...so you get x^2-1...make sense? I hope I helped...God bless.
615045
[x/(x+2)]\n= [1/(1+(2/x))] --- divide the numerator & denominator by x.\n\nAs x approaches +ve infinity, 2/x approaches 0.\nSo [x/(x+2)] approaches 1.
901793
x/(x-8) + 7 = 8/(x-8)\nMultiply everything by (x-8)\nx + 7(x-8) = 8\nDistribute the 7\nx + 7x - 56 = 8\nCombine x and 7x\n8x - 56 = 8\nAdd 56 to each side\n8x = 64\nDivide both sides by 8.\nx = 8\n\nNote that x-8 cannot equal zero, but we just found x to be 8.\nThat means x-8 = 8-8 = 0\nThus, we have no solution.
394528
You mean\nx^(-7)x^6/x^7\nAdd the exponents on top (you add exponents whenever you multiply)\n-7+6 = -1\nThus,\nx^(-7)x^6/x^7\n= x^(-1)/x^7\nSubtract exponents when dividing\n-1-7 = -8\nThus, the answer is\nx^(-8) which is d
329632
{x: x inequal to 2}\nor\n{x: x=/= 2}
223472
x - 1 / x + 3 = x + 2 / x - 4\nx-1/x-x-2/x = -3-4\n-1/x-2/x = -7\n-3/x = -7\n7x = 3\nx=3/7
715822
Limit (X^3 -8)/(X^2 -4)\nX ->2\n-----------------------------\n\n(X^3 -8)/(X^2 -4) = (2^3 - 8)/ (2^2-4)= 0/0, which mean you can use L' Hospitals rule which is take the derivative of numerator and denominator separately\n\n3x^2/2x, and you plug in 2 and you get 12/4 = 3\n\n------------------------------------------------------------------\nLimit (3lnX)/(5(x-1))\nX ->1\n----------------------------------\n(3ln1)/(5(1-1)), that is the same as 0/0. (ln(1)) = 0 fyi\n\nso take the derivative just like the first one and you get \n\n(3/x)/5\n(3/1)/5= 3/5\n-------------------------------------------------------------------------\n\nLimit (e^x)/(X^3)\nX ->infinity\n--------------------------\n\n(e^(inf))/((infin)^3) = inf/inf\n\nThis one is undefined because as you keep on taking the derivative, the numerator is going to stay e^x and the bottom is going to be infinity until it is equal to 0 on the fourth derivative.\n\nSo the answer is undefined or does not exist
561262
u jammed the equation together.\ndo u mean\n6x^3 + ( 6/x - 2 * x^2 - x/3x2 - 3x + 3)\nits really hard to see ur question. can u write it out with spaces...\nhurry up i am waiting... lol