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lynn555

consider the point (x,y) lying on the graph of y=(x+3)^1/2. let L be the distance between the points (x,y) and (4,0). write l as a function of y

  • one year ago
  • one year ago

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  1. Jemurray3
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    \[L = \sqrt{ (x-4)^2 + y^2 } \] and \[ y = \sqrt{x+3} \implies x = y^2 - 3\] so substitute that in.

    • one year ago
  2. lynn555
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    so the answer would be y^5-46+x?

    • one year ago
  3. Jemurray3
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    erm, no. Literally substitute y^2-3 in for x in the equation that starts with L = ...

    • one year ago
  4. Jemurray3
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    there should be no x involved.

    • one year ago
  5. LeeYeongKyu
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    |dw:1347425169078:dw|

    • one year ago
  6. LeeYeongKyu
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    so given function is y=(x+3)^1/2 so y^2=x+3, so, x = y^2-3. plugging x to the L=(x-4)^2 + y^2 then, L=(y^2-6)^2 + y^2

    • one year ago
  7. LeeYeongKyu
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    you can reduce it easily

    • one year ago
  8. Jemurray3
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    That should be y^2-7, not y^2-6, and the whole thing should be under a square root. But yeah.

    • one year ago
  9. lynn555
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    ok thanks

    • one year ago
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