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Re: [math_club] proportional

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  • mahbubul hasan
    well i dont know but i think u r not clr about the prop main concept... anyway, actually the Fs used in mine or nasa vi s isnt actually F alover... it is
    Message 1 of 31 , May 1 10:19 AM
      well i dont know but i think u r not clr about the prop main concept...
       
      anyway, actually the Fs used in mine or nasa vi's isnt actually F alover... it is changing...
       
      if u have diff to und it, then u may use F1, F2... and use F1/F2 = a1/a2...
       
      anyway, i don't know yet whether the proofs below r OK or not, but (actually i donnt have sufficient time to go over them) i think the 3D method is ok
       
      Mahbub

      Fahim <fahim_com@...> wrote:
      Nasa vai,
      apnar mail er >> and g(a=a2) is a fn of a.<< line er mane bughlam na. would u plz expand it?
      ar ekta bapar, >> F is prop to a provided b is const. that means k is k(b1), ie a fn of b<< keno hobe bughlam na?

      Shanto vai,
      apnar soln ta thik moto bughlam na. ami apnar graph theory niye soln ta paisi. kintu ei soln tao bughte chai.
      amar somossa gula dekhen:

      1.   >>so F is prop to a => F = ka. so by increasing k times it came to a2 state say.<<
      apni "F is prop to a" bolsen kintu "a" er jaygay "a1" thakar kotha.

      2.   jodi F prop to a1 hoy tahole F=ka1. ekhon  a1 ke k times increase korle ta a2 te keno asbe thik bughlam na.
      a1 ke k times increase manei k*a1. tahole apnar kotha mote k*a1=a2 hobe.
      orthat F=a2 hobe. eita ki somvob? (nasa vaio mone goy eita bolte chaisen).

      3.   >>
      now, keep a2 fixed that means a is const. move from b1 to b2.<<
      a2 fixed naki a1 fixed?

      4.  
      >>so from(a1,b1) state it came to (a2,b2) state by k*g times. that means by const and these. so, F is prop to ab.<<
      kintu tara orthat "a" ar "b" gun akare thakbe ken?

      hope u would ans.
      take care.
      good bye.

      FAHIM


      Nasa <nasarouf@...> wrote:
      correction: you *cannot* assume the equality since no such thing was implied.

      On 4/30/06, Nasa <nasarouf@... > wrote:
      i think you are forgetting something
      F is prop to a provided b is const.
      that means k is k(b1), ie a fn of b. there may be different k's at different values for b. k(b1) and k(b2) may not necessarily be equal. you assume the equality since no such thing was implied.
      and g(a=a2) is a fn of a.
      now do the proof from start.

      nasa


      On 4/29/06, mahbubul hasan < shanto86@...> wrote:
      assa, dhoro, F is prop to a when b const, and prop to b when a const.
       
      Now, say u will change a from a1 to a2, and similarly b1, b2.
       
      so say at first keep the system const at b1, and move a1 to a2.
       
      so F is prop to a => F = ka. so by increasing k times it came to a2 state say.
       
      now, keep a2 fixed that means a is const. move from b1 to b2.
       
      F = gb. it came to this is state by g times.
       
      so from(a1,b1) state it came to (a2,b2) state by k*g times. that means by const and these. so, F is prop to ab.
       
      Mahbub

      Fahim <fahim_com@... > wrote:
      shanto vai,
      ami onek chesta korei math_club e ei somossa ta pathaisi. dekhseni to ekhon porjonto keu mathematical proof dite pare ni. sobai antorikotar sathe chesta korsen. so, would u plz give me the soln?
      thank u all.
      FAHIM

      mahbubul hasan < shanto86@...> wrote:
      are eta k to oi vabei proman kora jay! valo moto chinta koro!
       
      Mahbub

      Fahim < fahim_com@...> wrote:
      i know the rule. but what i dont know is the mathematical prove.
      FAHIM

      mahbubul hasan < shanto86@...> wrote:
      well... think in this way, u will change both a and b.
       
      but first u will change a keeping b fixed, say u moved from a1 to a2.
       
      then u will change from b1 to b2, keeping a fixed.
       
      it will obey that rule.
       
      Mahbub

      Fahim < fahim_com@...> wrote:
      amar somossatar somadhan ki keu i dibe na?

      >>i know it is really ridiculous ask this kind of question. but really cant understand it. look at my prb:
      can u prove that mathematically:
      if A is_proportional_to a                     [when b is constant]
      and A is_proportional_to b                 [when a is constant]

      then A is_proportional_to (a*b)            [when a & b are not constant]<<

      it is the most common thing in proportion. it has used in the newton's law of gravitation to get the equation. aro onek jaigay tei er babohar ase. kintu keu somadhan dilona ken bughlam na.
      Hope u'll b interested.
      FAHIM

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    • Fahim
      If, A is proportional to b [when c constant] A is proportional to c [when b constant] then we get, A is proportional to bc [when nothing is constant] long
      Message 31 of 31 , Nov 6, 2006
        "If,
        A is proportional to b  [when c constant]
        A is proportional to c  [when b constant]
        then we get,
        A is proportional to bc [when nothing is constant]"

        long ago i asked for proofing this. but didnt get satisfactory ans. now i have a beautiful soln. try to prove it.
        FAHIM


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