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P is partly constant and partly varies inversely as Q.lf Q=9 when P=3 and Q=18 when P=9 find P when Q=12

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asked by anwari
  1. P = k/Q + C

    given:
    P=3, Q=9
    3 = k/9 + C
    27 = K + 9C ***

    P=9, Q = 18
    9 = k/18 + C
    162 = k + 18C **

    subtract *** from **
    135 = 9C
    C = 15
    in ***
    k+135 = 27
    k = -108

    P = -108/Q + 15

    so when Q = 12
    P = -108/12 + 15 = 6

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    posted by Reiny
  2. P=-108\12+ 15 =6
    Multiply both side /no. With 12.
    6 = 15 -108\12, you get
    72 = 180 - 108
    72 = 72, cross over 72 to the other side.
    72-72 = 0

    So, P = 0

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    posted by Nao

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