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anonymous

  • one year ago

I WILL GIVE MEDAL AND FAN: Find the center, vertices, and foci of the ellipse with equation ((x^2/)/(100)) + ((Y^2)/(64))= 1

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  1. anonymous
    • one year ago
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    @nincompoop

  2. anonymous
    • one year ago
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    @ash2326

  3. anonymous
    • one year ago
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    @ganeshie8

  4. UsukiDoll
    • one year ago
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    standard ellipse equation For a wider-than-tall ellipse with center at (h, k), having vertices a units to either side of the center and foci c units to either side of the center, the ellipse equation is: \[\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1\] For a taller-than-wide ellipse with center at (h, k), having vertices a units above and below the center and foci c units above and below the center, the ellipse equation is: \[\frac{(y-h)^2}{a^2}+\frac{(x-k)^2}{b^2}=1\] standard foci equation \[a^2-c^2=b^2\] standard center \[(h,k)\]

  5. UsukiDoll
    • one year ago
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    based on your equation \[\frac{x^2}{100}+\frac{y^2}{64}=1\] a > b there's a horizontal major axis

  6. UsukiDoll
    • one year ago
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    EXAMPLE: suppose we have the equation \[\frac{x^2}{4}+\frac{y^2}{1}=1\]

  7. UsukiDoll
    • one year ago
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    since a > b we have a horizontal major axis so we re-write this equation as where a = 2 and b = 1 \[\frac{x^2}{2^2}+\frac{y^2}{1^2}=1\]

  8. UsukiDoll
    • one year ago
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    before we start graphing, we want the center. Recall that the standard ellipse equation is \[\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1 \] For this equation we just have \[x^2 \] and \[y^2 \] \[\frac{x^2-0}{2^2}+\frac{y^2-0}{1^2}=1\] which means that our center \[(h,k) \rightarrow (0,0)\]

  9. UsukiDoll
    • one year ago
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    |dw:1437650242909:dw|

  10. UsukiDoll
    • one year ago
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    for vertices we need (-a,0), (a,0), (-b,0),(0,b) since a = 2 and b = 1 our vertex points are (-2,0) (2,0) (-1,0)(0,1)

  11. UsukiDoll
    • one year ago
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    |dw:1437650445276:dw|

  12. UsukiDoll
    • one year ago
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    Foci : \[a^2-c^2=b^2 \] since a = 2 and b = 1 \[2^2-c^2=1^2 \] \[4-c^2=1 \] \[-c^2=1-4 \] \[-c^2=-3 \] \[c^2=3 \] \[\sqrt{c^2}=\sqrt{3} \] \[c=\sqrt{3} \]

  13. anonymous
    • one year ago
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    lol nm i never saw that someone replied.

  14. anonymous
    • one year ago
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    @phi can you actually help a little more on explaining this a bit better

  15. phi
    • one year ago
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    I would hope that is enough info to answer the question

  16. phi
    • one year ago
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    try reading through the posts. 1) can you figure out the center? 2) the vertices?

  17. UsukiDoll
    • one year ago
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    don't forget the foci

  18. anonymous
    • one year ago
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    yeah hold on

  19. UsukiDoll
    • one year ago
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    alright. :)

  20. anonymous
    • one year ago
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    the center is (0,0) right?

  21. UsukiDoll
    • one year ago
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    yes

  22. anonymous
    • one year ago
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    for the foci: 10^2 - 8^2 = b^2 100 - 64 = b^2 36 = b^2 b = 6, so the foci is (-6,0) and (6,0)?

  23. anonymous
    • one year ago
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    how's that?

  24. UsukiDoll
    • one year ago
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    errr. \[a^2-c^2=b^2 \] we are looking for c a = 10 and b = 6

  25. anonymous
    • one year ago
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    oops, so: 100 - c^2 = 36 so c is 8?

  26. UsukiDoll
    • one year ago
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    but the vertices for b is (-6,0) (6,0)

  27. UsukiDoll
    • one year ago
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    we need vertices for a though (-a,0), (a,0)

  28. UsukiDoll
    • one year ago
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    yes c is 8

  29. anonymous
    • one year ago
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    vertices for a = (-10,0) and (10,0)

  30. UsukiDoll
    • one year ago
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    yes. now we can use that information to graph. though...if we draw it with os tools it's gonna look ugly. I'll attach a file

  31. UsukiDoll
    • one year ago
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  32. anonymous
    • one year ago
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    ok so is this right? all together: center: (0,0) vertices: (-10,0); (10,0) foci: (-6,0); (6,0)

  33. UsukiDoll
    • one year ago
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    yeah

  34. anonymous
    • one year ago
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    awesome thanks, help on another one?

  35. UsukiDoll
    • one year ago
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    sure

  36. anonymous
    • one year ago
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    Find the center, vertices, and foci of the ellipse with equation 3x^2 + 6y^2 = 18

  37. UsukiDoll
    • one year ago
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    alright... we don't have the ellipse in standard form. so we need to divide 18 all over the equation

  38. anonymous
    • one year ago
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    ok so...: (3/18)x^2 + (6/18)y^2 that looks weird

  39. UsukiDoll
    • one year ago
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    yeah but we need a 1 and dividing 18 all over the equation makes that =1 happen

  40. anonymous
    • one year ago
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    oh so i did it right?

  41. UsukiDoll
    • one year ago
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    yeah

  42. anonymous
    • one year ago
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    so now what

  43. UsukiDoll
    • one year ago
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    I'm just trying to think what's going on... I don't have perfect squares in here.

  44. anonymous
    • one year ago
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    ok

  45. UsukiDoll
    • one year ago
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    find the center.. that's the easiest part >_<

  46. anonymous
    • one year ago
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    actually not sure how, i know it is (h,k) but where is that in our equation

  47. anonymous
    • one year ago
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    is it (0,0) again. i went on wolphram alpha and it simplifies the equation to \[\frac{ x^2 }{ 6 } + \frac{ y^2 }{ 3 } = 1\]

  48. UsukiDoll
    • one year ago
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    If our equation was similar to \[7x^2 +16(y+2)^2=112\] and then divide 112 all over then our center would be (0,-2)

  49. anonymous
    • one year ago
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    ok actually, looking at my answer choices, the center can only be (0,0) lol so moving on to the vertices.

  50. UsukiDoll
    • one year ago
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    but when we reduced that fraction from earlier \[\frac{ x^2 }{ 6 } + \frac{ y^2 }{ 3 } = 1 \] there is no k and no h so center is (0,0) the vertices is next

  51. anonymous
    • one year ago
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    ok

  52. UsukiDoll
    • one year ago
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    and I just figured it out in my noggin we have square roots this time

  53. UsukiDoll
    • one year ago
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    \[\frac{ x^2 }{ 6 } + \frac{ y^2 }{ 3 } = 1 \] \[\frac{ x^2 }{ (\sqrt{6})^2 } + \frac{ y^2 }{ (\sqrt{3})^2 } = 1\]

  54. anonymous
    • one year ago
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    so i got vertices: \[(-\sqrt6,0); (\sqrt6,0)\] foci: \[(-\sqrt3,0); (\sqrt3,0)\]

  55. UsukiDoll
    • one year ago
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    yes!

  56. anonymous
    • one year ago
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    is that right?

  57. anonymous
    • one year ago
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    yay!

  58. anonymous
    • one year ago
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    one last one i have to work backwards on i thin. can you help?

  59. UsukiDoll
    • one year ago
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    I guess x.x ok y not

  60. anonymous
    • one year ago
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    Find an equation in standard form for the ellipse with the vertical major axis of length 18, and minor axis of length 16. No clue how to work this one. Sorry for being such a pain...

  61. UsukiDoll
    • one year ago
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    vertical major axis -> 18 minor axis of length -> 16

  62. UsukiDoll
    • one year ago
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    minor would be the foci and the major would be the vertices since we have vertical major axis, our equation should be in the form of \[\frac{(y-h)^2}{a^2}+\frac{(x-k)^2}{b^2}=1 \]

  63. anonymous
    • one year ago
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    ok so 18 ^2 for a^2? and 16 ^2 for b^2

  64. UsukiDoll
    • one year ago
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    yeah.....

  65. anonymous
    • one year ago
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    so how do I get the equation plug in these numbers so I get: \[(\frac{ x^2 }{ 18 })+(\frac{ y^2 }{ 16}) = 1\]is that right?

  66. UsukiDoll
    • one year ago
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    I haven't done this in a while, but these sites are confusing me... this vertical formula is switched \[\frac{(x-k)^2}{b^2}+\frac{(y-h)^2}{a^2}=1\]

  67. UsukiDoll
    • one year ago
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    our center is still (0,0) that's one good part

  68. anonymous
    • one year ago
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    ok so we are going to have x^2 and y^2 by themselves on the top but what about the bottom.

  69. UsukiDoll
    • one year ago
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    vertical major axis -> 18 minor axis of length -> 16

  70. UsukiDoll
    • one year ago
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    a = length of semi-major axis b = length of semi-minor axis vertices: (h, k + a), (h, k - a) co-vertices: (h + b, k), (h - b, k) [endpoints of the minor axis]

  71. UsukiDoll
    • one year ago
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    now I'm confusing myself.. but .. a =18 and b = 16

  72. UsukiDoll
    • one year ago
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    yeah if the number under the y part is larger then it's going to be vertical

  73. UsukiDoll
    • one year ago
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    so a has to be 18 and b has to be 16

  74. UsukiDoll
    • one year ago
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    but ... we're still not done we have to write those numbers to the second power

  75. UsukiDoll
    • one year ago
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    \[\frac{(x-k)^2}{b^2}+\frac{(y-h)^2}{a^2}=1 \]

  76. anonymous
    • one year ago
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    (x^2/16^2) + (y^2/18^2) = 1?

  77. UsukiDoll
    • one year ago
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    yup I think you may have to expand like 16 x 16 and 18 x 18

  78. anonymous
    • one year ago
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    do you want me to give the answer choices? doesn't look like any of them are working.

  79. UsukiDoll
    • one year ago
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    yeah what are they

  80. anonymous
    • one year ago
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    \[\frac{ x^2 }{64 }+\frac{ y^2 }{ 81 } = 1\] \[\frac{ x^2 }{ 9 }+\frac{ y^2 }{ 8 } = 1\] \[\frac{ x^2 }{ 81 }+\frac{ y^2 }{ 64 } = 1\] \[\frac{ x^2 }{ 8 }+\frac{ y^2 }{ 9 } = 1\]

  81. anonymous
    • one year ago
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    that took a while lol. so how do we get one of those.

  82. UsukiDoll
    • one year ago
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    what the heck that doesn't make sense.

  83. anonymous
    • one year ago
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    did we do anything wrong?

  84. UsukiDoll
    • one year ago
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    on one choice that would mean a = 9, b = 8 and on the other a = 8 and b = 9 last choice won't work

  85. UsukiDoll
    • one year ago
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    second choice won't work either.

  86. UsukiDoll
    • one year ago
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    did you copy the right numbers?

  87. anonymous
    • one year ago
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    yep

  88. anonymous
    • one year ago
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    that's why i'm confused

  89. UsukiDoll
    • one year ago
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    then that would mean that you were given the wrong numbers

  90. anonymous
    • one year ago
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    here i will get phi back in here maybe: @phi

  91. UsukiDoll
    • one year ago
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    @Astrophysics help!

  92. Astrophysics
    • one year ago
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    What's up?

  93. UsukiDoll
    • one year ago
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    this doesn't match... I mean @twistnflip tell her about the last problem

  94. anonymous
    • one year ago
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    ok so Find an equation in standard form for the ellipse with the vertical major axis of length 18, and minor axis of length 16. But we have done the work and the answers don't match. I posted them with the equation editor a few comments ago. What do we do?

  95. phi
    • one year ago
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    remember that a and b are the "semi-major" and "semi-minor" axes. (comparable to the radius) in other words, divide the given axes lengths by 2

  96. UsukiDoll
    • one year ago
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    *facepalm* missed that part.

  97. phi
    • one year ago
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    thus you want a=9 and b=8

  98. UsukiDoll
    • one year ago
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    k... I see it now 18/2 = 9 16/2 = 8

  99. UsukiDoll
    • one year ago
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    and since it's vertical that means the denominator in the y portion has to be larger

  100. anonymous
    • one year ago
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    so i get the equation: \[\frac{ x^2 }{ 8} + \frac{ y^2 }{ 9 }\]?

  101. Astrophysics
    • one year ago
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    Haha, well here is a good reference on the standard form of an ellipse, and will help you from not making these kinds of mistakes :P |dw:1437655357843:dw|

  102. UsukiDoll
    • one year ago
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    getting there. there's still one more step

  103. anonymous
    • one year ago
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    which is?

  104. phi
    • one year ago
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    you want a^2 and b^2 in your equation (not just a and b)

  105. Astrophysics
    • one year ago
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    |dw:1437655407458:dw|

  106. UsukiDoll
    • one year ago
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    since a = 9 and that formula has a^2 9^2 = ?

  107. UsukiDoll
    • one year ago
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    umm @Astrophysics it's vertical

  108. UsukiDoll
    • one year ago
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    \(\color{#0cbb34}{\text{Originally Posted by}}\) @twistnflip ok so Find an equation in standard form for the ellipse with the vertical major axis of length 18, and minor axis of length 16. But we have done the work and the answers don't match. I posted them with the equation editor a few comments ago. What do we do? \(\color{#0cbb34}{\text{End of Quote}}\) vertical major axis

  109. Astrophysics
    • one year ago
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    I know, I just put that to show it's squared, didn't really pay attention to the vertical/ horizontal :P

  110. anonymous
    • one year ago
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    81 and 8 ^2 = 64 so i have \[\frac{ x^2 }{ 64} + \frac{ y^2 }{81 } = 1\]

  111. UsukiDoll
    • one year ago
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    that's better.

  112. anonymous
    • one year ago
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    so that's it?

  113. phi
    • one year ago
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    you want it too be harder?

  114. anonymous
    • one year ago
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    haha no thanks so much guys! you da bomb

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