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![[Post New]](/templates/default/images/icon_minipost_new.gif) 1 May 2007 08:40:02 IST
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1. [ x] [ 1] [(2x -3)(  x -1)]/[2x 2+x-3] 2. [ x][0 ][(9+5x+4x 2) -3]/x 3.lim x-->1 [ (x+3) -- (8x+1)]/[ (5-x)- (7x-3)] 4.limx--->2 [2 x+2 3-x--6] / [ 2 -x--2 1-x] |
sathya prabha girish |
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1 May 2007 08:57:22 IST
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Answer 1 . [ x][ 1] [(2x -3)(x -1)]/[2x2+x-3] =[ x][ 1] [(2x -3)(x -1)] (2x + 3) (x-1) =[ x][ 1] (2x - 3) (2x +3)(x +1) = - 1/ 10 
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Manasi....
NIT-Allahabad...
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Challenges are High, Dreams r New..
The World out thr is waiting for U !!
Dare to dream, Dare to Try..
No Goal is distant, no Star is too high !!! |
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1 May 2007 09:10:51 IST
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HEY ur not supposed 2 post more than 1 question per message! ANSWERS 1) -1/10 2) 5/6 3) limit does not exist 4) 8 rate me if i am correct
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1 May 2007 09:13:07 IST
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[ x][0 ][(9+5x+4x2) -3]/x = [ x][0 ] 8x + 5 [Using L' hospital rule] 2 (9+5x+4x2) = 5 / 6
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Manasi....
NIT-Allahabad...
............................................................
Challenges are High, Dreams r New..
The World out thr is waiting for U !!
Dare to dream, Dare to Try..
No Goal is distant, no Star is too high !!! |
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1 May 2007 11:32:11 IST
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hi akku ,
there is nothing like that as u said !! u can ask more that 1 ....
The rule u said applys only when u r asking to "EXPERTS " , by spending ur nickels !
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Science is vision multiplied!
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