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anonymous

  • 5 years ago

A large tank contains 100 liters of a salt solution which has a concentration of 0.25 kilograms per liter. Pure water is added to the solution at a rate of 80 liters per minute. At the same time, the well-mixed solution drains from the tank at 80 liters per minute. Find the amount S of the salt in the solution as a function of t: kilograms At what time is there only 12 kilograms of salt remaining in the tank? t= minutes

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  1. anonymous
    • 5 years ago
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    Hi dichalao, Do you still need a solution?

  2. anonymous
    • 5 years ago
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    Yes! Or just give me some clues first :)

  3. anonymous
    • 5 years ago
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    What i have tried is to set up the first differential equation as : dS/dt= 25/(100+80t)(1-80t)

  4. anonymous
    • 5 years ago
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    You need to consider the physics of what's going on in a little bit of time,\[\delta t\]

  5. anonymous
    • 5 years ago
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    and you need to consider what's actually changing. You're told 80 L/min enters the tank, and 80 L/min leaves the tank, so volume's the same. What's changing is the concentration, or even better, the mass, of salt in the solution.

  6. anonymous
    • 5 years ago
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    You need to come up with an expression for the change in mass of salt in the tank as a function of time, then use that to get an expression for concentration and solve the system.

  7. anonymous
    • 5 years ago
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    Yea i see the solution is diluting

  8. anonymous
    • 5 years ago
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    So, let's consider what happens in that little bit of time, delta_t. No salt is entering the system, since what's coming in is pure water. Salt *is* leaving the system in the outflow. I said we need to consider mass, so, let the mass at time t be\[m(t)\] and that at time t+delta_t be\[m(t+\delta t)\]The difference between the two will be\[\delta m = m( t + \delta t) - m(t)\]

  9. anonymous
    • 5 years ago
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    Hello

  10. anonymous
    • 5 years ago
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    Hi Lok!!

  11. anonymous
    • 5 years ago
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    Yea

  12. anonymous
    • 5 years ago
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    The change in mass is the negative of the amount leaving (because these delta quantities are always read (final)-(initial), and since you have less at the end, the change will be negative). That is,\[\delta m = -Q \delta t C(t)\]where Q is the flow rate (liters/min) and C(t) is concentration at time t (this is why delta_t needs to be small, otherwise C(t) won't be valid).

  13. anonymous
    • 5 years ago
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    Notice the units on the left-hand side and the right-hand side. LHS is (mass), and right-hand side is (volume)/(time) x (time) x (mass)/(volume). The units equate.

  14. anonymous
    • 5 years ago
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    So what we have now is\[\delta m = m(t+ \delta t)-m(t)\]\[-Q \delta t C(t)=m(t+\delta t)-m(t)\]Divide both sides by the constant volume in the container, and by delta_t to get\[-\frac{QC(t)}{V}=\frac{C(t+\delta t)-C(t)}{\delta t}\]

  15. anonymous
    • 5 years ago
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    because \[C(t) = \frac{m(t)}{V}, C(t+\delta t)=\frac{m(t+\delta t)}{V}\]

  16. anonymous
    • 5 years ago
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    That expression above (above what I *just* wrote) should look familiar. Does it?

  17. anonymous
    • 5 years ago
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    Yea

  18. anonymous
    • 5 years ago
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    Sorry, my computer's acting up.

  19. anonymous
    • 5 years ago
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    You can now take the limit as delta_t approached zero on both sides to get,

  20. anonymous
    • 5 years ago
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    \[\lim_{\delta t \rightarrow 0}-\frac{QC(t)}{V}=\lim_{\delta t \rightarrow 0}\frac{C(t+ \delta t)-C(t)}{\delta t}\]\[\rightarrow -\frac{QC(t)}{V}=\frac{dC}{dt}\](the left-hand side is independent of delta_t, so it's just a constant in terms of taking the limit).

  21. anonymous
    • 5 years ago
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    You now have a separable differential equation:\[\frac{dC}{C}=-\frac{Q}{V}dt \rightarrow \int\limits_{}{}\frac{dC}{C}=\int\limits_{}{}-\frac{Q}{V}dt\]

  22. anonymous
    • 5 years ago
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    Thanks i can take it from here on

  23. anonymous
    • 5 years ago
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    \[\log C=-\frac{Q}{V}t+c\]

  24. anonymous
    • 5 years ago
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    ok

  25. anonymous
    • 5 years ago
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    I have one more question

  26. anonymous
    • 5 years ago
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    ok

  27. anonymous
    • 5 years ago
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    Suppose taht a_n>0, and lim n->infinity n^2a_n=0. Prove that sigma a_n converges

  28. anonymous
    • 5 years ago
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    Since lim n^2a_n->0, so i assume n^2a_n behave like 1/n^p where p >0, if so a_n should be have like (1/n^(2+p), a_n is a p-series, with p >1, so a_n converges. Do u think this approach is reasonable?

  29. anonymous
    • 5 years ago
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    You might have to leave it with me. I just got up and need to get ready for uni/work. I'll get back to you.

  30. anonymous
    • 5 years ago
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    Sure :) Have a good one :)

  31. anonymous
    • 5 years ago
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    Before I go, I want to confirm the sequence is\[n^2a_n\]

  32. anonymous
    • 5 years ago
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    yes

  33. anonymous
    • 5 years ago
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    ok

  34. anonymous
    • 5 years ago
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    i have a question regarding C(t) is concentration at time t (this is why delta_t needs to be small, otherwise C(t) won't be valid).

  35. anonymous
    • 5 years ago
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    Do you mean C(t)*delta_t wont be valid? Or just C(t)

  36. anonymous
    • 5 years ago
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    Well, C(t) is the concentration at time t, but that's always changing. We're looking at times t and t+delta_t, but only using C(t) to approximate the amount of mass that's leaving. This is valid in the limit, but if you leave everything as a finite difference (which is what you have before taking the limit), what you're really saying is\[-QC(t)/V \approx m(t+ \delta t)-m(t)\]

  37. anonymous
    • 5 years ago
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    But because our increment in time was small, the CHANGE in C(t) over that time is assumed very small too.

  38. anonymous
    • 5 years ago
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    So then all you need to do to find the mass is multiply it by volume flow rate.

  39. anonymous
    • 5 years ago
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    I don't know how well this is being explained. We could have used differentials instead of delta notation, but I think understanding the approximations before taking limits is useful.

  40. anonymous
    • 5 years ago
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    This is a good explanation, i understand everything you did

  41. anonymous
    • 5 years ago
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    -Q is -80L/min?

  42. anonymous
    • 5 years ago
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    No, Q=-80L/min

  43. anonymous
    • 5 years ago
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    I took Q as the magnitude of flow, that's why I introduced the minus sign when setting up the equation.

  44. anonymous
    • 5 years ago
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    You should end up with\[C(t)=\frac{1}{4}e^{-\frac{4}{5}(\min^{-1})t}kg/L\]

  45. anonymous
    • 5 years ago
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    which makes sense, since as t approaches infinity, C(t) approaches zero...the salt is being drained away.

  46. anonymous
    • 5 years ago
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    and at time 0, C(0) = 1/4 kg, which is what you've stated as an initial condition.

  47. anonymous
    • 5 years ago
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    1/4 kg/L, I mean.

  48. anonymous
    • 5 years ago
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    if i neeed to find the amount S of the salt in the solution as a function of t: i just need to multiply both side by 100? since V is constant?

  49. anonymous
    • 5 years ago
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    right side*

  50. anonymous
    • 5 years ago
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    Yes, (concentration) x (volume) = (mass) here.

  51. anonymous
    • 5 years ago
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    got it~

  52. anonymous
    • 5 years ago
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    If the volume flow rates were different either side, you'd have a very different situation to look at.

  53. anonymous
    • 5 years ago
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    lol it will be super complicated?

  54. anonymous
    • 5 years ago
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    Kind of...

  55. anonymous
    • 5 years ago
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    where are you from btw

  56. anonymous
    • 5 years ago
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    So, in these types of problems, look at what's actually changing, and try to come up with an expression that focuses on that. Try to formulate something you can turn into a differential equation in the end.

  57. anonymous
    • 5 years ago
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    UK citizen in Sydney.

  58. anonymous
    • 5 years ago
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    Where are you from? Are you male or female?

  59. anonymous
    • 5 years ago
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    I am from US, a kid

  60. anonymous
    • 5 years ago
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    :P

  61. anonymous
    • 5 years ago
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    So is this school stuff?

  62. anonymous
    • 5 years ago
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    yea, i am learning sequence/series and my friend is learning 1st degree differential equation

  63. anonymous
    • 5 years ago
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    You are really good at problem solving

  64. anonymous
    • 5 years ago
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    Is this for the international baccalaureate?

  65. anonymous
    • 5 years ago
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    haha, thanks

  66. anonymous
    • 5 years ago
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    international baccalaureate?

  67. anonymous
    • 5 years ago
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    hey, I just found the answer to your other problem

  68. anonymous
    • 5 years ago
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    haha do you have tips on how to approach a solution?

  69. anonymous
    • 5 years ago
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    AMAZING !!

  70. anonymous
    • 5 years ago
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    the stuff you're studying is covered in the international baccalaureate - it's an international diploma that can be used to get into universities around the world.

  71. anonymous
    • 5 years ago
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    actually, hang on, I'll need to check it. I was looking to apply the Dirichlet test with a series 1/n^2...but I want to see if I can get around one of the assumptions

  72. anonymous
    • 5 years ago
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    No it is just normal school work

  73. anonymous
    • 5 years ago
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    I'll think about tips on how to approach a solution and get back to you. Really, all you have to do is determine what you're being asked to find, put down the information you have before you and build a bridge.

  74. anonymous
    • 5 years ago
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    are you told that \[a_n \ge a_{n+1}>0?\]

  75. anonymous
    • 5 years ago
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    No, what i posted is everything given

  76. anonymous
    • 5 years ago
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    i wouldn't have thought about setting -Qdelta_t(Ct)=m(delta_t+t)-m(t) haha

  77. anonymous
    • 5 years ago
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    Did you ask yourself what was actually changing in the system?

  78. anonymous
    • 5 years ago
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    Concentration's harder to come to terms with than mass, since concentration is a ratio of quantities, whereas mass is something we have an intuitive feel for.

  79. anonymous
    • 5 years ago
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    experience helps too

  80. anonymous
    • 5 years ago
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    yea i know that the amount of salt in the container is lessening and the amount of salts going out is also lessening

  81. anonymous
    • 5 years ago
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    How old are you ~

  82. anonymous
    • 5 years ago
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    if you dont mind :)

  83. anonymous
    • 5 years ago
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    20s

  84. anonymous
    • 5 years ago
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    You are a genius i guess ..

  85. anonymous
    • 5 years ago
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    like I said, experience helps

  86. anonymous
    • 5 years ago
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    You should pass some experiences to me haha

  87. anonymous
    • 5 years ago
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    I have been! =D

  88. anonymous
    • 5 years ago
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    LOL True.. Are you still trying to find the solution to the problem?

  89. anonymous
    • 5 years ago
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    yes, but I should be showering...I have an idea, but not much time to check it out.

  90. anonymous
    • 5 years ago
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    take your time buddy, i have to go to class right now.. Thank you for helping out :) see you

  91. anonymous
    • 5 years ago
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    ok

  92. anonymous
    • 5 years ago
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    Damn, I figured it out just after you left. You can use the limit comparison test.

  93. anonymous
    • 5 years ago
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    The version we need says, If \[\lim_{n \rightarrow \infty} \frac{a_n}{b_n}=0\]then\[\sum_{n=0}^{\infty}b_n\]convergent implies\[\sum_{n=0}^{\infty}a_n\]is also convergent.

  94. anonymous
    • 5 years ago
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    Now, you need to show that the sum of a_n is convergent, and we can consider for out 'b' series, \[\sum_{n=0}^{\infty}b_n=\sum_{n=0}^{\infty}\frac{1}{n^2}\]which is convergent (p-series, p=2>1). Therefore, taking the ratio of a_n and b_n and forming the limit gives,\[\lim_{n \rightarrow \infty}\frac{a_n}{b_n}=\lim_{n \rightarrow \infty}\frac{a_n}{1/n^2}=\lim_{n \rightarrow \infty}n^2a_n=0\]by construction (i.e. by the setup of the problem). By the theorem above, your series is convergent.

  95. anonymous
    • 5 years ago
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    HAHA got it !!!!

  96. anonymous
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    Yeah. Whenever I see a polynomial in the variable, I think ratio test, limit comparison or Dirichlet first.

  97. anonymous
    • 5 years ago
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    A large tank contains 450 liters of a salt solution which has a concentration of 0.65 kilograms per liter. A salt solution which has a concentration of 0.8 kilograms per liter is added to the solution at a rate of 20 liters per minute. At the same time, the solution drains from the tank at 20 liters per minute.

  98. anonymous
    • 5 years ago
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    It's going to be similar. Volume in the tank won't change again, which is useful. Try to form a mass balance like we did before.

  99. anonymous
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    I learned the limit comparison and ratio test just now haha

  100. anonymous
    • 5 years ago
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    Ah..well, that may have been why you couldn't do it ;)

  101. anonymous
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    There are a couple of versions, depending on how the limit a_n/b_n behaves. I had to pick the right one.

  102. anonymous
    • 5 years ago
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    So this time, there is outflow and inflow of salt.

  103. anonymous
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    Yes

  104. anonymous
    • 5 years ago
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    What does your mass balance look like?

  105. anonymous
    • 5 years ago
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    i am trying to figure it out

  106. anonymous
    • 5 years ago
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    There's a general formula for changes of 'stuff' in a system (for further reference):\[\Delta = i-o+ \gamma - \delta\]

  107. anonymous
    • 5 years ago
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    Change = in - out + generated - destroyed

  108. anonymous
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    This isn't a chemical or nuclear reaction, so nothing is generated or destroyed.

  109. anonymous
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    So gamma = delta = 0

  110. anonymous
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    inflow=Qdelta_t(C(t)) outflow=-Q(deltat)(C(t+deltat)?

  111. anonymous
    • 5 years ago
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    Nearly...

  112. anonymous
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    Your inflow isn't changing with time - it's fixed at 0.8kg/L * 20 L/min

  113. anonymous
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    First, in a unit of time, delta t, how much salt is entering the system?

  114. anonymous
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    0.8(20)(delta t) (delta t in min of time)?

  115. anonymous
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    Yes :)

  116. anonymous
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    Just set it equal to some constant, a, or something.

  117. anonymous
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    You don't have to, but when you have numbers floating around, it can stuff the algebra if you're not careful.

  118. anonymous
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    so it will just be a-outflow=m(t+delta t)-m(t) and then everything goes the same way?

  119. anonymous
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    Yep

  120. anonymous
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    give a round of applause to loki haha

  121. anonymous
    • 5 years ago
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    but what about the outflow...what's your expression?

  122. anonymous
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    LOL ouchie

  123. anonymous
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    might want to save the round of applause for a bit =D

  124. anonymous
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    You might want to write your inflow as\[a \delta t Q\]where a is your constant concentration and Q is the inflow rate.

  125. anonymous
    • 5 years ago
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    outflow= 20(delta t)(Constantly changing concentration) :(

  126. anonymous
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    You might want to write your inflow as\[a \delta t Q\]where a is your constant concentration and Q is the inflow rate.

  127. anonymous
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    Your outflow is write. Don't be :(

  128. anonymous
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    right

  129. anonymous
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    not 'write'...

  130. anonymous
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    LOL you need more sleep

  131. anonymous
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    i know

  132. anonymous
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    concentration = mass over volume mass=(mass initial+mass inflow-mass outlfow)?

  133. anonymous
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    Now, what does\[m(t+ \delta t)-m(t)\]equal?

  134. anonymous
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    (constant inflow)a - (changing outflow), b delta t(C(t+delta t) ?

  135. anonymous
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    lol, you're giving me a headache...I'll write it out

  136. anonymous
    • 5 years ago
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    sorry.. i am not floating with you :(

  137. anonymous
    • 5 years ago
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    \[m(t+\delta t)-m(t)=a \delta t Q-C(t) \delta t Q=(a-C(t))Q \delta t\]

  138. anonymous
    • 5 years ago
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    the change is just *in* - *out*, as per that Delta equation I put up.

  139. anonymous
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    give me some time to digest haha

  140. anonymous
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    a is constant concentration for inflow, while C(t) is changing concentration

  141. anonymous
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    Yep.

  142. anonymous
    • 5 years ago
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    delta t is intended to cancel the time component in Q?

  143. anonymous
    • 5 years ago
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    Yes :)

  144. anonymous
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    got it now i think .. haha

  145. anonymous
    • 5 years ago
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    i got to go grab food now...otherwise i have to eat myself tonite .see you ~~thanks !!!

  146. anonymous
    • 5 years ago
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    Ok. I'll finish it.

  147. anonymous
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    Divide both side by V, the volume of the system to obtain\[C(t+\delta t)-C(t)=\frac{Q}{V}(a-C(t))\delta t\]and divide both sides by delta t\[\frac{C(t+\delta t)-C(t)}{\delta t}=\frac{Q}{V}(a-C(t))\]

  148. anonymous
    • 5 years ago
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    Take the limit as delta_t approaches zero on both sides to get\[\frac{dC}{dt}=\frac{Q}{V}(a-C(t)) \rightarrow \frac{dC}{a-C}=\frac{Q}{V}dt \rightarrow \int\limits_{}{}\frac{dC}{a-C}=\int\limits_{}{}\frac{Q}{V}dt\]

  149. anonymous
    • 5 years ago
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    You should get\[C(t)=a-C(0)e^{-\frac{Qt}{V}}\]

  150. anonymous
    • 5 years ago
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    Oh wait...I forgot about the salt already in there! This is what happens when you do it on the fly. Go eat.

  151. anonymous
    • 5 years ago
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    I'll write something later.

  152. anonymous
    • 5 years ago
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    Okay, this is much easier when you do it on paper. This site needs something like a whiteboard. I'll write it up and scan.

  153. anonymous
    • 5 years ago
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    1 Attachment
  154. anonymous
    • 5 years ago
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    Wow! you are the most amazing person in the world haha! So for the first question do i also need to add the original amount of salt to the equation?

  155. anonymous
    • 5 years ago
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    You know, I've forgotten the first question already. I don't think so since if you put t=0 into the solution, you should get the initial amount in the tank. Just check.

  156. anonymous
    • 5 years ago
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    Oh yea. it is right

  157. anonymous
    • 5 years ago
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    Always check the initial and terminating conditions (t=0, t-> infinity) to see if your answer makes sense.

  158. anonymous
    • 5 years ago
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    LOL i am too lazy to do so when there is a smart guy in the house :)

  159. anonymous
    • 5 years ago
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    was*

  160. anonymous
    • 5 years ago
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    Yeah, flattery will get you everywhere.

  161. anonymous
    • 5 years ago
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    LOL

  162. anonymous
    • 5 years ago
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    I am just fooling around ~~

  163. anonymous
    • 5 years ago
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    I know :P

  164. anonymous
    • 5 years ago
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    Is there a way that i can use to find square root super fast

  165. anonymous
    • 5 years ago
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    What do you mean? By hand?

  166. anonymous
    • 5 years ago
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    yea or better in mind

  167. anonymous
    • 5 years ago
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    Newton-Raphson method is a numerical method for extracting solutions from equations. It has something called, "quadratic convergence" which makes it very fast. Plus, it's easy to use.

  168. anonymous
    • 5 years ago
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    I explained it to someone last week. I'll see if I can find it. If not, the result for square root is \[x_{n+1}=\frac{x_n}{2}+\frac{1}{x_n}\]

  169. anonymous
    • 5 years ago
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    You start with an initial guess, x_0, and the number it spits out on the LHS will be x_1, which will be closer to the solution. But that means that this is a BETTER guess than before, so we can plug the result into the RHS to get an EVEN BETTER guess on the LHS and so on...you repeat for infinity, but in reality, it only takes a couple of cycles to get something close.

  170. anonymous
    • 5 years ago
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    Oh, that's only for the square root of 2.

  171. anonymous
    • 5 years ago
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    LHS= left hand sum?

  172. anonymous
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    For\[y=\sqrt{a}\]it would be\[x_{n+1}=\frac{x_n}{2}+\frac{a}{2x_n}=\frac{1}{2}\left( x_n+\frac{a}{x_n} \right)\]

  173. anonymous
    • 5 years ago
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    LHS = left-hand side.

  174. anonymous
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    So, start with an initial guess for sqrt(2) of 1.5. Plug 1.5 into the RHIS of the equation and you should get an answer, 1.4166666...

  175. anonymous
    • 5 years ago
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    How you got a/(2X_n)

  176. anonymous
    • 5 years ago
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    Plug that answer into the RHS again and you get on the LHS: 1.414215686...

  177. anonymous
    • 5 years ago
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    Already it's getting close.

  178. anonymous
    • 5 years ago
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    is this something similar to newton's method?

  179. anonymous
    • 5 years ago
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    Yes, it's the same.

  180. anonymous
    • 5 years ago
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    You've used it?

  181. anonymous
    • 5 years ago
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    My professor talked a little bit about it last quarter

  182. anonymous
    • 5 years ago
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    Oh do you have any cool math tricks that you can show me haha

  183. anonymous
    • 5 years ago
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    I'd have to think... Some tricks, you can add zero or multiply by 1 and not change a number...so you can extend this idea when it comes to solving a lot of problems since 0 and 1 can be written in many convenient forms. Example \[\int\limits_{}{}\frac{u}{1+u}du=\int\limits_{}{}\frac{u+0}{1+u}du \int\limits_{}{}\frac{u+1-1}{1+u}du\]\[=\int\limits_{}{}1-\frac{1}{1+u}du=u-\log (1+u) + c\]

  184. anonymous
    • 5 years ago
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    O.o magic~

  185. anonymous
    • 5 years ago
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    Why multiplying two negatives gives a positive...

  186. anonymous
    • 5 years ago
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    cuz ..mathematicians say so LOL

  187. anonymous
    • 5 years ago
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    \[-2 \times -3 \]\[= -2 \times -3 +0 \]\[=-2 \times -3 + 2 \times 0 \]\[=-2 \times -3+2 \times (3+-3)\]\[=-2 \times -3 + 2 \times 3 + 2 \times -3\]\[=-3 \times (2+-2)+ 2 \times 3\]\[=-3 \times 0 + 2 \times 3\]\[=0+2\times 3\]\[=2 \times 3\]

  188. anonymous
    • 5 years ago
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    Everyone takes it for granted. Hardly anyone ever checks or bothers to figure out why.

  189. anonymous
    • 5 years ago
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    Yea this is how education been conducted these days

  190. anonymous
    • 5 years ago
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    Were you math major? i hear that only math majors actually get into this kind of proofs, for example proof 1+1=2

  191. anonymous
    • 5 years ago
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    yes. pure mathematics and physics.

  192. anonymous
    • 5 years ago
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    COOL so you can help me with physics too haha!!

  193. anonymous
    • 5 years ago
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    why do you have classes so late?

  194. anonymous
    • 5 years ago
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    I had no choice, i picked my class on the last day of classes registration because i didn't have many credits when i entered this school.

  195. anonymous
    • 5 years ago
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    Hassle.

  196. anonymous
    • 5 years ago
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    did you go to University of Sydney ?

  197. anonymous
    • 5 years ago
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    Partly.

  198. anonymous
    • 5 years ago
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    Postgrad. stuff at Oxford

  199. anonymous
    • 5 years ago
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    yea i had to fight for classes

  200. anonymous
    • 5 years ago
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    Isn't it a school though?

  201. anonymous
    • 5 years ago
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    LOL that explained why you have 129 fans here haha

  202. anonymous
    • 5 years ago
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    Have you heard of University of California, Davis?

  203. anonymous
    • 5 years ago
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    Ah, yes.

  204. anonymous
    • 5 years ago
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    I don't get the American system.

  205. anonymous
    • 5 years ago
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    yea i am attending Davis ..since i couldnt get into UC berkerley

  206. anonymous
    • 5 years ago
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    It is annoying and difficult to deal with..

  207. anonymous
    • 5 years ago
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    Can't you try and transfer, or do postgraduate stuff...or even try for another course?

  208. anonymous
    • 5 years ago
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    I can transfer after my sophmore year , but it is really hard to transfer from one college to another college at the same level

  209. anonymous
    • 5 years ago
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    It is much easier if you try to transfer from a city college to a University

  210. anonymous
    • 5 years ago
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    Your system's quite weird. It seems almost punitive. In the Commonwealth, it's so much easier. We can transfer easily and get exchange placements.

  211. anonymous
    • 5 years ago
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    brb

  212. anonymous
    • 5 years ago
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    Yea, i had an internship at a company last summer. One of my colleagues is an immigrant from UK. He told me that the school system here is much more complicated than the one in UK

  213. anonymous
    • 5 years ago
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    Hell yeah.

  214. anonymous
    • 5 years ago
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    I think it has to do with the fact all universities in the Commonwealth (well, most) are public, and there are links between schools that act as 'feeder schools', although you can choose where you go.

  215. anonymous
    • 5 years ago
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    The governments ensure as much as possible there is enough money to allow people places in courses they want (assuming the academics think they can handle the work).

  216. anonymous
    • 5 years ago
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    So you actually go to lower division schools, elementary, middle and high school, that help you to get into the university you want?

  217. anonymous
    • 5 years ago
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    We have nothing like that here. NOw they are evening cutting classes because of the recent budget cut

  218. anonymous
    • 5 years ago
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    No, it's not like that. Because the government funds everything, the standards of all the universities have to be the same, which means most people are happy studying at a university that is close geographically to their home. There's no (as far as I'm aware) discrimination in selection.

  219. anonymous
    • 5 years ago
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    If you want to do research, then you have to get yourself into one of the major ones here. At Sydney, I was taught and worked with people who were at Oxford, Cambridge, Yale, etc.

  220. anonymous
    • 5 years ago
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    I think there was a mathematician who worked at Stanford too.

  221. anonymous
    • 5 years ago
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    Where did you do your undersgraduate

  222. anonymous
    • 5 years ago
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    Sydney.

  223. anonymous
    • 5 years ago
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    There's a guy in my old physics school there who worked at NASA after graduating. He's back now. I think he discovered something to do with black holes.

  224. anonymous
    • 5 years ago
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    What are you going to do when you finish your studies? I get the impression with the US system that you guys study for a general degree and then use it (if you want) to get into a professional school.

  225. anonymous
    • 5 years ago
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    OMG. You are surrounded by ...genius man so jealous

  226. anonymous
    • 5 years ago
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    I dont know...what i am gonna do.. I am majoring in mechanical engineering.. i dont know what i can do with it

  227. anonymous
    • 5 years ago
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    Well here, mechanical engineers make a lot of dosh.

  228. anonymous
    • 5 years ago
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    All engineers do.

  229. anonymous
    • 5 years ago
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    Maybe you could focus on making moolah?

  230. anonymous
    • 5 years ago
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    LOL you mean i design and produce my own moolah?

  231. anonymous
    • 5 years ago
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    I mean money ;)

  232. anonymous
    • 5 years ago
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    yea that s what i meant

  233. anonymous
    • 5 years ago
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    LITTERALLY producing moolah

  234. anonymous
    • 5 years ago
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    Sorry, subtleties are lost online.

  235. anonymous
    • 5 years ago
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    Why not, your Government does.

  236. anonymous
    • 5 years ago
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    *?*

  237. anonymous
    • 5 years ago
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    LOL

  238. anonymous
    • 5 years ago
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    How about you .what you going to do after post grad

  239. anonymous
    • 5 years ago
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    Well, I'm taking a break and teaching mathematics at a grammar school in Sydney. I'll get back into research a bit later.

  240. anonymous
    • 5 years ago
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    I C

  241. anonymous
    • 5 years ago
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    Do you think the earth's magnetic field is gonna turn 180% around

  242. anonymous
    • 5 years ago
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    i asked about the international baccalaureate before because that's part of what I'm teaching.

  243. anonymous
    • 5 years ago
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    Re. magnetic field, I think it's supposed to flip, and can do so anytime 'soon', but 'soon' is in astronomical terms, so it could be like, forever.

  244. anonymous
    • 5 years ago
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    Oh

  245. anonymous
    • 5 years ago
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    What kind of research are you working on

  246. anonymous
    • 5 years ago
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    gtg i will ttyl man..Thanks for all the help~:)

  247. anonymous
    • 5 years ago
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    No worries :D

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