tutoring thread

Discussion in 'Off-Topic' started by o.iatlhawksfan, Oct 17, 2007.

  1. Brand New

    Brand New so wavy

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    Thanks for your help MNFL!
     
  2. #1_War_Poet_ForLife

    #1_War_Poet_ForLife The Baker of Cakes

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  3. bbwchingy0007

    bbwchingy0007 BBW Member

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    x^3+5x^2=1

    (x^2)(x+5)=1

    That might be factorised better than before. (could take 1 from both sides obviously)
     
  4. Brand New

    Brand New so wavy

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  5. JCB

    JCB The Savage Nation

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    Can someone explain 4d?

    I've looked at the solution online but can't figure out what was done to arrive at the answer. Necesito help, por favor.
     
  6. bbwchingy0007

    bbwchingy0007 BBW Member

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    I think you will need to integrate f' to get f. Then plug in the two values.

    By my reckoning f'(x):
    y=-x-2 where -3\<x\<0
    (x-2)^2+(y+2)^2=4 where 0\<x\<4 and y>/-2.

    So try integrating those, I think [​IMG]
     
  7. Brand New

    Brand New so wavy

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    Out of
    Laos
    Thailand
    North Vietnam
    South Vietnam
    China

    which ones were neutral, communist and the other one i forgot the name.
     
  8. lukewarmplay

    lukewarmplay Hired Goons

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    <div class='quotetop'>QUOTE (Bynumite @ Apr 1 2008, 10:42 PM) <{POST_SNAPBACK}></div><div class='quotemain'>Out of
    Laos
    Thailand
    North Vietnam
    South Vietnam
    China

    which ones were neutral, communist and the other one i forgot the name.</div>

    www.wikipedia.org
     
  9. lukewarmplay

    lukewarmplay Hired Goons

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    <div class='quotetop'>QUOTE (chingy0007 @ Mar 14 2008, 11:43 AM) <{POST_SNAPBACK}></div><div class='quotemain'>I think you will need to integrate f' to get f. Then plug in the two values.

    By my reckoning f'(x):
    y=-x-2 where -3\<x\<0
    (x-2)^2+(y+2)^2=4 where 0\<x\<4 and y>/-2.

    So try integrating those, I think [​IMG]</div>

    nice. i guess they gave you f(0) because that way you can get both constants after the integrations.
     

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