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  1. u (t).v(t) = <-sqrt(t) sint, t, t^(2/3)>.<-sqrt(t) sint, cos^2 t, -t^(1/3)>. u (t).v(t) = t*sin^2(t) + t*cos^2(t) - t = t - t = 0 (d/dt)[ u (t).v(t)] = 0 u (t...

    1 Answers · Science & Mathematics · 19/07/2009

  2. u ^2 + 8u + 7 = 0 ( u + 7)( u + 1) = 0 Then u has to equal what's in the parentheses set equal to zero: u + 7 = 0 and u + 1 = 0 u = -7 and u = -1

    3 Answers · Science & Mathematics · 06/04/2011

  3. [( u + h)²/( u + h + 2) - u²/( u + 2)]/h = [(u² + 2uh + h²)/( u + h + 2) - u²/( u + 2)]/h ...

    1 Answers · Science & Mathematics · 01/06/2013

  4. ( u +6)² - 64 = 0 ( u + 6)² = 64 ( u + 6) = ± 8 u = 2, -14 Answer: u = 2 or u = -14

    4 Answers · Science & Mathematics · 07/05/2011

  5. Hello, ∫ √{[2 /(1 - u²)] + [1 /(1 - u )²] + [1 /(1 + u )²]} du = let's rearrange the radicand... to the rule a² + 2ab + b² = (a + b)²) ∫ √{[1 /(1 + u )] + [1 /(1 - u )]}² du = ∫ | [1 /(1 + u )] + [1 /(1 - u )] | du = (distinguishing...

    1 Answers · Science & Mathematics · 06/10/2016

  6. U be lernin not to speek and rite like a ghetto thug then u be getun sum rank 4 it wen u figer da yeers u do in da JROTC. dat b it

    2 Answers · Politics & Government · 02/05/2012

  7. u = u ^1; u * u ^(1/2) = u ^(1+1/2)= u ^(3/2) x^2 - 1 = u ; du = 2xdx x² = u +1 Int X^5 sqrt(x^2 - 1) dx Int x^4 √(x²-1) x dx = (1/2) Int ( u +1)²√ u du .. x dx = du/2 = (1/2) ...

    1 Answers · Science & Mathematics · 21/06/2008

  8. u just need to go to a website called gifszone.com then u look for wat u like then u copy the code then u paste in on your profile by cliking edit profile

    3 Answers · Computers & Internet · 16/09/2006

  9. you should replace the x in the numerator with x = ( u -1) /2 from u =1+2x your integral becomes (1/4) .∫ { ( u -1) / sqrt...

    2 Answers · Science & Mathematics · 11/03/2008

  10. u a freaking wierd i think u have to man sick thoughts on your head i bet u are looking for some freaking 12 year old to rap git a life ok they are freaking 12 why do u care!!!

    3 Answers · Business & Finance · 27/10/2007

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