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Babynini

  • one year ago

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  1. Babynini
    • one year ago
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  2. Babynini
    • one year ago
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    @freckles

  3. Babynini
    • one year ago
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    @rational

  4. Babynini
    • one year ago
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    oh I know where I calculated wrong. in that c/a and forgot that a was =4 and not 2

  5. Babynini
    • one year ago
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    would it be b^2 = 2?

  6. Babynini
    • one year ago
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    nope that's not right either.

  7. Babynini
    • one year ago
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    @rvc do you know how to do these?

  8. Babynini
    • one year ago
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    @jim_thompson5910

  9. rvc
    • one year ago
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    bad at circle and conics sorry :(

  10. Babynini
    • one year ago
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    Thanks for looking at it though. Ah well.

  11. rvc
    • one year ago
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    @divu.mkr help please :)

  12. Babynini
    • one year ago
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    first off we find a from the major axis a= 4 then we find c by looking at the eccentricity c/a = sq7/2

  13. anonymous
    • one year ago
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    focii on y axis means that the major axis is on Y aix and you are given with the length of major axis and ecc.. you can find minor axis from that data

  14. anonymous
    • one year ago
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    axis**

  15. alekos
    • one year ago
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    It would have a vertical major axis so (x-h)^2/b^2 + (y-k)^2/a^2 =1 where 2b=8

  16. Babynini
    • one year ago
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    how do I find the minor axis in there?

  17. alekos
    • one year ago
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    eccentricity = sqrt(a^2-b^2)/a

  18. anonymous
    • one year ago
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    b^2 = a^2(1-e^2)

  19. alekos
    • one year ago
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    so from this you can find the minor axis

  20. Babynini
    • one year ago
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    hmm

  21. alekos
    • one year ago
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    focii are on the y-axis so h=0

  22. Babynini
    • one year ago
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    so x^2/8 + y^2/16 = 1

  23. Babynini
    • one year ago
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    is that the answer?

  24. Babynini
    • one year ago
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    @iambatman save me. lol

  25. anonymous
    • one year ago
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    Hollllllaaaa

  26. Babynini
    • one year ago
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    aloo

  27. anonymous
    • one year ago
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    major axis = 8 ye?

  28. Babynini
    • one year ago
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    I just need to find an equation :P

  29. Babynini
    • one year ago
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    yeah

  30. Babynini
    • one year ago
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    major axis = 2a so a = 4

  31. anonymous
    • one year ago
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    Yes good :P \[e = \frac{ c }{ a } ~~~\text{and}~~~ c^2 = a^2-b^2\] yes yes this looks good, it will work out nicely

  32. anonymous
    • one year ago
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    Ok so notice we're given the eccentricity (e) and it tells us the Foci is on the y - axis, this is important because the foci are ALWAYS on the major axis. So will the major axis run vertically or horizontally?

  33. Babynini
    • one year ago
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    vertically?

  34. anonymous
    • one year ago
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    That sounds good, so what should we do next

  35. anonymous
    • one year ago
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    You should already realize an error in your work with that information

  36. Babynini
    • one year ago
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    a should be the bigger one?

  37. anonymous
    • one year ago
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    You're too good :)

  38. Babynini
    • one year ago
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    Aw, shucks.

  39. Babynini
    • one year ago
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    so what numbers do i flip around?

  40. anonymous
    • one year ago
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    Well lets find b first

  41. anonymous
    • one year ago
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    To do that we have \[e = \frac{ c }{ a }\] and \[c^2 = a^2 - b^2\]

  42. Babynini
    • one year ago
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    e = sq7/2 but we're given that a = 4 so do we double that whole thing?

  43. anonymous
    • one year ago
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    What was that cute game we were playing earlier? I had a name for it

  44. Babynini
    • one year ago
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    lol substitution?

  45. anonymous
    • one year ago
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    Yeah that's it ;)

  46. Babynini
    • one year ago
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    but we're given the e!

  47. anonymous
    • one year ago
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    So what's stopping you from solving for c :)

  48. Babynini
    • one year ago
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    no b?

  49. anonymous
    • one year ago
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    We have a and e!

  50. Babynini
    • one year ago
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    ooh oh

  51. Babynini
    • one year ago
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    sq7/2=c/4

  52. anonymous
    • one year ago
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    :))))

  53. anonymous
    • one year ago
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    So what will be our c value?

  54. Babynini
    • one year ago
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    sq14?

  55. Babynini
    • one year ago
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    \[2\sqrt{7}\]

  56. anonymous
    • one year ago
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    Looks good, now what's our next step :)

  57. Babynini
    • one year ago
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    find b!

  58. anonymous
    • one year ago
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    Right, and how will we do that

  59. Babynini
    • one year ago
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    so that's just c, not c^2

  60. anonymous
    • one year ago
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    Yes good observation!

  61. Babynini
    • one year ago
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    so now c^2=a^2+b^2

  62. anonymous
    • one year ago
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    Stop right there for a second!

  63. Babynini
    • one year ago
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    o.o yes sir.

  64. anonymous
    • one year ago
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    Lets look at the equation you said, what is that for :)

  65. Babynini
    • one year ago
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    c

  66. anonymous
    • one year ago
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    ircle

  67. Babynini
    • one year ago
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    lol

  68. anonymous
    • one year ago
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    That's where a lot of people make mistakes, for an ellipse it's \[c^2 = a^2 - b^2\]

  69. anonymous
    • one year ago
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    Note the negative sign

  70. Babynini
    • one year ago
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    o.o gasp. that is no bueno.

  71. Babynini
    • one year ago
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    b^2=a^2-c^2 then

  72. anonymous
    • one year ago
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    :), now remember as I always say it's a game of substitution

  73. Babynini
    • one year ago
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    \[b^2 = 16 - (2\sqrt{7})^2\]

  74. Babynini
    • one year ago
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    \[b^2 = 16-28\] \[b^2=-12\]

  75. Babynini
    • one year ago
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    \[b=\sqrt{12}\]

  76. anonymous
    • one year ago
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    No, the negative is there, something is wrong

  77. anonymous
    • one year ago
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    Err, \[e = \frac{ c }{ a } \implies c = e \times a \implies c = \frac{ \sqrt{7} }{ 4 } \times 4 \implies \sqrt{7}\] bleh :)

  78. Babynini
    • one year ago
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    ..oh hehe

  79. anonymous
    • one year ago
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    I guess I should be doing the math as well haha

  80. Babynini
    • one year ago
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    sorry!

  81. Babynini
    • one year ago
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    ai ai ai I need to go to bed soon. waay to early on a school night =.= and you do too.

  82. anonymous
    • one year ago
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    No it's alright we just had mistaken the eccentricity we had sqrt(7)/2 when it's 4 :P

  83. Babynini
    • one year ago
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    oh xD

  84. anonymous
    • one year ago
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    Don't worry about me :), I'm the dark knight, night is where I work muahah!

  85. Babynini
    • one year ago
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    right, right, how could I forget!

  86. anonymous
    • one year ago
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    Alright so lets find b again bleh :)

  87. anonymous
    • one year ago
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    \[b^2 = a^2-c^2\]

  88. anonymous
    • one year ago
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    \[a = 4~~~c = \sqrt{7}\]

  89. Babynini
    • one year ago
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    b = 3

  90. Babynini
    • one year ago
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    b^2 = 9

  91. anonymous
    • one year ago
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    Now that looks good

  92. Babynini
    • one year ago
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    so..the answer I had at the very beginning. Look at the screenshot thing =.=

  93. anonymous
    • one year ago
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    major axis runs vertically :P

  94. Babynini
    • one year ago
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    ooh myy goooshh aspidghpaihgpi

  95. Babynini
    • one year ago
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    fricken hour later hahahah

  96. anonymous
    • one year ago
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    It's ok, the struggle is more important than the final answer, you will remember how to do it now :)

  97. Babynini
    • one year ago
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    we're geniuses! :)

  98. Babynini
    • one year ago
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    for sure though!

  99. anonymous
    • one year ago
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    Not me, you are a genius! You deserve a fields medal hehe!

  100. Babynini
    • one year ago
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    noo no bro the answers wrong still D:

  101. anonymous
    • one year ago
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    What did you put

  102. Babynini
    • one year ago
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  103. anonymous
    • one year ago
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    Nope.

  104. anonymous
    • one year ago
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    a = 4, a^2 = 16

  105. Babynini
    • one year ago
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    ...can you write out the whole answer just so i make sure to put it right?

  106. anonymous
    • one year ago
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    and y runs vertically hence \[\frac{ x^2 }{ 9 }+ \frac{ y^2 }{ 16 } = 1\]

  107. Babynini
    • one year ago
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    That little green mark though :')

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  108. anonymous
    • one year ago
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    Yes, sorry about that I did not check if you put a^2

  109. Babynini
    • one year ago
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    it's all good :) so it was indeed my first answer just forgot to ^2 the a

  110. anonymous
    • one year ago
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    Yes, we just had to switch it around, but in a test you would have to show your work, and you now know how to do that! :) It's a game of substitution as always!

  111. Babynini
    • one year ago
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    haha cuz we hate the players (not really) but love the game ;)

  112. Babynini
    • one year ago
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    yes indeed!

  113. Babynini
    • one year ago
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    *hug hug hug. you are a trooper. haha and with that. I am off to bed.

  114. anonymous
    • one year ago
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    Take care :)

  115. Babynini
    • one year ago
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    Night, Batman.

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