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12 July, 14:24

Find an equation that meets the conditions p varies inversely as d, and p = 2 when d = 3

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  1. 12 July, 14:46
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    pd = 6

    Step-by-step explanation:

    p ∝ 1/d

    change ∝ to = by adding constant k or c

    p = k * 1/d

    p = k/d

    Cross multiply

    pd = k

    k = pd

    k = 2 * 3

    k = 6

    From pd = k

    pd = 6
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