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    dbeaudry's Avatar
    dbeaudry Posts: 2, Reputation: 1
    New Member
    		  
    #1

    Jan 27, 2009, 07:27 PM
    Transposing Formulas
    How do I solve the following:
    A=3C+4C for C

    5/9X - 13 = 27

    Thanks
    ROLCAM's Avatar
    ROLCAM Posts: 1,420, Reputation: 23
    Ultra Member
    		  
    #2

    Jan 27, 2009, 07:46 PM

    A=3C+4C
    A = 7C
    A/7 = C

    C = 1/7 A

    _____________________________________




    5/9X - 13 = 27

    5/9x = 27 +13 when 13 moves sides it changes its sign.

    5/9x = 40

    x = 40* 9/5 when 5/9 changes side it reverses itself.

    x = 40*9/5
    x = 8*9
    x = 72
    bones252100's Avatar
    bones252100 Posts: 253, Reputation: 29
    Full Member
    		  
    #3

    Jan 27, 2009, 07:52 PM

    Add the two C multipliers then divide both sides by the sum.

    Add 13 to both sides then divide both sides by the fraction. Remember the inversion factor. The 9 is a divisor which becomes a multiplier on the other side. The five is a multiplier which becomes a divisor on the other side.
    shegun4u's Avatar
    shegun4u Posts: 5, Reputation: 1
    New Member
    		  
    #4

    Nov 4, 2009, 05:16 AM

    E=mgh+1-/2mv(squared) for h
    shegun4u's Avatar
    shegun4u Posts: 5, Reputation: 1
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    #5

    Nov 4, 2009, 05:16 AM
    E=mgh+1/2mv(squared) for h
    Nhatkiem's Avatar
    Nhatkiem Posts: 120, Reputation: 9
    Junior Member
    		  
    #6

    Nov 4, 2009, 12:08 PM

    There is no need to post twice here and post 2 different topics in the homework section. Also if this is homework Shegun4u, you are not allowed to post here for answers.
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