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anonymous

  • 5 years ago

Could someone tell me if my answer was correct? I will attach the equation and available answers in a comment! Its just a algebra question but im evidently unable to focus rite now >.<

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  1. anonymous
    • 5 years ago
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    here :)

  2. anonymous
    • 5 years ago
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    Which one is the question?

  3. anonymous
    • 5 years ago
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    the one that says "ok" ^_^

  4. anonymous
    • 5 years ago
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    I see.

  5. anonymous
    • 5 years ago
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    The process is confusing me im pretty sure this is one of the lessons that i missed :P

  6. anonymous
    • 5 years ago
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    So the question is to simplify: \[{4 \over m+2}+{2 \over m-2}?\]

  7. anonymous
    • 5 years ago
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    The m + 2 and m - 2 is confusing me :P And it says to Add and Simplify.

  8. anonymous
    • 5 years ago
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    Good. First, what is the LCM of (m+2) and (m-2).

  9. anonymous
    • 5 years ago
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    Thats the part that was confusing me ^_^ lol

  10. anonymous
    • 5 years ago
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    Haha I see. Do (m+2) and (m-2) have any common factors?

  11. anonymous
    • 5 years ago
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    Yep lol >.< i think urghh i should of paid attention :P and i just didnt know how to write it bc of the - and + signs >.< lol

  12. anonymous
    • 5 years ago
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    Well they don't have a common factor, and because of that the LCM of them is their multiplication.

  13. anonymous
    • 5 years ago
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    Ahhhhhh alright i get it so far. ^_^

  14. anonymous
    • 5 years ago
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    so (m+2)(m-2)=m^2-4. That's would be our LCM.

  15. anonymous
    • 5 years ago
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    That's good. Now, we have to apply this to our equation.

  16. anonymous
    • 5 years ago
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    expression*

  17. anonymous
    • 5 years ago
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    ohhhh haha wow ya bc since they dont have a common factors u'd multiply them together. alright.

  18. anonymous
    • 5 years ago
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    Exactly.

  19. anonymous
    • 5 years ago
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    That means, we will multiply the first term (top and bottom) by (m-2). and multiply the second term (top and bottom by (m+2).

  20. anonymous
    • 5 years ago
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    Does that make sense?

  21. anonymous
    • 5 years ago
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    Yes so far so good so then the answer would be the "ya" one? lol like my labels? :D

  22. anonymous
    • 5 years ago
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    >.< i think... :P

  23. anonymous
    • 5 years ago
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    \[{4 \over m+2}+{2 \over m-2}={4(m-2) \over m^2-4}+{2(m+2) \over m^2-4}={4m-8+2m+4 \over m^2-4}={6m-4 \over m^2-4}\]

  24. anonymous
    • 5 years ago
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    OR as in the choices: \[{6m-4 \over (m+2)(m-2)}\]

  25. anonymous
    • 5 years ago
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    WOOOOOOOOOOOOO!!!! so i was rite huh?!?! YAY YAY YAY!!

  26. anonymous
    • 5 years ago
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    lol ^_^

  27. anonymous
    • 5 years ago
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    Haha Yeah you were :)

  28. anonymous
    • 5 years ago
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    :) Thanks, helped my bunches! :D

  29. anonymous
    • 5 years ago
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    np

  30. anonymous
    • 5 years ago
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    Oh! and do you think you could help me on part of this other question? :) plesss? lol

  31. anonymous
    • 5 years ago
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    This is it. do u think u will help meh? :)

  32. anonymous
    • 5 years ago
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    so ok is the original expression?

  33. anonymous
    • 5 years ago
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    You make strange names? :P

  34. anonymous
    • 5 years ago
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    My question is about the denominators again so would there be common factors? >.< i got confused again :P lol and ya i am making strange names ^^

  35. anonymous
    • 5 years ago
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    The same first question again, do they have common factors?

  36. anonymous
    • 5 years ago
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    And yea this is the original expression .. pretty sure lol

  37. anonymous
    • 5 years ago
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    and i thought they did >.< like since 3x can go into 6x.. but then i got lost arghh

  38. anonymous
    • 5 years ago
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    Okay can you factor any of the two denominators?

  39. anonymous
    • 5 years ago
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    No i dont think so

  40. anonymous
    • 5 years ago
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    Wait which version of factoring? >.<

  41. anonymous
    • 5 years ago
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    3x+5 can't be factored.. (6x+10)=2(3x+5).. right?

  42. anonymous
    • 5 years ago
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    Right.

  43. anonymous
    • 5 years ago
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    Okay now your LCD should be each one of the factor to be taken only one. the factors we have are (3x+5) and 2. so our LCD would 2(3x+5).

  44. anonymous
    • 5 years ago
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    I don't think that was clear enough, was it?

  45. anonymous
    • 5 years ago
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    Not really im sitting here repeating it over in my head going wth.... ^_^

  46. anonymous
    • 5 years ago
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    LOL

  47. anonymous
    • 5 years ago
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    Well, you can see that (3x+5) occur in both denominators. right?

  48. anonymous
    • 5 years ago
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    after factorization*

  49. anonymous
    • 5 years ago
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    Yep i see that.

  50. anonymous
    • 5 years ago
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    \[{2 \over 3x+5}-{1 \over 6x+10}={2 \over 3x+5}-{1 \over 2(3x+10)}\] Right?

  51. anonymous
    • 5 years ago
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    Now list all factors in both denominators.

  52. anonymous
    • 5 years ago
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    I mean 2(3x+5) :)

  53. anonymous
    • 5 years ago
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    Ahhhh so then i see why you would do that. But couldnt you just multiply the 3x + 5 expression by 2 and then subtract?

  54. anonymous
    • 5 years ago
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    the two equations with the same denominator i mean ^^

  55. anonymous
    • 5 years ago
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    Yes you could, but then you have to simplify after that.

  56. anonymous
    • 5 years ago
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    that wouldn't be the simplest form.

  57. anonymous
    • 5 years ago
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    Ahhh rite, ^_^ ok then go on lol

  58. anonymous
    • 5 years ago
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    Well you're saying that you multiply the (3x+5) by 2, that's right. And that's what I was trying to say when I said that 2(3x+5) is the LCD. So then we keep the 6x+10 and multiply the (3x+5) term (top and bottom) by 2.

  59. anonymous
    • 5 years ago
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    :)

  60. anonymous
    • 5 years ago
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    ohh so the answer would be B haha wow i rly need to stop doubting myself argh :P

  61. anonymous
    • 5 years ago
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    LOL thanks for the help! AGAIN ^_^

  62. anonymous
    • 5 years ago
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    But you know why we can do that. That's because the denominator of the second term is nothing but 2 times the denominator of the first term.

  63. Yuki
    • 5 years ago
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    AnwarA, how did you make the rational expression look like that ?

  64. anonymous
    • 5 years ago
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    It's 2(3x+5).. just typo :P

  65. Yuki
    • 5 years ago
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    all I can do is\[{a+bi}/{10a-c}\]

  66. Yuki
    • 5 years ago
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    how did you make it go under a long bar ?

  67. anonymous
    • 5 years ago
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    lol think person means like how do u ya long bar ^^

  68. anonymous
    • 5 years ago
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    Oh.. {a+bi over 10a=c}

  69. anonymous
    • 5 years ago
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    And hello yuki.

  70. anonymous
    • 5 years ago
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    :)

  71. anonymous
    • 5 years ago
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    >.< i will never understand computer tech/ math peoples smartness :P

  72. anonymous
    • 5 years ago
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    haha. I think I am the only one who does it this way :P

  73. Yuki
    • 5 years ago
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    \[\sqrt(b^2-4ac) \over 2a\]

  74. Yuki
    • 5 years ago
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    that is sooo cooooool !! I am goint to use it from now on, thanks!

  75. anonymous
    • 5 years ago
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    LOL obviously not anymore!! :D and i see someone ik viewing the question but he isnt in our meeting place!! >:P lol (inside joke cant rly explain it ^^)

  76. anonymous
    • 5 years ago
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    \[{\sqrt{b^2-4ac} \over 2a} \] :P

  77. Yuki
    • 5 years ago
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    can you do "plus or minus" as well ?

  78. anonymous
    • 5 years ago
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    Haha

  79. anonymous
    • 5 years ago
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    >.< smart people arg lol

  80. anonymous
    • 5 years ago
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    Yep anything

  81. anonymous
    • 5 years ago
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    \[{-b \pm \sqrt{b^2-4ac} \over 2a}\]

  82. Yuki
    • 5 years ago
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    how did you make the sqrt top part long ?! I have so much to learn lol

  83. anonymous
    • 5 years ago
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    Haha try that yourself :P

  84. Yuki
    • 5 years ago
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    is there a place that have all of that written down ?

  85. anonymous
    • 5 years ago
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    *sigh* so much to learn to little time lmao im jk :)

  86. anonymous
    • 5 years ago
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    just use {}, with anything that you want to make long :)

  87. Yuki
    • 5 years ago
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    \[\sqrt{10x+b}\]

  88. anonymous
    • 5 years ago
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    Hah, i think its funnay how i have no idea how the heck u guys are doing that ^^

  89. anonymous
    • 5 years ago
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    \[e^{ANWAR}={1 \over \sqrt{yuki}^{dolly}}\] :P that does not mean anything by the way

  90. Yuki
    • 5 years ago
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    \[\int\limits _{a,b}_{b,a}\]

  91. anonymous
    • 5 years ago
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    lol im above yuki hee hee!!!!

  92. anonymous
    • 5 years ago
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    I am in the top, that's what matters :D

  93. Yuki
    • 5 years ago
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    \[\ln(\ln (x)) = 0\]

  94. anonymous
    • 5 years ago
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    :P looks more like im the glue that holds everything together mwuahahaha!!!

  95. anonymous
    • 5 years ago
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    take ln of both sides to bring me down :(

  96. Yuki
    • 5 years ago
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    \[+,- \pm \]

  97. Yuki
    • 5 years ago
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    Ha ! found it lo

  98. anonymous
    • 5 years ago
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    You're learning :P

  99. Yuki
    • 5 years ago
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    \[\cos(a \pm b) = \cos(a)\cos(b) \mp \sin(a)\sin(b)\]

  100. anonymous
    • 5 years ago
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    im not ^_^ LOL

  101. anonymous
    • 5 years ago
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    You want to?

  102. anonymous
    • 5 years ago
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    But you can't use your weird names there :P

  103. anonymous
    • 5 years ago
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    yuki, whenever you use these things, rmr to thank me first

  104. anonymous
    • 5 years ago
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    awww yep sure the person i am waiting for isnt talking so sure :) lol

  105. Yuki
    • 5 years ago
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    \[\int\limits_{a,b \int\limits_{1}^{2}}\]

  106. anonymous
    • 5 years ago
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    Haha

  107. anonymous
    • 5 years ago
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    whom are you waiting for? xD

  108. Yuki
    • 5 years ago
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    I definitely will AnwarA

  109. anonymous
    • 5 years ago
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    his name starts with an I and he is rlyyyy sweet and hot and all tht jaz oh and did i mention it was u ? xD lmao

  110. anonymous
    • 5 years ago
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    haha anwar is all like wat the hell.... LOL

  111. anonymous
    • 5 years ago
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    Thanks buddy!! I was just kidding. I did nothing, anyone who knows that would help do that. :)

  112. anonymous
    • 5 years ago
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    @dolly: LOL

  113. Yuki
    • 5 years ago
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    \[{a+bi \over a-bi} *{a+bi \over a-bi}\]

  114. anonymous
    • 5 years ago
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    ^_^

  115. anonymous
    • 5 years ago
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    why do you think I was "all like what the hell"? :)

  116. anonymous
    • 5 years ago
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    haha to what i said lol jeezus anwar pay attention!! >:D im jk :)

  117. anonymous
    • 5 years ago
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    Haha, I read that and I wasn't "all like what the hell" at all :P

  118. anonymous
    • 5 years ago
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    yuki, what grade are you at?

  119. anonymous
    • 5 years ago
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    LOL then i was wrong i thought u would be ^_^ lol and prolly college or high school senior!! lol

  120. anonymous
    • 5 years ago
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    I think wherever I need info about anyone in this site, I should just come straight to you. You got it all :O

  121. anonymous
    • 5 years ago
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    Do you work for the CIA?

  122. anonymous
    • 5 years ago
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    haha its easy i just notice the kind of math they know how to do, the way they type.. etc :) nm too it :D

  123. anonymous
    • 5 years ago
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    so you are not here to gather informations about us for some agencies?

  124. anonymous
    • 5 years ago
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    Uhh, no. lol That would only be extremely creepy lol

  125. anonymous
    • 5 years ago
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    :) Anyway, I am off to bed. Yuki would be more of help than I am, specially that he now knows how to do the equations thing :D.

  126. anonymous
    • 5 years ago
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    Bye all!!

  127. anonymous
    • 5 years ago
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    LOL okie dokie nite buddaaayyyy!!! :)

  128. Yuki
    • 5 years ago
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    sigma

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