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anonymous

  • one year ago

Will Medal, just really need help!! Find an equation in standard form for the hyperbola with vertices at (0, ±4) and asymptotes at y = ±1 divided by 3.x.

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  1. anonymous
    • one year ago
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    looks like you are stuck with analytic geometry :) now lets start with how standard form looks like\[\frac{ y^2 }{ a^2 } - \frac{ x^2 }{ b^2 } = 1\]

  2. anonymous
    • one year ago
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    Ok @saseal

  3. anonymous
    • one year ago
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    |dw:1439304877703:dw|

  4. anonymous
    • one year ago
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    My answer choices are: A. y squared over 144 minus x squared over 16 = 1 B. y squared over 16 minus x squared over 36 = 1 C. y squared over 16 minus x squared over 144 = 1 D. y squared over 36 minus x squared over 4 = 1 I'm between B and C

  5. anonymous
    • one year ago
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    now we have the gradient y=a/b which is 1/3

  6. anonymous
    • one year ago
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    and we know a is 4 so b gotta be 12 since the gradient have to match up

  7. anonymous
    • one year ago
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    you can guess which one it is by now :)

  8. anonymous
    • one year ago
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    C? @saseal

  9. anonymous
    • one year ago
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    yes

  10. anonymous
    • one year ago
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    Can you help me with a few others?

  11. anonymous
    • one year ago
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    ok

  12. anonymous
    • one year ago
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    Thanks! @saseal Find the center, vertices, and foci of the ellipse with equation x squared divided by 400 plus y squared divided by 625 = 1.

  13. anonymous
    • one year ago
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    \[\frac{ x^2 }{ 400 } + \frac{ y^2 }{ 625 } = 1\] this should be easy by now

  14. anonymous
    • one year ago
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    \[\frac{ x^2 }{ 20^2 } + \frac{ y^2 }{ 25^2 } = 1\]

  15. anonymous
    • one year ago
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    where do you think is the major axis of this egg?

  16. anonymous
    • one year ago
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    major axis is the longer side

  17. anonymous
    • one year ago
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    The y-axis

  18. anonymous
    • one year ago
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    correct

  19. anonymous
    • one year ago
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    now think where's the center

  20. anonymous
    • one year ago
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    theres no h and k here

  21. anonymous
    • one year ago
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    0,0

  22. anonymous
    • one year ago
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    there we have the both the center and vertices now

  23. anonymous
    • one year ago
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    What's the vertices though?

  24. anonymous
    • one year ago
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    center (0, 0) vertices (0, ±25)

  25. anonymous
    • one year ago
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    Oh ok!

  26. anonymous
    • one year ago
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    |dw:1439306628656:dw|

  27. anonymous
    • one year ago
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    now we need to find c to get the foci

  28. anonymous
    • one year ago
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    \[c=\sqrt{a^2-b^2}\]

  29. anonymous
    • one year ago
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    Ok, so C = 15? @saseal

  30. anonymous
    • one year ago
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    +/- 15

  31. anonymous
    • one year ago
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    yes

  32. anonymous
    • one year ago
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    Ok so would my answer be: Center: (0, 0); Vertices: (-25, 0), (25, 0); Foci: (-15, 0), (15, 0)

  33. anonymous
    • one year ago
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    Or no. It would be: Center: (0, 0); Vertices: (0, -25), (0, 25); Foci: (0, -15), (0, 15

  34. anonymous
    • one year ago
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    Could you help me with one more? @saseal

  35. anonymous
    • one year ago
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    yea

  36. anonymous
    • one year ago
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    sry im not looking at the pc all the time

  37. anonymous
    • one year ago
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    No problem. And I'm in a new chapter so: Find the derivative of f(x) = 6 divided by x at x = -2.

  38. anonymous
    • one year ago
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    And also: Find the derivative of f(x) = 4x + 7 at x = 5. @saseal

  39. anonymous
    • one year ago
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    sry i fell asleep

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