find the product of the denominators...?
340071
The product is the answer in a multiplication equation.  The denominator is the bottom number in a fraction.  The simple way to find your answer is to just multipy the bottom numbers (5x4x7).  After adjusting the numerators, you can usually add the fractions.\nGood question. \nGood luck with your homework
984538
(2/6) / (4/9)\n1/3 * 9/4\n3/4
165476
How about (a+c)/(b+d)  ? more interesting!!\n\nbtw this only work when a/b != c/d
167162
I do agree with Anniemc46.\nIf the denominator is 1+(tanhx)^2 then it is easy to do.\n\nI have done and uploaded to my web site.\nPlease check the url I provided in the sourse.\n\nThanking U,\nThuvalpakshi.
1162826
Least common denominator is (x-3)(x+3).\n\n2/(x-3)(x+3) + 3*(x-3)/(x-3)(x+3)\n\n(2+3x-9)/(x-3)(x+3)\n\n(3x-7)/(x-3)(x+3).
87883
First write y as a product instead of a quotient:\n\n(2+x)*(2-3x)^(-1)\n\nNow you should be able to do this. \nUse 2+x as your f and (2-3x)^(-1) as your g. \nYou will have to use the chain rule to find g'.
511329
Start by multiplying both sides by D this gives you\n(D*P) = (2*s*t)\n\nthe divide both sides by 2t which gives you\n(D*P)/(2*t) = s\n\nor\n\ns = (D*P)/(2*t)
557964
Wouldn't it be something like:\n\n2x(8) + 8s     to get rid of the denominator\n\n16x + 8s\n\nThat is as far as I think you could simplify it, because each has different variables and you can't combine something with different variables.
55738
3/(x+1)-1/2             =  1/(3x+3)\n3/(x+1)-1/2             =  1/[3(x+1)]          \n\nfactor 3\n\n[3*2-1*(x+1)]/[2*(x+1)] =  1/[3(x+1)]     \n\ncancel common denomenator   (x+1)\n\n[3*2-1*(x+1)]/2         =  1/3\n6-x-1                   =  2*(1/3)\n-x+5                    =  2/3\n-x                      =  2/3-5\n-x                      =  -13/3\nx                       =  13/3\n\ntherefore x equals   13/3
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
672754
Take out the five so you have:\nx^2 + 8 + 16\nthen you can change that to\n(x + 4)^2 b/c you have to find a number that is the product of the a and c and the sum of b\nwhere you have: ax^2 + bx + c
890923
To find the common denominator between fractions simply multiply the denominators and this is the common denominator.\n\nHowever this number may be large so try and find a number that all denominators will divide into evenly.\n\nHope this helps.
541370
Multiplying the denominators of all the fractions would be the easiest method.\n\nHowever, if the numbers are really big and you prefer working with small ones (e.g. if you do not have a calculator), you can first find the lowest common multiple (LCM).
224288
You will need a common denominator.  To find that, each denominator needs to be factored completely.\nExample\nx/(x^2 - 9) + 4/(x-3)\nThe denominators are x^2 - 9 and x-3\nx^2 - 9 factors into (x+3)(x-3)\nso now the denominators are\n(x+3)(x-3) and x-3\nStart with first denominator and list each different factor.  Repeat this process with each denominator.  After you have each different factor written down, go back and take the highest exponent of that factor.\nIn this problem there are two different factors:\n(x+3)(x-3)  The highest exponent for x+3 is 1 and 1 is also the highest exponent on x-3 so the lowest common denominator(LCD) is (x+3)(x-3)\n(IF the denominators had been (x+3)(x-3) and (x-3)^2 then the common denominator would be (x+3)(x-3)^2 because the highest exponent on the factor (x+3) is 1 and the highes exponent on (x-3) is 2.)\nNow back to the example above:\nThe LCD is (x+3)(x-3)\nx/(x^2 - 9) + 4/(x-3) = x/((x+3)(x-3)) + 4/(x-3)\nThe first term already has the LCD, so nothing has to be changed.\nThe second term has denominator x-3.  It is missing x+3 so we multiply the top and bottom by x+3\n4/(x-3) = (4(x+3))/((x-3)(x+3))  Multiply the top out\n= (4x+12)/((x-3)(x+3))  \nNow the terms have a common denominator so you can just add the numerators.\nx/((x+3)(x-3)) + (4x+12)/((x-3)(x+3))\n= (x + 4x+12)/((x+3)(x-3))\n= (5x+12)/((x+3)(x-3)) \nThe numerator cannot be factored so this problem is now in simplest form.\n\nHope this helps.
696212
find the common denominator\nmultipy the top and bottom of the fraction so the denominators are equal and then add the numerator
787158
y = xtan(x)\n\nUse product rule:\ndy/dx = x/cos²(x) + tan(x)\n= x/cos²(x) + sin(x)/cos(x) \n= [x + sin(x)·cos(x)] / cos²(x)\n\nUse product and chain rules:\nd²y/dx² = 1/cos²(x) + (-2)·(x/[cos(x)]^3)·(-sin(x)) + 1/cos²(x)\n= 2/cos²(x) + [2·x·sin(x)] / [cos(x)]^3\n= [2·cos(x) + 2·x·sin(x)] / [cos(x)]^3
1183491
Lim { [ sin(x-1)  ] / [ x^2 + x - 2] }\nx-->1\n\nAdd the following to the denominator\n-2x + 2x +1 -1\n\nLim { [ sin(x-1)  ] / [ x^2 + x - 2 -2x +2x +1 -1] }\nx-->1\n\nLim { [ sin(x-1)  ] / [ x^2 -2x + 1 +3x -3] }\nx-->1\n\nNote: (x^2-2x+1) =(x-1)^2    and  (3x-3) = 3(x-1)\n\nThen\nLim { [ sin(x-1)  ] / [ (x-1)^2 +3(x-1)] }\nx-->1\n\nLim { [ sin(x-1)  ] / [ (x-1)^2 +3(x-1)] }\nx-->1\n\nLim { [ sin(x-1)  ] / [ (x-1)[ (x-1)+3 ] }\nx-->1\n\nLim { [ sin(x-1) / (x-1) ] * [ 1/ (x-1+3 )] }\nx-->1\n\nLet u= x-1 then as x-->1, u--> 0\nand\nLim { [ sin u / u ] * [ 1/ (u+3 )] }\nu-->0\n\nSince\n\nLim  [ sin(u) / (u) ]=1\nu-->0\n\nThen \nLim { [ sin u / u ] * [ 1/ (u+3 )] } = 1 * [1/(0+3)] = 1*1/3 = 1/3\nu-->0