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pablobegins Group Title

Really annoying trigonometric graphing, help!

  • 3 years ago
  • 3 years ago

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  1. pablobegins Group Title
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    • 3 years ago
  2. pablobegins Group Title
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    I need 1,2,5, and 10-24

    • 3 years ago
  3. pablobegins Group Title
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    I'm just not grasping the concept.

    • 3 years ago
  4. amistre64 Group Title
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    f(x) = tan πx/4 g(x) =1/2 sec πx/4 Approximate the interval where f < g. lets set this up then, we know f and g tan (πx/4) < 1/2 sec(πx/4)

    • 3 years ago
  5. amistre64 Group Title
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    one thing that might help is to rewrite sec, or maybe even tan. into equivalent terms

    • 3 years ago
  6. pablobegins Group Title
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    how so?

    • 3 years ago
  7. amistre64 Group Title
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    well, tan = sin/cos, right? and sec = 1/cos .... that might help us see the resemblence

    • 3 years ago
  8. amistre64 Group Title
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    \[\frac{sin (πx/4)}{cos (πx/4)} < \frac{1}{2}\frac{1}{cos(πx/4)}\] we can get rid of that 1/2 by multiplying by 2 \[\frac{(2)sin (πx/4)}{cos (πx/4)} < \frac{1(2)}{2}\frac{1}{cos(πx/4)}\] \[\frac{2sin (πx/4)}{cos (πx/4)} < \frac{1}{cos(πx/4)}\] now it looks better

    • 3 years ago
  9. amistre64 Group Title
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    since the denominators are the same, lets equate numberators

    • 3 years ago
  10. amistre64 Group Title
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    2sin(pi x/4) < 1

    • 3 years ago
  11. pablobegins Group Title
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    so now we graph the equation to find the point?

    • 3 years ago
  12. amistre64 Group Title
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    id stick to the analysis rather than a graph; the graph can be used to dbl chk the results; but i doubt it will give a definitive result

    • 3 years ago
  13. amistre64 Group Title
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    sin (t) < 1/2; well, when does sin(t) = 1/2?

    • 3 years ago
  14. pablobegins Group Title
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    how would we go about the analysis?

    • 3 years ago
  15. amistre64 Group Title
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    divide each side by 2 to get: sin(t) < 1/2 im using "y" to help clean up the argument

    • 3 years ago
  16. amistre64 Group Title
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    "t" that is ... cant type ;)

    • 3 years ago
  17. pablobegins Group Title
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    that's not one of the choices though :/ ,

    • 3 years ago
  18. amistre64 Group Title
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    why would it be? we are only in the middle of it

    • 3 years ago
  19. GT Group Title
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    Hang in there and understand the steps.

    • 3 years ago
  20. pablobegins Group Title
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    OH! Gotcha. I just have a terrible teacher who goes about everything explaining really fast, and then expecting us to know all this. I'm sorry.

    • 3 years ago
  21. amistre64 Group Title
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    it would be good to remember the basic angles of trig; i believe this is one of them

    • 3 years ago
  22. amistre64 Group Title
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    |dw:1323627720794:dw|

    • 3 years ago
  23. amistre64 Group Title
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    sin(t) = 1/2, when t = 30 degrees, or pi/6

    • 3 years ago
  24. pablobegins Group Title
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    is that the special right triangle?

    • 3 years ago
  25. amistre64 Group Title
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    it is

    • 3 years ago
  26. amistre64 Group Title
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    i have to remember the interval: -pi/4 to pi/4, that is between 45 and -45 degrees if i recall it correctly

    • 3 years ago
  27. amistre64 Group Title
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    now i would see about equating pi x/4 and t pi x/4 = pi/6 divide off the pis x/4 = 1/6 and multiply off the 1/4 x = 4/6 , and simplify x = 2/3 so, if i did it right, it should be (-1, 2/3) but if you have questions as to why, it would be good to ask. The numbers tend to be unimportant and it is the process that matters.

    • 3 years ago
  28. pablobegins Group Title
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    why the -1?

    • 3 years ago
  29. amistre64 Group Title
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    its part of our initial interval; and sin(pi *-1/4) is less than 1/2

    • 3 years ago
  30. pablobegins Group Title
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    i see. so for number 2 i would just multiply our original interval?

    • 3 years ago
  31. amistre64 Group Title
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    it would be nice if we could, but no. trig functions are periodic, they do not act like linear functions. so its best to retrace the steps to make sure

    • 3 years ago
  32. pablobegins Group Title
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    or would it not change?

    • 3 years ago
  33. amistre64 Group Title
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    2f just changes how high or low the graph would go; so id guess that its the same intervals. But i would still have to dbl chk the results

    • 3 years ago
  34. pablobegins Group Title
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    its not it.

    • 3 years ago
  35. pablobegins Group Title
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    like, number 1 was right, but 2f changed it in some way, because now it's not the same

    • 3 years ago
  36. amistre64 Group Title
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    \[2*\frac{sin (πx/4)}{cos (πx/4)} < 2*\frac{1}{2}\frac{1}{cos(πx/4)}\] \[\frac{2sin (πx/4)}{cos (πx/4)} < \frac{1}{cos(πx/4)}\] \[2sin (πx/4) < 1\] \[sin (πx/4) < 1/2\] it looks to be the same

    • 3 years ago
  37. pablobegins Group Title
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    must be a problem with the software. now for number 5,i cant tell the graphs apart

    • 3 years ago
  38. amistre64 Group Title
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    which one is number 5? this thing is hard to follow

    • 3 years ago
  39. pablobegins Group Title
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    graph for y=csc (x)

    • 3 years ago
  40. amistre64 Group Title
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    well, I know csc is the humpbacks of sine|dw:1323628829369:dw|

    • 3 years ago
  41. amistre64 Group Title
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    the closest one I see is A

    • 3 years ago
  42. amistre64 Group Title
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    we can draw the sin(x) in A and see that csc rides the humpbacks

    • 3 years ago
  43. pablobegins Group Title
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    yeah, me too! thanks so much

    • 3 years ago
  44. pablobegins Group Title
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    number 10?

    • 3 years ago
  45. amistre64 Group Title
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    id ditch the graph... see that sqrt(3)? that comes about in the 30-60-90 triangle

    • 3 years ago
  46. amistre64 Group Title
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    and since cot = cos/sin; this is negative when either cos is negative OR sin is negative; but not both. that occurs in q2 and q4

    • 3 years ago
  47. pablobegins Group Title
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    so we would relate the degrees with a sin fuction?

    • 3 years ago
  48. amistre64 Group Title
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    |dw:1323629102467:dw|

    • 3 years ago
  49. amistre64 Group Title
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    almost; lets draw the 30-60-90 tri again and see what cot comes from

    • 3 years ago
  50. amistre64 Group Title
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    |dw:1323629160526:dw| the basic angle is then cot(30) do you agree? 30 = pi/6

    • 3 years ago
  51. pablobegins Group Title
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    yeah, i see where you're coming from

    • 3 years ago
  52. amistre64 Group Title
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    q2: 180-30 = 150 degrees q4: 360 - 30 = 330 degrees

    • 3 years ago
  53. amistre64 Group Title
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    hmmm, 5pi/6 and 11pi/6 seem to fit in there somewhere. but i do get these mixed aroung at times :)

    • 3 years ago
  54. amistre64 Group Title
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    i see the interval now; it one full revolution forward and one full revolution backwards

    • 3 years ago
  55. pablobegins Group Title
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    ignore that

    • 3 years ago
  56. pablobegins Group Title
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    it sent a previous message

    • 3 years ago
  57. pablobegins Group Title
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    would the pi/6 be negative? since were dealing with cot?

    • 3 years ago
  58. amistre64 Group Title
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    |dw:1323629494595:dw|

    • 3 years ago
  59. amistre64 Group Title
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    -pi/6, 5pi/6, -7pi/6, 11pi/6 is my best assessment

    • 3 years ago
  60. amistre64 Group Title
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    thats option 4 right?

    • 3 years ago
  61. pablobegins Group Title
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    for 11, is the period 16/

    • 3 years ago
  62. pablobegins Group Title
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    YES, YOU WERE RIGHT ONCE AGAIN! :)

    • 3 years ago
  63. amistre64 Group Title
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    tan and cot have a normal period of pi; this speeds that period up; by a factor of 8.; so id say its a period of 1/8

    • 3 years ago
  64. amistre64 Group Title
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    that doesnt work out right ....

    • 3 years ago
  65. pablobegins Group Title
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    if we divide it by 2pi, you get 16

    • 3 years ago
  66. amistre64 Group Title
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    that does sound familiar :)

    • 3 years ago
  67. amistre64 Group Title
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    but the 2pi is wrong

    • 3 years ago
  68. amistre64 Group Title
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    divide by pi; since this is a pi period 2pi is for sin and cos

    • 3 years ago
  69. pablobegins Group Title
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    so it would be 1/8?

    • 3 years ago
  70. amistre64 Group Title
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    pi/(pi/8) = pi * 8/pi = 8

    • 3 years ago
  71. amistre64 Group Title
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    normal period/w

    • 3 years ago
  72. pablobegins Group Title
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    and the normal period is 8??

    • 3 years ago
  73. amistre64 Group Title
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    look again

    • 3 years ago
  74. amistre64 Group Title
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    tan/cot has a normal pf pi everything else is 2pi

    • 3 years ago
  75. amistre64 Group Title
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    pi/(w) = periodicity

    • 3 years ago
  76. amistre64 Group Title
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    since w = pi/8 .... we plug it in

    • 3 years ago
  77. pablobegins Group Title
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    we end up with 1/8.

    • 3 years ago
  78. amistre64 Group Title
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    \[\frac{pi}{pi/8}=\frac{pi*8}{pi}=8\]

    • 3 years ago
  79. pablobegins Group Title
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    oh, i was solving it incorrectly. can you explain 12 to me please? i know it's similar to 10, but i want to grasp the concept.

    • 3 years ago
  80. amistre64 Group Title
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    since csc rides the humpbacks of sin; lets equate this to an angle of sin. csc = 1/sin is a good thing to remember

    • 3 years ago
  81. amistre64 Group Title
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    \[csc(x)=\frac{2\sqrt{3}}{3}\] \[\frac{1}{sin(x)}=\frac{2\sqrt{3}}{3}\] \[\frac{sin(x)}{1}=\frac{3}{2\sqrt{3}}\] and draw a triangle to help visualise it

    • 3 years ago
  82. amistre64 Group Title
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    |dw:1323630627950:dw|

    • 3 years ago
  83. amistre64 Group Title
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    this doesnt seem to match any of our basic triangles does it?

    • 3 years ago
  84. amistre64 Group Title
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    maybe .... divide off sqrt(3) 2sqrt(3)/sqrt(3) = 2 3/sqrt(3) = 3sqrt(3)/3 = sqrt(3) its our 30-60-90 in disguise

    • 3 years ago
  85. amistre64 Group Title
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    |dw:1323630843700:dw|

    • 3 years ago
  86. pablobegins Group Title
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    'disguise' lol

    • 3 years ago
  87. pablobegins Group Title
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    so we're just rearranging the equation so the triangle makes sense, in a way

    • 3 years ago
  88. amistre64 Group Title
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    \[csc(60) = \frac{2}{\sqrt{3}}\] \[csc(60) = (\frac{2}{\sqrt{3}}*\frac{\sqrt{3}}{\sqrt{3}})\] \[csc(60) = \frac{2\sqrt{3}}{3}\] so we are good :)

    • 3 years ago
  89. amistre64 Group Title
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    60 = pi/3 .... so we need to adapt that to the interval

    • 3 years ago
  90. amistre64 Group Title
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    |dw:1323631158475:dw|

    • 3 years ago
  91. amistre64 Group Title
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    |dw:1323631248664:dw|

    • 3 years ago
  92. amistre64 Group Title
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    thatss what, option 7?

    • 3 years ago
  93. pablobegins Group Title
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    seven is incorrect :-(nt and so is 3

    • 3 years ago
  94. amistre64 Group Title
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    trig takes alot of mental gymnastics :)

    • 3 years ago
  95. amistre64 Group Title
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    hmmm

    • 3 years ago
  96. amistre64 Group Title
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    i think I know where I went astray at

    • 3 years ago
  97. amistre64 Group Title
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    csc is only positive in q1 and q2, i confused it with a tan ....

    • 3 years ago
  98. amistre64 Group Title
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    |dw:1323631597639:dw|

    • 3 years ago
  99. amistre64 Group Title
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    -5,-4,1,2 ---------, might be better 3

    • 3 years ago
  100. pablobegins Group Title
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    what choice is that?

    • 3 years ago
  101. amistre64 Group Title
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    option 4, since that aint in the choices

    • 3 years ago
  102. amistre64 Group Title
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    youre right tho, these ARE really annoying ;)

    • 3 years ago
  103. pablobegins Group Title
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    finally correct! this thing penalizes me for every wrong answer too... now for the period of number 13, is 7?

    • 3 years ago
  104. amistre64 Group Title
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    the value attached to x is out "w" and sec is a 2pi normal \[\frac{2pi}{-7pi/2}\] \[\frac{2pi*2}{-7pi}\] \[\frac{2*2}{-7}=-4/7\] id go with option 1

    • 3 years ago
  105. pablobegins Group Title
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    ow do we find the frequency?

    • 3 years ago
  106. amistre64 Group Title
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    frequency is how fast it moves along .... id have to look it up :)

    • 3 years ago
  107. amistre64 Group Title
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    http://regentsprep.org/Regents/math/algtrig/ATT7/graphvocab.htm

    • 3 years ago
  108. amistre64 Group Title
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    freq = 1/period

    • 3 years ago
  109. pablobegins Group Title
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    initial phase?

    • 3 years ago
  110. amistre64 Group Title
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    i dont recall ever doing an initial phase ...

    • 3 years ago
  111. amistre64 Group Title
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    is that a phase shift perhaps?

    • 3 years ago
  112. pablobegins Group Title
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    it says initial phase, it might be referring to the same thing

    • 3 years ago
  113. amistre64 Group Title
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    i cant make any sense of the online stuff. I doubt that it is the same thing then

    • 3 years ago
  114. amistre64 Group Title
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    initial phase is when t=0 ....

    • 3 years ago
  115. amistre64 Group Title
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    but i got no idea at the moment what the means

    • 3 years ago
  116. pablobegins Group Title
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    what about the period of 1 7? i got 10/9, but that's apparently wrong

    • 3 years ago
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