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trumpetmaster7777 Group Title

Graph a hyperbola with foci at (0, ±10) and vertices at (0, ±8) using graphing technology.

  • one year ago
  • one year ago

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  1. hba Group Title
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    @brookester6 can help you on this :P

    • one year ago
  2. brookester6 Group Title
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    umm yeah JAS!

    • one year ago
  3. blossomemerald Group Title
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    wow a lot of people on one question lol party

    • one year ago
  4. trumpetmaster7777 Group Title
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    well, I did tag a lot of people. I really need help.

    • one year ago
  5. trumpetmaster7777 Group Title
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    :( everyone gone.

    • one year ago
  6. Loser66 Group Title
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    I witnessed many people stopped by to help you, why did they go all?

    • one year ago
  7. trumpetmaster7777 Group Title
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    they, left, It's too hard of a question

    • one year ago
  8. trumpetmaster7777 Group Title
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    @brookester6 said just a second, and then left.

    • one year ago
  9. Loser66 Group Title
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    not that girl!! I think because you called too many people. so, this one thinks that the other's duty. hahahaa

    • one year ago
  10. trumpetmaster7777 Group Title
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    :( aw man, i just need an equation

    • one year ago
  11. trumpetmaster7777 Group Title
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    Let's start from scratch again

    • one year ago
  12. Loser66 Group Title
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    ok, my question is " what does graphing technology?" I always graph by hand, don't know how to use "technology"

    • one year ago
  13. trumpetmaster7777 Group Title
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    lol using a website, but I just need an equation, I already have a website to use

    • one year ago
  14. Loser66 Group Title
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    so, you can solve it by yourself? no need to stay ? right? actually, I am a super lazy guy. If you don't need help anymore, I glad to hear that. hehehe

    • one year ago
  15. trumpetmaster7777 Group Title
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    no, i need to have the numbers put into an equation, I don't know how to do that

    • one year ago
  16. Loser66 Group Title
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    shoot.what is your equation? for hyperbola?

    • one year ago
  17. trumpetmaster7777 Group Title
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    foci at (0, ±10) and vertices at (0, ±8) Make this into an equation

    • one year ago
  18. Loser66 Group Title
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    remember that foci (0,+-c) and vertices (0,+-a) and c^2 = a^2 + b^2( use this form to find out b) and put everything into the standard form of hyperbola \[\frac{y^2}{a^2}-\frac{x^2}{b^2}=1\]

    • one year ago
  19. Loser66 Group Title
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    and draw the asymptotes by \[y= (+-)\frac{a}{b}*x\]

    • one year ago
  20. Loser66 Group Title
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    that's it

    • one year ago
  21. trumpetmaster7777 Group Title
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    10^2=8^2+b^2

    • one year ago
  22. Loser66 Group Title
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    remember that if your foci is (0,+- c) your hyperbola has the shape like|dw:1369325453179:dw|

    • one year ago
  23. Loser66 Group Title
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    so b^2 = 100 - 64 = 36 so ?

    • one year ago
  24. trumpetmaster7777 Group Title
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    100=64+b^2 36=b^2 6=b

    • one year ago
  25. Loser66 Group Title
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    \[\frac{y^2}{64}-\frac{x^2}{36}=1\] your equation form is a^2 and b^2 , you need a, b when you have to figure out the asymptotes only. to get the form of hyperbola, no need to get that far

    • one year ago
  26. Loser66 Group Title
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    got what i mean?

    • one year ago
  27. trumpetmaster7777 Group Title
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    ok, but how do i find x or y?

    • one year ago
  28. Loser66 Group Title
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    why you have to know x, y?

    • one year ago
  29. trumpetmaster7777 Group Title
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    oh wait, that's it

    • one year ago
  30. Loser66 Group Title
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    yeap!!!

    • one year ago
  31. Loser66 Group Title
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    if you have to sketch the graph, you have to sketch the asymptotes, too. but it's quite easy when you have a, b, just draw the lines out.

    • one year ago
  32. trumpetmaster7777 Group Title
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    I got the graph

    • one year ago
  33. trumpetmaster7777 Group Title
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    But, I have online class

    • one year ago
  34. trumpetmaster7777 Group Title
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    girl?

    • one year ago
  35. trumpetmaster7777 Group Title
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    no you say girl at the end of sentence, I CONFUSE

    • one year ago
  36. Loser66 Group Title
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    i think you are a girl

    • one year ago
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