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ajprincess

  • 2 years ago

Plzzz can someone help me? Thanx.:)

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  1. ajprincess
    • 2 years ago
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  2. timo86m
    • 2 years ago
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    I understand the q and I know it is true I jsut dont know how to prove it. maybe something about f(x) and shifting it f(x-a)+b :)

  3. dpaInc
    • 2 years ago
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    sorry... i don' have open office or word... how 'bout a pdf?

  4. timo86m
    • 2 years ago
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    if you have hotmail it has a free viewer :) via skydrive

  5. ajprincess
    • 2 years ago
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    jst a moment let me convert it.

  6. dpaInc
    • 2 years ago
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    it says Advance.. i hope it's not too advanced otherwise you're doing this for nothing.. :)

  7. ajprincess
    • 2 years ago
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  8. ajprincess
    • 2 years ago
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    @dpaInc hw abt a screenshot? I hav posted a screenshot.

  9. dpaInc
    • 2 years ago
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    hmmm.. yep.. too advanced for me... sorry.... :(

  10. ajprincess
    • 2 years ago
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    it's k. Thanxx a lot for tryng to help me. :)

  11. dpaInc
    • 2 years ago
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    when ur done with this i can help with the times table... :)

  12. ajprincess
    • 2 years ago
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    :)

  13. ajprincess
    • 2 years ago
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    Thanxx a lot @timo86m for the suggestion.

  14. shubhamsrg
    • 2 years ago
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    r1,r2,r3,r4 are vectors a1r1 + a2r2 + a3r3 + a4r4 =0 now r1 is position vector w.r.t O,,for any other point O' ,,vector is r1 - OO' and the others become r2 - OO' , r3 -OO', r4-OO' we need to find a1(r1 - OO') + a2(r2-OO') + a3(r3-OO') + a4(r4 -OO') => ( 0 ) - OO' ( a1 + a2 +a3 +a4) clearly is becomes 0 for a1 +a2 + a3 + a4 =0 hence proved..

  15. ajprincess
    • 2 years ago
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    if u dnt mind can u please tell me hw u got r1-OO'? That part is nt clear to me.

  16. shubhamsrg
    • 2 years ago
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    |dw:1338704382266:dw| we can see easily that r1 = OO' + (required vector) thus req. vector is r1 - OO'

  17. ajprincess
    • 2 years ago
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    Oh k. Thanxxxx a lotttt.

  18. shubhamsrg
    • 2 years ago
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    glad to help ^_^

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