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Math Help - how value of d is calculated in this equation

  1. #1
    Member moonnightingale's Avatar
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    how value of d is calculated in this equation

    kindly help me to solve this equation
    how he calculated vlue of d to be 16.1
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  2. #2
    MHF Contributor harish21's Avatar
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    RHS=0.022\times10^{-6}\times 3\times10^8=6.6

    Therefore, you have:

    2\sqrt{\bigg(\dfrac{d}{2}\bigg)^2+8^2}-d= 6.6

    2\sqrt{\bigg(\dfrac{d^2}{4}\bigg)+64}=6.6+d

    squaring both sides gives you:

     \bigg[2\sqrt{\bigg(\dfrac{d^2}{4}\bigg)+64}\bigg]^2=(6.6+d)^2

     4 \cdot \bigg(\dfrac{d^2}{4}+64\bigg) = 6.6^2+2d(6.6)+ d^2

    d^2+256=43.56+13.2d+d^2

    the d^2 terms cancel each other out and simplification gives you the value of d.
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