find the domain rational expression x+1 over x^2-9?
305609
Division by zero is not allowed.\nSet the denominator equal to zero and solve for x.\nx^2 - 9 = 0\ngives x = -3 and x = 3.\nThus, the domain is\nAll real numbers except -3 and 3.
931054
Group all elements with y together:\n\ny(x^2 + 6x +9) -x^2 + 2x + 3 = 0\n\nMove the rest to the other side:\n\ny(x+3)^2 = x^2 - 2x + 3 (1)\n\nx = -3 is obviously not the solution of this equation therefore we can divide both sides of (1) by (x+3)^2.\n\n(1) is equivalent to:\n\ny = (x^2 - 2x + 3) / (x+3)^2
373987
of cos!\n\n9*9 + 9 + 9 + 9/9 + (9-9)9
727286
you basically have to factor the whole problem out, i.e:\n\nthe given equation factors out to:\n\n(x-4)(x+4)\n\nSo the domain is all real numbers except 4 and -4. This is so, because when x is either 4 or -4 the equation is equal to zero.
729449
It will not factorise in the rational number field so there are no rational solutions\n\nIrrational roots are -1.72107 and 0.269209 (approximately) using Quickmath
445985
if by this you mean\n\n((3x^-2 * y)^-2)/((4xy^-2)^-1)\n\nthis becomes\n\n((3^-2) * x^(-2 * -2) * y^-2)/((4^-1) * x^-1 * y^(-2 * -1))\n\n((1/9)x^4 * y^-2)/((1/4) * x^-1 * y^2)\n\n((1/9)/(1/4)) * x^(4 - (-1)) * y^(-2 - 2)\n\n((1/9)*(4/1)) * x^(4 + 1) * y^(-2 + (-2))\n\n(4/9)x^5y^-4\n\nwithout using negative exponents, the answer would be\n\n(4x^5)/(9y^4)
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
387682
This expression is defined for every real number, domain is the whole set R, because it wouldn't be defined only for x^2+3=0, but x^2+3 is always greater than zero.
782919
Domain (D):\nx can be any real number except any x that causes the radicand (expression inside the radical symbol) to be negative. Also, any x that causes division by zero is out of the domain. So, we can immediately throw out x = 1.\n\nx + 5\n――– ≥ 0\nx - 1\n\nSo, how do you figure out when the fraction is positive or negative? One way is to make a chart. Values of interest include x = -5 and x = 1. Let's look at the entire real number line...\n\n-∞ ←――――――― -5 ―――――1 ―――――→ +∞\n\nx+5→- - - - - - - - - - - - 0 + + + + + + + + + + + + +\n\nx-1 →- - - - - - - - - - - - - - - - - - - - - - 0+ + + + + + +\n\nx + 5\n――– →+ + + + + + + 0 - - - - - - - - ∞+ + + + + + +\nx - 1\n\nTherefore, it appears that your radicand is non-negative from negative infinity up to and including -5. Also, from 1 to positive infinity the radicand is non-negative. However, as mentioned earlier, bad things happen at +1 so we'll exclude +1. So, in interval notation, your domain is given by the following:\nD = (-∞, -5] U (1, ∞).\n\nRange (R):\nWhen you take the square root of a non-negative number, you get a non-negative number. Notice that your radicand will never equal 1. There is a rule for determining this. It has something to do with (ax+b)/(cx+d)→a/c as x→±∞. You'll need to look that up in your textbook or discover it on your own by thinking about it. \n\nTherefore, R = [0, 1) U (1, +∞).
722836
I think you are asking to simplify the rational expression:\n[(x^3+27)/(9x+27)]/\n[(3x^2-9x+27)/(4x)]\n\nIf this is what you are asking, here's what you do.\n\nWe are dividing by a fraction, so we must flip the second fraction and multiply.  Our problem now becomes:\n[(x^3+27)/(9x+27)]*\n[(4x)/(3x^2-9x+27)]\n\nNow we must factor each expression individually.\n\nThe first expression--x^3+27--is a sum of two cubes and factors as:\n(x+3)(x^2-3x+9)\n\nThe second expression--9x+27--has 9 as a common factor and factors as:\n9(x+3)\n\nThe third expression--4x--cannot be factored, so we leave it alone.\n\nThe fourth expression--3x^2-9x+27--has 3 as a common factor and factors as:\n3(x^2-3x+9)\n\nUsing substitution, our expression:\n[(x^3+27)/(9x+27)]*\n[(4x)/(3x^2-9x+27)]\nbecomes:\n{[(x+3)(x^2-3x+9)]/[9(x+3)]}*\n{(4x)/[3(x^2-3x+9)]}\n\nMultiplying these expressions into one large expression, we get:\n[(x+3)*(x^2-3x+9)*(4x)]/\n[27*(x+3)*(x^2-3x+9)]\n\nCancelling the (x+3)'s and the (x^2-3x+9)'s that are common to the numerator and denominator leaves us with our final answer:\n\n(4x)/27\n\nGood luck.
893324
domain:  from - infinity to + infinity  (that's for x)\n\nrange: 0 to infinity (for y)\n\ny = |x+4|
122527
1 - 64x^2 = (1 + 8x)(1 - 8x)\n\ny^2 - 12y + 36 = (y - 6)(y - 6)  or (y - 6)^2
730130
Domain = x... should be all real numbers\n\nRange = y... 2 and up... x can be a negative number, but will always end up being postive when added to two, therefore the lowest number is 2
491386
Just substitute x+1 into f(x)\n\nf(x+1) = [ 6(x+1)^3 ] - [ 3(x+1)^2 ] + 4(x+1) - 9
468001
x+2 must not be 0\nso x must not be -2\ndomain is \nall real numbers - (-2)\nall real no except -2
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)
873410
1.  (x + 9)(x - 9)\n2.  (t + 2)(t - 2)\n3.  (b + 12)(b - 12)