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anonymous

  • one year ago

Free medal, lolz simple question. Cow+Bull=?

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  1. LoganH
    • one year ago
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    Cow

  2. anonymous
    • one year ago
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    Wrong, try again

  3. LoganH
    • one year ago
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    longhorn?

  4. anonymous
    • one year ago
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    x3 what r de babies called.

  5. LoganH
    • one year ago
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    calf

  6. anonymous
    • one year ago
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    Yay x3

  7. LoganH
    • one year ago
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    lol

  8. LoganH
    • one year ago
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    plural calves?

  9. anonymous
    • one year ago
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    I do this out of randomness

  10. LoganH
    • one year ago
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    QUICK! what is 13-5*2?

  11. anonymous
    • one year ago
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    Pemdas!

  12. LoganH
    • one year ago
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    YAY!

  13. LoganH
    • one year ago
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    :D

  14. anonymous
    • one year ago
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    x3 do I win?

  15. LoganH
    • one year ago
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    what is the square root of calf

  16. LoganH
    • one year ago
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    idku

  17. idku
    • one year ago
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    Suppose that there is such a vector X, that: \(\Large \vec{X}=\vec{\rm Cow}+\vec{\rm Bull}\) And suppose that \(\Large \vec{\rm Cow}\) and \(\Large \vec{\rm Bull}\) are two vertical and horizontal velocity components respectively.

  18. anonymous
    • one year ago
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    meat

  19. LoganH
    • one year ago
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    Dear god

  20. LoganH
    • one year ago
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    idku XD

  21. LoganH
    • one year ago
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    I have to undo my best response XD

  22. anonymous
    • one year ago
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    lol

  23. idku
    • one year ago
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    So, you can conclude that: \(\Large \vec{\rm Cow}=\vec{X}\cos(\theta)\) \(\Large \vec{\rm Bull}=\vec{X}\sin(\theta)\)

  24. anonymous
    • one year ago
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    Lol sin

  25. idku
    • one year ago
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    So, \(\Large \vec{\rm X}=\vec{X}\cos(\theta)+\vec{X}\sin(\theta)\)

  26. LoganH
    • one year ago
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    Hmm, enlightenment

  27. idku
    • one year ago
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    So if I wanted to find the angle: \(\Large \vec{\rm X}=\vec{X}\cos(\theta)+\vec{X}\sin(\theta)\) \(\Large 1=\cos(\theta)+\sin(\theta)\) \(\Large 1=\cos(\theta)+\sqrt{1-\cos^2(\theta)}\) \(\Large 1-\cos(\theta)=\sqrt{1-\cos^2(\theta)}\) \(\Large 1-2\cos(\theta)+\cos^2(\theta)=1-\cos^2(\theta)\) then, \(\Large 2\cos^2(\theta)=2\cos(\theta)\)

  28. LoganH
    • one year ago
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    But what if \[\sin (\theta) \int\limits_{bull}^{cow}\xi \Theta _{1}^{cow*bull}\]

  29. LoganH
    • one year ago
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    changes completely

  30. idku
    • one year ago
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    So either \(\cos\theta=0\) or \(\cos(\theta )\)

  31. idku
    • one year ago
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    i meant cos theta is 1 for the second option

  32. LoganH
    • one year ago
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    But look

  33. idku
    • one year ago
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    hey, why do you have a a reimann zeta function in it:)

  34. idku
    • one year ago
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    |dw:1441818273057:dw|defined for x greater than 1 only

  35. LoganH
    • one year ago
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    \[\cos 1bull ^{2}if \sum_{cow}^{bull}\left\{ cow*bull \right\}\]

  36. idku
    • one year ago
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    ajnd you forgot the differential in your integral LOL

  37. LoganH
    • one year ago
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    OH! of course! :D

  38. LoganH
    • one year ago
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    because \[cow \approx bull\]

  39. LoganH
    • one year ago
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    but \[cow+bull \neq cow+cow\]

  40. idku
    • one year ago
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    we know that shi(t) - shi(t)=0 (which is true for shi(t) just like for any number) So, this is what we can do: bull = bull bull = bull + 0 bull = bull + shi(t) - shi(t) bull = bullshi.t - shi.t bull+shi.t=bullshi.t And the: \(\displaystyle\lim_{n\rightarrow \infty}\left(b.u.l.l.s.h.i.t\right)^n\) diverge to infinity

  41. idku
    • one year ago
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    and same if you have \(\displaystyle\lim_{n\rightarrow \infty}\left(b.u.l.l{~~+~~}s.h.i.t\right)^n\) diverge to infinity

  42. idku
    • one year ago
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    but the bullshi.t one will diverge quicker than the bull+shi.t

  43. LoganH
    • one year ago
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    hmm, but how do you solve for angle 1? of bullshi(t)

  44. idku
    • one year ago
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    We solved theta to be 1....with this condition vector cow and vector give a sum of vector X - initial speed. But if we added pellet to each in direction of each: Such that: \(\Large \vec{Xpellet}\ne\vec{\rm Cow+pellet}+\vec{\rm Bull+pellet}\)

  45. idku
    • one year ago
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    pellet = shi.t

  46. LoganH
    • one year ago
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    i would think that angle 1 would be congruent with pellet. am I correct?

  47. LoganH
    • one year ago
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    as \[pellet=shi.t so \angle 1 \neq \angle bull\]

  48. idku
    • one year ago
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    Why?

  49. idku
    • one year ago
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    well it is not an angle of 1 really, rather cos(\(\theta\))=1 or 0

  50. LoganH
    • one year ago
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    Because \[Bullshi.t ^{2^{cow}}+bull=cow*copulation\]

  51. idku
    • one year ago
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    So, \(\pi/2\) or \(\pi\)

  52. idku
    • one year ago
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    And at that angle only is it true thatSuppose that there is such a vector X, that: \(\Large \vec{\rm Cow}\) = \(\Large \vec{\rm Bull}\)

  53. LoganH
    • one year ago
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    But, would we also consider the size of said bull? because \[bull21 \neq bull 13\]

  54. idku
    • one year ago
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    And thus follows that: \(\Large \rm CowShi.t\) = \(\Large\rm BullShi.t\) and since both sides are equivalent you can remove the vectors as I did.

  55. idku
    • one year ago
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    yes, true, because bull\(\ne\)=0

  56. LoganH
    • one year ago
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    |dw:1441819171326:dw|

  57. idku
    • one year ago
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    And \(\large {shi.t}=\displaystyle \int x^{shi.t~~-1}e^{-x}dx=\Gamma({shi.t})\)

  58. idku
    • one year ago
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    so pellet=1

  59. LoganH
    • one year ago
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    There we go

  60. LoganH
    • one year ago
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    and pellet=shi.t right?

  61. idku
    • one year ago
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    Nice

  62. idku
    • one year ago
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    pellet is shi.t

  63. idku
    • one year ago
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    I have to go now... se you

  64. LoganH
    • one year ago
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    So we have solved it?

  65. LoganH
    • one year ago
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    Thanks! :D

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