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anonymous

  • 5 years ago

find the volume of the region bounded by the graphs y = sqrt(x), y = 0, and x = 6, revolved about the line x = 6

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  1. amistre64
    • 5 years ago
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    :) again :)

  2. anonymous
    • 5 years ago
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    hello! This problem is similar to the one we did together yesterday, but for some reason I am still doing this wrong

  3. anonymous
    • 5 years ago
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    I know that I am supposed to do the disk method, and that I need to solve for each equation in terms of y

  4. amistre64
    • 5 years ago
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    rewrite y= sqrt(x) in terms of x = f(y) y = sqrt(x) ; ^2 both side y^2 = x this we can solve for easily right?

  5. anonymous
    • 5 years ago
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    so I have v = pi (the integral from 0 to 6) (y+9)^2 - (9)^2 dx

  6. amistre64
    • 5 years ago
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    graph this and see that it is the same

  7. anonymous
    • 5 years ago
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    yes

  8. amistre64
    • 5 years ago
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    x = 6 is a pain, we want to move it to the x=0 line to examine it; how do we move x = 6 to x=0? x= 6 ; -6 from both sides x-6 = 0 do the same for the x=y^2 to keep it in the same position

  9. anonymous
    • 5 years ago
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    minus 6 from both sides in the equation I wrote?

  10. amistre64
    • 5 years ago
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    x = y^2 ; -6 both sides x-6 = y^2 -6

  11. anonymous
    • 5 years ago
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    whoops---I am revolving around x = 9, sorry about that

  12. amistre64
    • 5 years ago
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    brb..

  13. anonymous
    • 5 years ago
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    okay

  14. amistre64
    • 5 years ago
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    back...

  15. anonymous
    • 5 years ago
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    hi

  16. amistre64
    • 5 years ago
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    since it 9 we just -9 fro itall

  17. anonymous
    • 5 years ago
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    so v = pi (intergral from 0 to 6) (y + 9)^2 - 9^2 ?

  18. amistre64
    • 5 years ago
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    move it all to the left by 9 to examine it at x=0; does that make sense? Its over on a shelf and we want to examine it at the table where we can see it clearly; the table is the origin of the graph where x=0 and y=0, so we move it

  19. anonymous
    • 5 years ago
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    So now I am looking at a graph that is shifted backwards by 9 on my calculator

  20. amistre64
    • 5 years ago
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    y = sqrt(x) ; change to an equal form to accomodate for the y axis y^2 = x ; now move it over to the x=0 by subtracting 9 y^2 -9 = x-9 ; the x-9 can be rewritten as just x

  21. anonymous
    • 5 years ago
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    sqrt(x +9)

  22. anonymous
    • 5 years ago
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    okay so it is y^2 -9 not +9

  23. amistre64
    • 5 years ago
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    sqrt(x+9) is good; now reform it y = sqrt(x+9) y^2 = x+9 y^2 -9 = x

  24. amistre64
    • 5 years ago
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    whats our bounds now: for y=0 to how far?

  25. anonymous
    • 5 years ago
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    x = 6....?

  26. anonymous
    • 5 years ago
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    I've never seen anyone do transformations of a graph this way..

  27. amistre64
    • 5 years ago
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    y = sqrt(9) y = 3

  28. anonymous
    • 5 years ago
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    we have to redo them when we solve for a different variable...keep forgetting

  29. amistre64
    • 5 years ago
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    integrate the discs from 0 to 3

  30. anonymous
    • 5 years ago
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    okay

  31. amistre64
    • 5 years ago
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    pi {S} y^2-9 dy ; [0,3] = volume of revolution

  32. amistre64
    • 5 years ago
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    forgot to square the function lol

  33. amistre64
    • 5 years ago
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    pi {S} [y^2-9]^2 dy ; [0,3]

  34. anonymous
    • 5 years ago
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    so you don't have to subtract another 9?

  35. amistre64
    • 5 years ago
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    why would you subtract two 9s?

  36. amistre64
    • 5 years ago
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    we only move the graph over to x=0 from x=9 right?

  37. anonymous
    • 5 years ago
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    one from the first eqn and one from eqn y = 0?

  38. amistre64
    • 5 years ago
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    if i tellyou to move 9 steps to the left; what do you do? move 18?

  39. anonymous
    • 5 years ago
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    ohhhh

  40. amistre64
    • 5 years ago
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    y=0 is already bounding the volume; no need to adjust that part lol

  41. anonymous
    • 5 years ago
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    answer is 648/5?

  42. amistre64
    • 5 years ago
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    dunno; lets check..

  43. anonymous
    • 5 years ago
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    no, its' not

  44. amistre64
    • 5 years ago
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    pi {S} (y^2-9)^2 dy pi {S} (y^4 -18y^2 +81) dy pi(y^5/5 -18y^3/3 +81y) at y=3

  45. anonymous
    • 5 years ago
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    had that part

  46. anonymous
    • 5 years ago
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    and got 648/5, but the homework grader marked it wrong

  47. amistre64
    • 5 years ago
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    i get 129.6pi

  48. anonymous
    • 5 years ago
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    same answer, and the wrong one. What did we do wrong?

  49. anonymous
    • 5 years ago
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    Hang on. You're rotating the graph of y= sqrt(x) from x=6 to x=9 about the line x=9 ? Is that right?

  50. amistre64
    • 5 years ago
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    243 27 243 pi( ---- - --- + ---) right? 5 3 1

  51. amistre64
    • 5 years ago
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    x = 0 to 9

  52. anonymous
    • 5 years ago
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    right, and we are rotating the region about the line x = 9

  53. amistre64
    • 5 years ago
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    ilove; retype the problem in its entirety so that we have a clear picture

  54. anonymous
    • 5 years ago
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    Find the volumes of the solids generated by revolving the regions bounded by the graphs of the equations about the given lines. y = sqrt(x) y = 0 x = 6 about the line x = 9

  55. amistre64
    • 5 years ago
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    ohhh... so its a donut shape; i assumed you meant y=0; x=9 about the x=9

  56. anonymous
    • 5 years ago
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    oh, nope sorry about that

  57. anonymous
    • 5 years ago
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    I don't think it's a doughnut

  58. amistre64
    • 5 years ago
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    its a torus alright :)

  59. anonymous
    • 5 years ago
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    More like a mountain.

  60. amistre64
    • 5 years ago
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    we have to subtract the middle of what we found to get the volume then

  61. anonymous
    • 5 years ago
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    oh wait. I'm looking at wrong graph

  62. anonymous
    • 5 years ago
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    Small hill. Not a doughnut. There is no hole in the middle.

  63. amistre64
    • 5 years ago
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    x=6 to x=9 is empty tho

  64. anonymous
    • 5 years ago
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    there is no hole it the middle? isn't the shape a bunt cake?

  65. anonymous
    • 5 years ago
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    no

  66. anonymous
    • 5 years ago
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    graph y = sqrt(x) from x = 6 to x=9 y = 0 to y = 3

  67. anonymous
    • 5 years ago
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    but the region is cut off at x = 6, and we have to get it to revolve about line x = 9

  68. anonymous
    • 5 years ago
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    that's where it starts

  69. anonymous
    • 5 years ago
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    not where it ends. it doesn't say it's bounded by x=0

  70. anonymous
    • 5 years ago
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    hmmm...then how to we caluclate where it begins and ends?

  71. anonymous
    • 5 years ago
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    http://www.wolframalpha.com/input/?i=graph+y+%3D+sqrt%28x%29+from+x+%3D+6+to+x%3D9+y+%3D+0+to+y+%3D+3 Rotate that around x=9

  72. amistre64
    • 5 years ago
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  73. anonymous
    • 5 years ago
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    If the problem said bounded by x=0, x=6 and y= 0 and y = sqrt(x) I would agree with that picture. It doesn't say x=0.

  74. anonymous
    • 5 years ago
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    so using what you just said, the boundries would be?

  75. anonymous
    • 5 years ago
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    x = 6, x = 9, y = 0, y = sqrt(x) and rotate around x = 9

  76. amistre64
    • 5 years ago
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    then we would double our results right?

  77. anonymous
    • 5 years ago
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    no, you have the wrong part of the graph I think.

  78. amistre64
    • 5 years ago
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    with this boundary; y = [0,sqrt(6)]

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  79. anonymous
    • 5 years ago
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    You want to integrate over the white part in your picture rather than the grey part.

  80. anonymous
    • 5 years ago
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    what wouuld the equation look like?

  81. anonymous
    • 5 years ago
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    \[v = \pi \int\limits_{0}^{3}...\]

  82. anonymous
    • 5 years ago
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    You're using disks or shells?

  83. anonymous
    • 5 years ago
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    We can only use the disk method, right?

  84. anonymous
    • 5 years ago
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    where did you get the boundaries 0 to 3 from , again?

  85. anonymous
    • 5 years ago
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    do you have the value for the answer so we can be sure we're doing the right region?

  86. anonymous
    • 5 years ago
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    nope----online h/w problem

  87. anonymous
    • 5 years ago
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    dumb.

  88. anonymous
    • 5 years ago
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    I'm pretty sure I'm reading it right though. For this I would use cylindrical shells

  89. amistre64
    • 5 years ago
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    y = sqrt(x); curve of sqrt(x) y = 0 ; the x axis is y=0 x = 6 ; the left boundary x=9; axis to spin around

  90. amistre64
    • 5 years ago
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    x=6 the right bound

  91. anonymous
    • 5 years ago
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    x = 6 is the left bound in my interpretation of the problem, and the axis of rotation would be the right bound.

  92. anonymous
    • 5 years ago
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    ???? what do you get as your answer?

  93. anonymous
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    I thought the entire region was left bound, and the axis is right bound.

  94. anonymous
    • 5 years ago
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    like left-right or top-bottom

  95. amistre64
    • 5 years ago
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    the region to cut from my original; if im right is: is the cylindar of r = 3, and height = sqrt(6) 216 sqrt(6) pi ------------ 5

  96. anonymous
    • 5 years ago
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    okay, this is the right answer. Can you go over it one more time?

  97. amistre64
    • 5 years ago
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    which answer is correct? :)

  98. anonymous
    • 5 years ago
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    So it was the doughnut?

  99. anonymous
    • 5 years ago
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    both are correct, but how do you do it correctly?

  100. anonymous
    • 5 years ago
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    They should have been more specific about the left boundary... grr..

  101. amistre64
    • 5 years ago
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    lol..... the only way we are both correct is if the volume of each version are identical :)

  102. anonymous
    • 5 years ago
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    I have not subtracted by 9sqrt(6) pi at all in what I have been doing though

  103. anonymous
    • 5 years ago
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    okay, so i should have\[\pi \int\limits_{0}^{3}(....?....)\]

  104. amistre64
    • 5 years ago
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  105. amistre64
    • 5 years ago
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    your interval of [0,3] is wrong, since we only need the the sqrt(6)

  106. anonymous
    • 5 years ago
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    how do I know when to remove a cylinder?

  107. amistre64
    • 5 years ago
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    when the spin axis is not the same is the boundary axis; there is a gap that is formed

  108. anonymous
    • 5 years ago
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    So you've confirmed the answer is what amistre has?

  109. amistre64
    • 5 years ago
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    that gap is not a part of the volume when revolved so it must be cut out; like makeing a solid donut then removing the middle

  110. anonymous
    • 5 years ago
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    it's what both of you said. Now I am trying to determine what is the best way to approach this problem

  111. anonymous
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    So I remove the middle when the region is bounded off by a number x = # that is smaller that the x = # that it is rotating about?

  112. anonymous
    • 5 years ago
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    it's not what I said if it's what he said. I have a very different answer.

  113. amistre64
    • 5 years ago
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  114. amistre64
    • 5 years ago
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    when the spin is different from the bound, you get an offset that needs to be accounted for....

  115. anonymous
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    .....

  116. anonymous
    • 5 years ago
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    polpak, with your graph, how did you attempt the problem?

  117. anonymous
    • 5 years ago
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    did you eliminate the middle part in the volume equation?

  118. amistre64
    • 5 years ago
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    polpak eliminated the outer part

  119. anonymous
    • 5 years ago
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    can I get this step by step? I am totally confused. I need to see an equation

  120. anonymous
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    well we're still not clear on what problem you're solving ;p

  121. amistre64
    • 5 years ago
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    for mine: we did the stuff above and then I subtracted the middle in an equation like this: \[\pi.\int\limits_{0}^{\sqrt{6}} (y^4 -18y^2+81) - (-3)^2.dy\]

  122. anonymous
    • 5 years ago
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    Okay,that i what I was looking for. I had every thing except the -3. where did that come from?

  123. amistre64
    • 5 years ago
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    that is the y = -3 line that result from the boundary between x=6 and x=9

  124. amistre64
    • 5 years ago
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    teh 3 and 6 in my circle graph should be switched :) and then you can see the "3"

  125. anonymous
    • 5 years ago
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    OHHHHHHH

  126. anonymous
    • 5 years ago
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    okay, so now I finally get it. Thanks everyone! :D Thanks for staying on so long amistre64

  127. amistre64
    • 5 years ago
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    hey, if you dont figure it out, neither do I lol

  128. anonymous
    • 5 years ago
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    see ya later!

  129. anonymous
    • 5 years ago
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    I thought it was this

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  130. anonymous
    • 5 years ago
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    and still do for that matter, but I could be wrong.

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