find the domain of each rational expressions x-3 over x+2?
468001
x+2 must not be 0\nso x must not be -2\ndomain is \nall real numbers - (-2)\nall real no except -2
893324
domain:  from - infinity to + infinity  (that's for x)\n\nrange: 0 to infinity (for y)\n\ny = |x+4|
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
665603
you put (x+2) in for x in the equation first\n\n((x+2)^3)+(2(x+2)^2)-4\n\nthen simplify \n\nx^3 + 8x^2 + 20x + 12\n\nand multiply the entire thing by 3\n\n3(x^3 + 8x^2 + 20x + 12)\n\n3x^3 + 24x^2 + 60x + 36
38332
(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))
942101
start by seperating y, u then have y=3x/2+3 then plug in the domain, or the numbers u are givin\nu'll have 0, 1.5, 3, 6, 7.5
652392
1/2 x^2 -3x\n- 2/3 x^2 +3x -5\n________________\n\n-1/6x^2 (3x's cancel out) -5\n______________________________\n\n-1/6x^2 -5
881529
fist domain\n\nthe number we search the log must be positive so x>-2 \n\ndomain -2<x<infinity \nfor infinity log (base3), infinite is infinite. for x=-2 we tend to x+2 =0 and log (x+2 ) = -infinite  Vertical asypmtote for x=-2 \n\nthe equation can be rewritten  h(x)  = log (base 3) (x+2/7^3)
256079
x(x^2 -9)= x (x-3)(x+3)
1059114
3x^2 - 9x + 6 - 3(x - 1)(x - 2)\n3x^2 - 9x + 6 - 3(x^2 - 2x - x + 2)\n3x^2 - 9x + 6 - 3(x^2 - 3x + 2)\n3x^2 - 9x + 6 - 3x^2 + 9x - 6\n\nEverything gets cancelled out, so it doesn't matter if i put 10 in for x or not.\n\n3(10)^2 - 9(10) + 6\n3(100) - 90 + 6\n300 - 90 + 6\n210 + 6\n216
944695
12 * (3-e^-3x)^3 *  e^-3x
207106
x^2-3=-2\nx^2=-2+3\nx^2= 1\nx=1\n\n1^2+-3+2=0
387063
X=[-(30y-433)]/1232\nY=[-(1232x-433)]/30
664975
Domain of the function is the value of x and the range is f(x)....\nfor f(-3) the domain is -3 and the range f(-3)=(-3)^2+2(-3)-48=9-6-48=-45  which is the range....\nsimilarly for f(a+1) the domain is a+1 and the range by substituting a+1 in the equation and factorising you get the range as (a+9)(a-5)...
751318
x^2-x-6>0\nor, x^2-3x+2x-6>0\nor, x(x-3)+2(x-3)>0\nor, (x-3)(x+2)>0\nor x>3 and x<2\n\nhope this helps :)
98350
limit as x tends to 3 ;( x^2 -3)/2 =(3^2 -3)/2 =(9-3)/2 = 3
831999
y = (x+3)(x-1)(x-2) = x^3 - 7x + 6