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anonymous

  • 5 years ago

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  1. anonymous
    • 5 years ago
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    \[\int\limits e^{(sinx)}sinx dx\]

  2. anonymous
    • 5 years ago
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    HEY LOKI

  3. anonymous
    • 5 years ago
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    Evening :)

  4. anonymous
    • 5 years ago
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    Make a substitution,\[u=\sin x\]

  5. anonymous
    • 5 years ago
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    du=cos x

  6. anonymous
    • 5 years ago
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    Yeah, hang on...

  7. anonymous
    • 5 years ago
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    haha i tried..by parts..and substitution..

  8. anonymous
    • 5 years ago
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    If you're looking for a solution in terms of standard integrals, I'm pretty sure it doesn't exist. You either have to write your answer in terms of the Bessel and Struve functions, or take a series solution, where you take the series expansion for sin(x), the expansion for e^(sin(x)) and then take the Cauchy product of the two series, and integrate term-by-term (which you can do since the series is uniformly convergent). Where did this problem come from?

  9. anonymous
    • 5 years ago
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    LOL umm someone asked this question yesterday, and i couldn't solve it.So i am asking you, the boss

  10. anonymous
    • 5 years ago
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    Haha, well, boss says that.

  11. anonymous
    • 5 years ago
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    oh dam be humble :P

  12. anonymous
    • 5 years ago
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    lol, I am ;)

  13. anonymous
    • 5 years ago
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    LOL sure boss :P

  14. anonymous
    • 5 years ago
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    Well, I'm signing out...have a *lot* of work to do ><

  15. anonymous
    • 5 years ago
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    I'll answer your ferris wheel question later :)

  16. anonymous
    • 5 years ago
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    You should chillax, it's Friday.

  17. anonymous
    • 5 years ago
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    question!!

  18. anonymous
    • 5 years ago
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    What is the difference between power and geometric series!

  19. anonymous
    • 5 years ago
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    I am chillaxing by chit chating with my roomate

  20. anonymous
    • 5 years ago
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    Take your time on the wheel question :)

  21. anonymous
    • 5 years ago
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    A geometric series is one where there's a constant ratio between each successive terms. A power series is different in that it is not always the case that the ratio between two successive terms is constant. In a geometric series, you'd have to successive terms,\[\frac{a_{n+1}(x-x_0)^{n+1}}{a_n(x-x_0)^n}=\frac{a_{n+1}}{a_n}(x-x_0)\]which is only constant if a_(n+1) and a_n give some number independent of n. In general, this isn't the case.

  22. anonymous
    • 5 years ago
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    Thank you. I will try to do a little bit more research on my ow :)

  23. anonymous
    • 5 years ago
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    This is research :)

  24. anonymous
    • 5 years ago
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    LOL just go do your work!

  25. anonymous
    • 5 years ago
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    Yeah I know...

  26. anonymous
    • 5 years ago
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    I have Lammy below trying to figure something out. When he/she's done, I'm out... -.-

  27. anonymous
    • 5 years ago
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    Alright i will ask questions later if have any :) Peace~

  28. anonymous
    • 5 years ago
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    Hey loki, i have a question, what test should i use to determine the convergence/divergence of the series sin(x/9)

  29. anonymous
    • 5 years ago
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    \[\sum_{1}^{\infty}\sin(9/x)\]

  30. anonymous
    • 5 years ago
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    I don't think this converges. Let me check.

  31. anonymous
    • 5 years ago
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    ok

  32. anonymous
    • 5 years ago
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    I'm distracted because some moron in my building has been playing the pellettest music I've ever heard, all day, all loud and I'm about to go break down doors ><

  33. anonymous
    • 5 years ago
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    Dude you are the boss, just order your followers to do it for you

  34. anonymous
    • 5 years ago
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    lol, i wish :(

  35. anonymous
    • 5 years ago
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    I'm thinking you can set the problem up as proving it divergent using something like a limit comparison test.

  36. anonymous
    • 5 years ago
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    Sorry, had to go talk to the neighbor.

  37. anonymous
    • 5 years ago
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    no problem :) i went through every single tests...but none of them worked

  38. anonymous
    • 5 years ago
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    To use the limit comparison, we need a series that is divergent (since we're suspicious this thing won't converge). When I think 'non-convergence', the first series I look to to test against is the harmonic series.

  39. anonymous
    • 5 years ago
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    Should I wait for you to finish typing?

  40. anonymous
    • 5 years ago
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    lol what if i am suspicious this thing converge... then i will be in trouble

  41. anonymous
    • 5 years ago
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    Lol, yes, you would be :D

  42. anonymous
    • 5 years ago
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    Form the ratio\[\frac{1/n}{\sin \frac{9}{n}}\]and note that, in the limit, you get an indeterminate form, \[\frac{0}{0}\]

  43. anonymous
    • 5 years ago
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    This is great, because you now have access to L'Hopital's rule.

  44. anonymous
    • 5 years ago
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    So\[\lim_{n \rightarrow \infty}\frac{1/n}{\sin 9/n}=\frac{\lim_{n \rightarrow \infty}1/n}{\lim_{n \rightarrow \infty}\sin 9/n}=\frac{\lim_{n \rightarrow \infty}-\frac{1}{n^2}}{\lim_{n \rightarrow \infty}\frac{-9\cos 9/n}{n^2}}=\frac{1}{9}\]

  45. anonymous
    • 5 years ago
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    since cos(9/n) goes to 1 as n goes to infinity.

  46. anonymous
    • 5 years ago
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    Now, since your ratio was that for a known divergent series with your own series, and because this ratio converged, you know that your series is divergent also.

  47. anonymous
    • 5 years ago
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    So at the beginning, you have to guess whether this series diverges or converges?

  48. anonymous
    • 5 years ago
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    Pretty much...it's okay, since you then turn towards something that can either prove or disprove your case.

  49. anonymous
    • 5 years ago
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    A lot of mathematics is experimental.

  50. anonymous
    • 5 years ago
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    But you yourself said you had a feeling that it was divergent...

  51. anonymous
    • 5 years ago
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    On that integral before, you'd have to find more than just a couple of series expansions. I over-simplified it. If you ever see something like that in practice (i.e. engineering) you would use an approximation rule, like Simpson's or something.

  52. anonymous
    • 5 years ago
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    I know it behaves like p-series as n->infinity, so if i had compared it to 1/n^2, it wouldn't follow any of the outcomes for the test?

  53. anonymous
    • 5 years ago
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    1/n^2 is convergent, whereas your suspicion is that sin(9/n) is divergent. It wouldn't have helped.

  54. anonymous
    • 5 years ago
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    i used the nth term test, and the limit goes to 0, so i thought it should be convergent.

  55. anonymous
    • 5 years ago
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    Ahhhhh....no....

  56. anonymous
    • 5 years ago
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    If you read the statement of the nth term test, it says:

  57. anonymous
    • 5 years ago
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    If \[\lim_{n \rightarrow \infty}a_n \ne0 \] or if this limit does not exist as n tends to infinity, then\[\sum_{n}^{\infty}a_n \]does not converge.

  58. anonymous
    • 5 years ago
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    The nth term test doesn't say that if your sequence tends to zero, your series converges.

  59. anonymous
    • 5 years ago
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    It says if it DOESN'T tend to zero, your series DIVERGES.

  60. anonymous
    • 5 years ago
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    Take 1/n for example. The limit is 0 as n goes to infinity, but the series itself does not converge.

  61. anonymous
    • 5 years ago
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    yea, so , as a normal college kid, my suspicion would be convergent

  62. anonymous
    • 5 years ago
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    SUSPICION

  63. anonymous
    • 5 years ago
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    Now we come to 'instinct', which is a function of experience. It means you just need practice. You're looking at good questions. Trust your teachers - they'll show you all sorts of things you need to consider.

  64. anonymous
    • 5 years ago
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    LOL what my math professors have taught me the most is HUMOR

  65. anonymous
    • 5 years ago
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    Well, that *can* help..? :~S

  66. anonymous
    • 5 years ago
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    1 Attachment
  67. anonymous
    • 5 years ago
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    i am stuck at number 3

  68. anonymous
    • 5 years ago
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    you mean 4?

  69. anonymous
    • 5 years ago
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    3..that answer is wrong

  70. anonymous
    • 5 years ago
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    Does this have to be submitted now?

  71. anonymous
    • 5 years ago
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    Nope, but soon~~->>>> tomorrow night

  72. anonymous
    • 5 years ago
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    Oh, okay. I need to go out, that's all. I'll take a look later.

  73. anonymous
    • 5 years ago
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    OH alright :) have fun!

  74. anonymous
    • 5 years ago
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    Bye..

  75. anonymous
    • 5 years ago
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    Did you figure it out?

  76. anonymous
    • 5 years ago
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    Nope haha you told me to chillax

  77. anonymous
    • 5 years ago
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    Slacker... ;p

  78. anonymous
    • 5 years ago
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    I'll look now.

  79. anonymous
    • 5 years ago
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    take your time. just look at it whenever u feel like it

  80. anonymous
    • 5 years ago
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    The first one can be 2/361. You have 1/361.

  81. anonymous
    • 5 years ago
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    Have you covered asymptotic equivalence?

  82. anonymous
    • 5 years ago
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    what is that? might have..not sure

  83. anonymous
    • 5 years ago
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    brb

  84. anonymous
    • 5 years ago
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    kay kay

  85. anonymous
    • 5 years ago
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    I know that as n-> infin, some terms become negligible

  86. anonymous
    • 5 years ago
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    Yeah, basically. The definition is this:

  87. anonymous
    • 5 years ago
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    But how you determine which term is going to infinity faster

  88. anonymous
    • 5 years ago
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    Ah, there's an order to it. You can see these orders if you make plots of the corresponding functions. Exponential functions will dominate polynomials at some point. You'll see, I'll write out how I worked out your question.

  89. anonymous
    • 5 years ago
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    can you please do number 4 also :P

  90. anonymous
    • 5 years ago
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    \[\frac{2^n+n^5+2}{19^{2n}+6^n+3} \iff \frac{2^n}{(19^2)^n}\]since in the numerator, the exponential dominates the quintic and 2, and in the denominator, (19^2)^n dominates 6^n+3 (19^2 > 6).

  91. anonymous
    • 5 years ago
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    The equivalence sign should be replaced with ~. For some reason it doesn't show up in the equation editor.

  92. anonymous
    • 5 years ago
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    brb

  93. anonymous
    • 5 years ago
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    i thought n^5 is going to infinity faster

  94. anonymous
    • 5 years ago
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    No, if you plot n^5 and 2^n, you'll see 2^n overtake pretty quickly.

  95. anonymous
    • 5 years ago
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    But listen, you can always check domination by using the definition. I'll scan in something.

  96. anonymous
    • 5 years ago
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  97. anonymous
    • 5 years ago
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    You asked for tricks in math - I love asymptotic equivalence when it comes to limits - along with L'Hopital's rule. It saves your butt.

  98. anonymous
    • 5 years ago
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    \[\frac{2^n/(19^2)^n}{r^n}=\left( \frac{2}{361r} \right)^n\]

  99. anonymous
    • 5 years ago
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    Then choose r = 2/361 to ensure a convergent limit that's non-zero.

  100. anonymous
    • 5 years ago
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    You can show that the original expression is asymptotically equivalent to the one we use in the end by taking the ratio between the two and showing the ratio approaches 1 as n approaches infinity.

  101. anonymous
    • 5 years ago
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    Does exponential function always dominate quadratic, quintic ?

  102. anonymous
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    \[a^n>n^m\] for all a >1 and positive exponent. You have positive exponent because of the problem, and here you have the numbers 2, 6, 19. It is the case,\[19^n>6^n>2^n\]

  103. anonymous
    • 5 years ago
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    Remember, when you're dealing with limits, the definition of limits makes use of "for *large* n...so you have to be thinking about the general properties of functions once they go through the motions and their domains start heading to infinity.

  104. anonymous
    • 5 years ago
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    :P can you prove a^n >n^m

  105. anonymous
    • 5 years ago
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    yes, but I'll show you a plot...

  106. anonymous
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    1 Attachment
  107. anonymous
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    They're all monotonic increasing functions, so when it gets to a point where the rate of one outstrips the other, the one with the slower rate of increase is dead in the water.

  108. anonymous
    • 5 years ago
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    is that a free program haha

  109. anonymous
    • 5 years ago
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    Yes it's free.

  110. anonymous
    • 5 years ago
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    GeoGebra

  111. anonymous
    • 5 years ago
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    http://www.geogebra.org/cms/

  112. anonymous
    • 5 years ago
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    thanks !!

  113. anonymous
    • 5 years ago
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    how do i input multiple graphs

  114. anonymous
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    Just type them in on the one page. y=x^2, y=x^4, etc. They should show up.

  115. anonymous
    • 5 years ago
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    Got it hohoho

  116. anonymous
    • 5 years ago
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    I don't know if you've done the last one, but I have \[r=\left( \frac{2^{7/10}}{19} \right)^3\]

  117. anonymous
    • 5 years ago
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    yea i did it haha

  118. anonymous
    • 5 years ago
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    what you up to right now

  119. anonymous
    • 5 years ago
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    I should be doing my work...

  120. anonymous
    • 5 years ago
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    Seriously going offline...it's getting dark and I've done nothing.

  121. anonymous
    • 5 years ago
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    http://www.youtube.com/watch?v=-C3v3aThfFA just when you have time loki, check out the electric car my friends from the university i wanna go built, it won xprize 2nd place

  122. anonymous
    • 5 years ago
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    LOL when is your presentation ? i forgot~~

  123. anonymous
    • 5 years ago
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    @ oktal Nice car ~ haha

  124. anonymous
    • 5 years ago
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    I'll check it out - I saved the link. And presentations on Wed and Fri >< And to both of you - there's a site you should check out if you haven't: wolframalpha.com. You can check out math questions, like integrals and such.

  125. anonymous
    • 5 years ago
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    yeah, when I found out they were builing it, i could not believe it, and when it won google x-prize 2nd place, I was stunned

  126. anonymous
    • 5 years ago
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    ok, i will see it, thanks bye

  127. anonymous
    • 5 years ago
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    lol you still have few days to prepare for it no worries. you should chill with us

  128. anonymous
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    no, let him work, he has a life to manage...

  129. anonymous
    • 5 years ago
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    LOL he has troublesome neighbors to manage as matter of fact :P

  130. anonymous
    • 5 years ago
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    Nice car, oktal

  131. anonymous
    • 5 years ago
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    not mine :(

  132. anonymous
    • 5 years ago
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    my friend's

  133. anonymous
    • 5 years ago
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    and actually it belongs to the university now, as it funded the building of it

  134. anonymous
    • 5 years ago
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    it's pretty cool. they should at least be able to take it out on weekends ;p

  135. anonymous
    • 5 years ago
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    "Hey ladies, check out my revolutionary design..." ;)

  136. anonymous
    • 5 years ago
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    :)

  137. anonymous
    • 5 years ago
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    you would haha

  138. anonymous
    • 5 years ago
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    lol. oktal, check out a nerd link above for free plotting software, GeoGebra.

  139. anonymous
    • 5 years ago
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    they did it few times, drove around town :)

  140. anonymous
    • 5 years ago
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    ok

  141. anonymous
    • 5 years ago
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    You could have used it to check out that absolute value problem.

  142. anonymous
    • 5 years ago
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    nice, will check it

  143. anonymous
    • 5 years ago
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    Okay, I'm out...seriously need to get sorted. Later!

  144. anonymous
    • 5 years ago
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    Laters ~~ i am going to to bed too

  145. anonymous
    • 5 years ago
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    nice meeting you okatal :)

  146. anonymous
    • 5 years ago
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    ok, see you, nice to meet you too dichalao

  147. anonymous
    • 5 years ago
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    Hi Loki, can you help me out with this problem

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  148. anonymous
    • 5 years ago
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    What did you enter before?

  149. anonymous
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    11.7

  150. anonymous
    • 5 years ago
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    brb...when's this due?

  151. anonymous
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    umm not soon :)

  152. anonymous
    • 5 years ago
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    oh, are these your physics questions that aren't due till the end of the week or something?

  153. anonymous
    • 5 years ago
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    it is due on tuesday before lecture

  154. anonymous
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    Your value is what you get when you work out magnitude, since \[a=\frac{v^2}{r}\]

  155. anonymous
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    11.7 is what i got from that equation..but it is wrong..probably becasue this is not uniform circular motion

  156. anonymous
    • 5 years ago
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    This is a vector problem, and you're going to have to subtract an initial vector from a final vector...no, this is uniform circular motion...

  157. anonymous
    • 5 years ago
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    So,\[a=\frac{\Delta v}{\Delta t}=\frac{v_2-v_1}{t_2-t_1}\]where the a and v's are vectors.

  158. anonymous
    • 5 years ago
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    The velocity in uniform circular motion is\[v=(-r \omega \sin \theta, r \omega \cos \theta)\]

  159. anonymous
    • 5 years ago
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    why is there a negative sign for V_x

  160. anonymous
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    and we know, from definition of angular speed and uniform circular motion that,\[\omega = \frac{\theta }{t} \rightarrow \theta= \omega t\]

  161. anonymous
    • 5 years ago
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    So \[v=(-r \omega \sin (\omega t), r \omega \cos (\omega t))\]

  162. anonymous
    • 5 years ago
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    There's a minus sign from when we take the derivative of the x-component of displacement with respect to time. That component is a cosine.

  163. anonymous
    • 5 years ago
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    Since\[v = r \omega \rightarrow \omega = \frac{v}{r}\]so\[V=(-v \sin \frac{vt}{r}, v \cos \frac{vt}{r})=v(\sin \frac{vt}{r}, \cos \frac{vt}{r})\]

  164. anonymous
    • 5 years ago
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    You should be able to work out two vectors at different times and then subtract.

  165. anonymous
    • 5 years ago
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    Is that making sense? I hope it is 'cause I just got up and I'm doing the on-the-fly.

  166. anonymous
    • 5 years ago
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    in fact, we haven't "learned" angular speed .

  167. anonymous
    • 5 years ago
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    There's a minus sign on the last expression outside sine...typo.

  168. anonymous
    • 5 years ago
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    It goes like this: let the arc of a circle of radius r have length (distance) l. Then that distance is\[l=r \theta\]when theta s in radians. Take the derivative with respect to time on both sides to get\[\frac{dl}{dt}=\frac{d(r \theta)}{dt}=r \frac{d \theta }{dt}+ \theta \frac{dr}{dt}=r \frac{d \theta }{dt}:=r \omega\]

  169. anonymous
    • 5 years ago
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    dr/dt is zero by construction of the problem -> the radius is fixed and therefore unchanging in time.

  170. anonymous
    • 5 years ago
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    Angular speed is the rate of change in the angle.

  171. anonymous
    • 5 years ago
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    brb

  172. anonymous
    • 5 years ago
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    kk

  173. anonymous
    • 5 years ago
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    Did you work it out?

  174. anonymous
    • 5 years ago
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    kk

  175. anonymous
    • 5 years ago
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    is there something wrong with this site..i didn't reply"kk"

  176. anonymous
    • 5 years ago
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    do i just plug in t=0,and t=0.6?

  177. anonymous
    • 5 years ago
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    it came through...and there's always something wrong with this site.

  178. anonymous
    • 5 years ago
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    I would plug 0 and 0.6.

  179. anonymous
    • 5 years ago
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    is vt/r in radians?

  180. anonymous
    • 5 years ago
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    Always.

  181. anonymous
    • 5 years ago
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    Let me check something before you waste a turn.

  182. anonymous
    • 5 years ago
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    is there a way solving this problem without using angular speed?

  183. anonymous
    • 5 years ago
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    Not off the top of my head.

  184. anonymous
    • 5 years ago
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    \[\Delta v = 5.3(-\sin (1.325), \cos (1.325)-1)\]

  185. anonymous
    • 5 years ago
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    \[a=\frac{\Delta v}{\Delta t}\approx \frac{1}{0.6}(-5.14,-4.01) \approx (-8.57, -6.68)\]

  186. anonymous
    • 5 years ago
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    Thanks!!!!

  187. anonymous
    • 5 years ago
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    What, did it work?

  188. anonymous
    • 5 years ago
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    yep~

  189. anonymous
    • 5 years ago
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    Good -.-

  190. anonymous
    • 5 years ago
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    i am going to ask my professor... see if there is another way to do it..since we haven't covered angular velocity....

  191. anonymous
    • 5 years ago
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    Okay. At least it worked. I hate online submission. I used to hate it as an undergrad.; you can never argue a point if it's wrong.

  192. anonymous
    • 5 years ago
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    Anyways, I have to sign out and get some work done. Happy mathing :)

  193. anonymous
    • 5 years ago
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    LOL you meant physicsing?

  194. anonymous
    • 5 years ago
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    All the same...

  195. anonymous
    • 5 years ago
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    alright thanks for helping!! have a great day...still have an essay to do..

  196. anonymous
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    At least they run you into the ground.

  197. anonymous
    • 5 years ago
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    LOL true~ see ya ~buddy

  198. anonymous
    • 5 years ago
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    cu ;)

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