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anonymous

  • 5 years ago

if a differential equations solution divides by x^2 and the initial condition is a point where x=0, what do you do? (problem: dy/dx=xyy or dy/dx=xy^2)

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  1. anonymous
    • 5 years ago
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    initial condition is (0,-2)

  2. anonymous
    • 5 years ago
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    You can use separation by variables here and then integrate. Once you have y after integration, you can use the initial condition.

  3. anonymous
    • 5 years ago
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    i did that, but my solution is y=(-2/x^2)+2 which doesnt technically do through the initial condition

  4. anonymous
    • 5 years ago
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    i think its a vertical asymptote which... definitely isint the point XD

  5. anonymous
    • 5 years ago
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    There's nothing wrong with that equation. After you put it into separable form and solve, you should get,\[\frac{x^2y}{2}+cy+1=0\]Initial condition,\[0-2c+1=0 \rightarrow c=\frac{1}{2}\]

  6. anonymous
    • 5 years ago
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    You can then factor out y and rearrange into the form y=y(x).

  7. anonymous
    • 5 years ago
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    brb

  8. anonymous
    • 5 years ago
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    i got c=2?

  9. anonymous
    • 5 years ago
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    When you separated the variable and tried to get y by itself did you include c and also multiply the y by it?

  10. anonymous
    • 5 years ago
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    .... -.-

  11. anonymous
    • 5 years ago
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    ima try that, how about: y'=1+yy

  12. anonymous
    • 5 years ago
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    You can do that problem the same way. I look at it as y' is dy/dx and then separate variables. Make sure when you do though, that the 1 goes with the y^2. So move the entire quantity.

  13. anonymous
    • 5 years ago
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    dang, the small things get by me... that should be something like arctany=x+c? so tan(x+c)=y?

  14. anonymous
    • 5 years ago
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    that explains why my condition is in radians

  15. anonymous
    • 5 years ago
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    Yes.

  16. anonymous
    • 5 years ago
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    alright... major confusion... \[y'=(e ^{x-y})\div(1+e ^{x})\]

  17. anonymous
    • 5 years ago
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    You want to solve that?

  18. anonymous
    • 5 years ago
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    sadly ><

  19. anonymous
    • 5 years ago
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    Just pull out the e^(-y) and it becomes separable ;)

  20. anonymous
    • 5 years ago
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    asoghaqoeraifoiafauirfhlanihusdfvnnfldanufd

  21. anonymous
    • 5 years ago
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    \[e^ydy=\frac{e^x}{1+e^x}dx\]

  22. anonymous
    • 5 years ago
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    differential equations are gay -.-

  23. anonymous
    • 5 years ago
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    lol

  24. anonymous
    • 5 years ago
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    dont tell me any more!!!

  25. anonymous
    • 5 years ago
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    tell you what? answers?

  26. anonymous
    • 5 years ago
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    yeah.. dont tell me any more lol

  27. anonymous
    • 5 years ago
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    no worries :D

  28. anonymous
    • 5 years ago
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    I think you have the best icon...if that helps...

  29. anonymous
    • 5 years ago
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    hahahaha it was annoyingly hard to make XD

  30. anonymous
    • 5 years ago
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    i have a pretty good feeling that your icon is not you... though there is a significant chance that i am wrong given by the circumstances

  31. anonymous
    • 5 years ago
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    it could be me, since i look like that

  32. anonymous
    • 5 years ago
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    lol, how did your frigging doughnut go?

  33. anonymous
    • 5 years ago
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    pretty good, i have a way of doing things when i need help: use the first sentence or line of the help, fail a thousand times before i use the second, so on and so forth. so i used quite a bit, but not all of your help in the process of doing the assignment, and in the end i was flowing through it quite nicely, i compared my results to yours and they were identical, i then compared my results to the assignment and apparently the last couple steps, which i had done by hand, i could take the integral with a calculator... and then later he gave us the hand written answer, so i felt proud that i did it without a calculator... but yeah your help is insanely apreciated

  34. anonymous
    • 5 years ago
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    alright im stuck at e^y=ln(1+e^x)+c initial condition is (0,1)

  35. anonymous
    • 5 years ago
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    Is that y supposed to be y' (y-prime)?

  36. anonymous
    • 5 years ago
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    nope, thats my solution

  37. anonymous
    • 5 years ago
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    to the above equation

  38. anonymous
    • 5 years ago
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    oh..

  39. anonymous
    • 5 years ago
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    Yeah. You're right. Leave it like that too.

  40. anonymous
    • 5 years ago
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    but i need to find c

  41. anonymous
    • 5 years ago
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    Oh, just plug your numbers in.

  42. anonymous
    • 5 years ago
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    \[e=\ln 2 +c \rightarrow c = e- \ln 2\]

  43. anonymous
    • 5 years ago
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    any way to do that without a calc?

  44. anonymous
    • 5 years ago
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    You mean simplify?

  45. anonymous
    • 5 years ago
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    No.

  46. anonymous
    • 5 years ago
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    What's wrong with Euler's number? :p

  47. anonymous
    • 5 years ago
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    ah god its 1... can you check these real quick?

  48. anonymous
    • 5 years ago
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    Check what?

  49. anonymous
    • 5 years ago
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    Do you ever sleep?

  50. anonymous
    • 5 years ago
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    1 Attachment
  51. anonymous
    • 5 years ago
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    not too much anymore lol

  52. anonymous
    • 5 years ago
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    I need paper...I've run out...hang on...

  53. anonymous
    • 5 years ago
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    but all my work is shown?

  54. anonymous
    • 5 years ago
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    I get something different for the first one. Just go to bed and I'll look at it and scan and you can look at it when you get up.

  55. anonymous
    • 5 years ago
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    do i have to multiply the c by 2 aswell?

  56. anonymous
    • 5 years ago
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    I'll do it here:

  57. anonymous
    • 5 years ago
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    my answer doesnt make sense anyways... when i find y' its all weird and quotienty

  58. anonymous
    • 5 years ago
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    \[\frac{dy}{dx}=xy^2 \rightarrow y^{-2}dy = x dx \rightarrow -y^{-1}=x+c\]

  59. anonymous
    • 5 years ago
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    Your boundary condition is (0,-2), so

  60. anonymous
    • 5 years ago
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    x+c or (1/2)xx+c?

  61. anonymous
    • 5 years ago
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    \[-2^{-1}=0+c \rightarrow c=-\frac{1}{2}\]

  62. anonymous
    • 5 years ago
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    So\[y^{-1}+x=\frac{1}{2} \rightarrow 2+2xy=y \rightarrow y(1-2x)=2 \rightarrow y=\frac{2}{1-2x}\]

  63. anonymous
    • 5 years ago
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    waitttt the integral of xdx is (1/2)xx+c though!

  64. anonymous
    • 5 years ago
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    Oh crap, I forgot the frigging x. Sorry Arman.

  65. anonymous
    • 5 years ago
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    nah dont be sorry, i should hit the sheets tho, i'll probably have more questions tomorrow lol

  66. anonymous
    • 5 years ago
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    I still get a different answer to you.

  67. anonymous
    • 5 years ago
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    basic differential equations (like these) arent usually part of calc II are they?

  68. anonymous
    • 5 years ago
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    Yours is close. I'll work through them - they won't take long - and scan and you can read it in the morning.

  69. anonymous
    • 5 years ago
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    different answer?

  70. anonymous
    • 5 years ago
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    I'm not sure about the American system.

  71. anonymous
    • 5 years ago
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    y=2/(1-x^2)

  72. anonymous
    • 5 years ago
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    did you use (0,-2) for the initial whatever?

  73. anonymous
    • 5 years ago
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    pellet, I wrote it wrong on my paper. I wrote (0,2). SORRY AGAIN :'(

  74. anonymous
    • 5 years ago
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    NOW I get what you got.

  75. anonymous
    • 5 years ago
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    >< nah man i hate when ppl tell me sorry

  76. anonymous
    • 5 years ago
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    lol XD

  77. anonymous
    • 5 years ago
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    my calc teacher, this morning, was like "oh flutter..." *covers mouth*... *whole class starts laughing caus we hear him say pellet sometimes but never the f word so it meant he was really screwed over XD

  78. anonymous
    • 5 years ago
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    lol, it's great when that happens. I got the same results as you for the other two. The *only* thing I would do differently is leave \[e-\log 2\]as is, and not use a numerical approximation (in the last one).

  79. anonymous
    • 5 years ago
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    thats probably better in the math world, where you need to be precise

  80. anonymous
    • 5 years ago
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    Yeah, exactly.

  81. anonymous
    • 5 years ago
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    is that prefered in college?

  82. anonymous
    • 5 years ago
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    preferred*

  83. anonymous
    • 5 years ago
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    Absolutely. You'd only use a numerical approximation for the time you actually need to extract a number. Other than that, we leave it because we may want to do further mathematics on it and we lose information/exactness if we start approximating beforehand.

  84. anonymous
    • 5 years ago
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    just like the AP exams lol. i think i asked you before but what are the remaining major math courses after differential eqas and linear algebra (when i say top i mean like... there should be less than 3 "top" ones)

  85. anonymous
    • 5 years ago
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    ?

  86. anonymous
    • 5 years ago
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    Yeah, I really don't know. Your system is different to the system in UK, Australia and Canada. We don't break mathematics up into sections like calculus and then apply a skill level to it. We apply a skill level to ALL of mathematics at particular points and you learn EVERYTHING at a particular level.

  87. anonymous
    • 5 years ago
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    oh... well what are the chances that i can attain a college grad level (in math) of mathematical understanding in the next year?

  88. anonymous
    • 5 years ago
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    given the rate that i covered calc 1 and 2 within 1.5 months

  89. anonymous
    • 5 years ago
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    I'd say you're pretty good.

  90. anonymous
    • 5 years ago
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    College grad level in calculus...is that what you're asking?

  91. anonymous
    • 5 years ago
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    You have to cover calculus of a complex variable as well. There's also tensor calculus.

  92. anonymous
    • 5 years ago
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    not just calculus, but all mathematics, can i attain that sort of knowledge, about 4 or more years of college math in the next 9 months?

  93. anonymous
    • 5 years ago
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    complex variable wont be too hard, what is tensor calc??

  94. anonymous
    • 5 years ago
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    It *might* take a little longer ;p

  95. anonymous
    • 5 years ago
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    They're devices we use to extend scalars, vectors and matrices to higher dimensions.

  96. anonymous
    • 5 years ago
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    They're very useful.

  97. anonymous
    • 5 years ago
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    my goal, i am currently 11 years behind

  98. anonymous
    • 5 years ago
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    It also depends on what areas you want to focus on in maths. There's discrete mathematics/number theory, mathematical logic, topology, algebra, mathematical statistics,...

  99. anonymous
    • 5 years ago
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    Yeah, I'be heard of him. I hope you get there (minus the autism).

  100. anonymous
    • 5 years ago
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    *I've*

  101. anonymous
    • 5 years ago
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    We 'know' general relativity is wrong - feel like fixing it, Arman?

  102. anonymous
    • 5 years ago
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    sounds like a plan

  103. anonymous
    • 5 years ago
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    What about bed?

  104. anonymous
    • 5 years ago
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    Might be the first step.

  105. anonymous
    • 5 years ago
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    oiudhvcapoufimnac

  106. anonymous
    • 5 years ago
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    Is that code?

  107. anonymous
    • 5 years ago
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    fine -.-

  108. anonymous
    • 5 years ago
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    lol

  109. anonymous
    • 5 years ago
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    nah, gibberish.... alright my chances of catching up to an ausbergers kid is highly unlikely, but i can at least try to match his pace...

  110. anonymous
    • 5 years ago
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    i think... idk. i will win -.-

  111. anonymous
    • 5 years ago
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    yeah, that's the right attitude

  112. anonymous
    • 5 years ago
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    Just ace your doughnuts.

  113. anonymous
    • 5 years ago
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    lol iight gnight yo

  114. anonymous
    • 5 years ago
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    nite

  115. anonymous
    • 5 years ago
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    hey arman

  116. anonymous
    • 5 years ago
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    I am becomeMyFan lol, got a new account because I got tired of my old name

  117. anonymous
    • 5 years ago
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    hey loki=D

  118. anonymous
    • 5 years ago
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    hello

  119. anonymous
    • 5 years ago
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    how are you doin?

  120. anonymous
    • 5 years ago
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    do you like my new name LOL

  121. anonymous
    • 5 years ago
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    ok. bit tired. should be doing my work but wasting time

  122. anonymous
    • 5 years ago
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    Yeah, it's cool...did BMF die?

  123. anonymous
    • 5 years ago
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    yeah

  124. anonymous
    • 5 years ago
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    sad

  125. anonymous
    • 5 years ago
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    RIP, BMF

  126. anonymous
    • 5 years ago
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    LOL =D, or as sstarica says it, ^_^

  127. anonymous
    • 5 years ago
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    Is oktalBlizzard meant to mean something?

  128. anonymous
    • 5 years ago
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    i just fanned you

  129. anonymous
    • 5 years ago
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    All that work you did collecting fans, gone!

  130. anonymous
    • 5 years ago
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    well, it is just my internet user name/ nick name.Oktal means base 8, and blizzard is a natural disaster LOL, thanx I fanned you too, the 132nd fan

  131. anonymous
    • 5 years ago
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    =D yeah

  132. anonymous
    • 5 years ago
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    As in base 8 for numbers?

  133. anonymous
    • 5 years ago
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    but now I am planning to help people more once I learn some more stuff and write my exams

  134. anonymous
    • 5 years ago
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    yes, for numbering system, like binary, octal, hexadecimal...

  135. anonymous
    • 5 years ago
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    Cool.../random

  136. anonymous
    • 5 years ago
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    well, I wanted it to be a bit diferent then my usual nicks

  137. anonymous
    • 5 years ago
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    Arman's gone to bed

  138. anonymous
    • 5 years ago
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    :)

  139. anonymous
    • 5 years ago
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    How's the studying?

  140. anonymous
    • 5 years ago
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    great

  141. anonymous
    • 5 years ago
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    wow, you're probably the first person i've even asked that question who's said that

  142. anonymous
    • 5 years ago
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    :) but really it is great, because I made up my mind to enjoy maths and physics and stuff like this

  143. anonymous
    • 5 years ago
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    it is all about the attitude, I think

  144. anonymous
    • 5 years ago
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    lol

  145. anonymous
    • 5 years ago
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    it is

  146. anonymous
    • 5 years ago
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    research backs it

  147. anonymous
    • 5 years ago
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    If you assume, when you get to something in maths that you don't understand, that you can understand it eventually, well, the research says those people are more successful.

  148. anonymous
    • 5 years ago
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    Took longer to say that than I had anticipated.

  149. anonymous
    • 5 years ago
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    Remains of the day.

  150. anonymous
    • 5 years ago
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    I just looked at my About me, and noticed something LOL, "...I have infinite interesting ideas but my limit is time..." coincidense but that actually reminds me of something I have learned in maths

  151. anonymous
    • 5 years ago
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    lol

  152. anonymous
    • 5 years ago
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    Should you be studying versus helping others?

  153. anonymous
    • 5 years ago
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    well, I am actually solving questions my self, but once every hour or something I come here to solve a problem or two, to change the mental environment.

  154. anonymous
    • 5 years ago
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    lol, okay. you seem sorted.

  155. anonymous
    • 5 years ago
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    \[\sum_{\lim_{n \rightarrow \infty}}^{n} n\] where n is ideas something like this? can you improve it?

  156. anonymous
    • 5 years ago
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    You could just say\[\left| \left\{ I_{oB} \right\} \right|=\infty\]

  157. anonymous
    • 5 years ago
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    The cardinality (size) of the set of oktalBlizzard's ideas is infinite.

  158. anonymous
    • 5 years ago
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    :) nice

  159. anonymous
    • 5 years ago
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    too bad we can include maths symbols in about me text

  160. anonymous
    • 5 years ago
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    You can refer people here to get a better idea.

  161. anonymous
    • 5 years ago
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    =D

  162. anonymous
    • 5 years ago
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    so, have you checked out the new study groups? Physics, Chemistry and Computer Science?

  163. anonymous
    • 5 years ago
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    I had a look at physics yesterday and I was the only one online

  164. anonymous
    • 5 years ago
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    me too

  165. anonymous
    • 5 years ago
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    I haven't looked at the others.

  166. anonymous
    • 5 years ago
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    I don't know anything else about the other subjects. Well, enough to offer any advice.

  167. anonymous
    • 5 years ago
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    Q: How does one insult a mathematician? A: You say: "Your brain is smaller than any ε > 0"

  168. anonymous
    • 5 years ago
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    lol, that's pretty good.

  169. anonymous
    • 5 years ago
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    Theorem: Every positive integer is interesting. Proof: By contradiction, assume that there exists an uninteresting positive integer. Then there must be a smallest uninteresting positive integer. But that's pretty interesting! Therefore a contradiction!

  170. anonymous
    • 5 years ago
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    Let epsilon be less than zero...

  171. anonymous
    • 5 years ago
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    Are you coming up with this random stuff?

  172. anonymous
    • 5 years ago
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    In some alley, a function meets up with a differential operator: "Get out of my way - or I'll differentiate you till you're zero!" "Try it - I'm \[e ^{x} \]..." "Too bad... I'm d/dy." no, I got this from a friend

  173. anonymous
    • 5 years ago
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    I have one.

  174. anonymous
    • 5 years ago
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    Pi had an argument with i. i told pi to be rational and pi told i to get real

  175. anonymous
    • 5 years ago
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    Nice! LOL, I think I heared something like that, Check this: Salary Theorem: The less you know, the more you make. Proof: Fact #1: Knowledge is Power Fact #2: Time is Money We know that: Power = Work / Time And since Knowledge = Power and Time = Money It is therefore true that Knowledge = Work / Money Solving for Money, we get: Money = Work / Knowledge Thus, as Knowledge approaches zero, Money approaches infinity, regardless of the amount of Work done

  176. anonymous
    • 5 years ago
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    Yeah, I think that's more sadly true than funny ;p

  177. anonymous
    • 5 years ago
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    aiight OB, I have to go. Have to do some work before the days runs out and I feel guilty ;)

  178. anonymous
    • 5 years ago
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    he also gave me this, but I dont get it at all, maybe you do: A SLICE OF PI ****************** 3.14159265358979 1640628620899 23172535940 881097566 5432664 09171 036 5 ok, bye, see you next time

  179. anonymous
    • 5 years ago
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    hehe..see you :)

  180. anonymous
    • 5 years ago
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    lolol

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