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anonymous

  • one year ago

someone help me! I have an assignment which requires me to write the equation that represents the image of an arch monument that i will post below. PLEASE HELP

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  1. anonymous
    • one year ago
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  2. anonymous
    • one year ago
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    the height of the arch is 630ft as well as the width

  3. anonymous
    • one year ago
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    @ganeshie8 ?

  4. anonymous
    • one year ago
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    @Hero ?

  5. anonymous
    • one year ago
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    this was the assignment. https://i.gyazo.com/ea2c7cf9f897771440fb99ec9074fb06.png

  6. Hero
    • one year ago
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    Start by making a sketch of it on graph paper.

  7. anonymous
    • one year ago
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    do i just place the vertex at the origin?

  8. Hero
    • one year ago
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    It's probably best to place one of the ends of the foundation at the origin.

  9. anonymous
    • one year ago
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    ok give me a second to sketch it

  10. anonymous
    • one year ago
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    |dw:1439572894166:dw|

  11. anonymous
    • one year ago
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    ???

  12. anonymous
    • one year ago
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    NO ONE?

  13. Hero
    • one year ago
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    That's a good start. Now, find the midpoint of the line segment containing endpoints (0,0) and (630,0)

  14. anonymous
    • one year ago
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    (0, 315)

  15. Hero
    • one year ago
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    (315, 0) actually. Plot that point on your graph and draw a vertical line segment from the vertex to the midpoint of the base segment.

  16. anonymous
    • one year ago
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    yea thats what i meant. and ok

  17. Hero
    • one year ago
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    That vertical line segment is essentially the axis of symmetry.

  18. anonymous
    • one year ago
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    |dw:1439574756537:dw|

  19. Hero
    • one year ago
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    You should probably use the vertex form of a parabola to write the equation for it and make sure to include the correct restrictions for the domain.

  20. Hero
    • one year ago
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    |dw:1439574890871:dw|

  21. anonymous
    • one year ago
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    so y = a(x-h)^2 +k ?

  22. Hero
    • one year ago
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    Yes that. You already know the values for (h,k).

  23. anonymous
    • one year ago
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    what do i plug in for a though? do i leave it as it is?

  24. Hero
    • one year ago
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    You have to solve for a

  25. Hero
    • one year ago
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    To do that, insert (h,k), then for (x,y) insert (0,0)

  26. Hero
    • one year ago
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    Then solve for a

  27. anonymous
    • one year ago
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    oh ok ill do that

  28. anonymous
    • one year ago
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    i got a= -2/315

  29. Hero
    • one year ago
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    Show the work you did to arrive at that result

  30. anonymous
    • one year ago
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    a(0-315)^2 + 630 =0 99225a + 630 = 0 99225a = -630 a = -630/99225 a = -2/315

  31. Hero
    • one year ago
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    Okay good.

  32. Hero
    • one year ago
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    So you should be able to express the equation of the parabola in vertex form. Is that all you're required to do for this problem?

  33. Hero
    • one year ago
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    Actually, that value for a is an approximation. It's not exact.

  34. anonymous
    • one year ago
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    yea thats pretty much it. My teacher just told us to do that for one of the pictures given.

  35. anonymous
    • one year ago
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    thanks for the help.

  36. Hero
    • one year ago
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    yw

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