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anonymous

  • 5 years ago

int(int(sqrt(x^2+y^2), x = y .. sqrt(8*y)), y = 0 .. 8) Please help me! :( this should be calculated using polar cor

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  1. Owlfred
    • 5 years ago
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    Hoot! You just asked your first question! Hang tight while I find people to answer it for you. You can thank people who give you good answers by clicking the 'Good Answer' button on the right!

  2. amistre64
    • 5 years ago
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    double inting on polars eh .... sounds delightful :)

  3. amistre64
    • 5 years ago
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    {SS} sqrt(x^2 +y^2) dr.dt

  4. amistre64
    • 5 years ago
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    be right back .....

  5. amistre64
    • 5 years ago
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    y <= x <= sqrt(8y) 0 <= y <= 8

  6. amistre64
    • 5 years ago
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    {SS} sqrt(x^2 +y^2) dx.dy seems more appropriate than dr.dt we gotta convert this to polar?

  7. anonymous
    • 5 years ago
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    yeah we need to convert it to polar,,what i dnt understand is..How I should change my boundaries

  8. amistre64
    • 5 years ago
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    id have to draw a picture..... :)

  9. amistre64
    • 5 years ago
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    i know the cap is a sphere ... but for the domain I need to see it

  10. anonymous
    • 5 years ago
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    This is the question though

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  11. amistre64
    • 5 years ago
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    nice :)

  12. amistre64
    • 5 years ago
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    this is my picture

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  13. amistre64
    • 5 years ago
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    as we sweep this across our dt doesnt really change does it?it goes for pi/4 to pi/2 if anything right?

  14. amistre64
    • 5 years ago
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    and our radius goes from 0 to the longest distance there is along the y=x line; or 8sqrt(2)

  15. anonymous
    • 5 years ago
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    yeah I dont understand what my radius should be..

  16. amistre64
    • 5 years ago
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    if i see it right; your radius goes from 0 to 8sqrt(2)

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  17. anonymous
    • 5 years ago
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    how do u say 8sqart2..i dnt understand that part

  18. amistre64
    • 5 years ago
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    your radius is a constant change; your theta depends on sqrt(8y) right?

  19. amistre64
    • 5 years ago
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    ok... the radius is the distance from the origin to the end of the line right? it travels up the y=x line; which is a 45 degree angle the total length of that line is from 0, the origin, up to (8,8) the common point

  20. amistre64
    • 5 years ago
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    the slant of a rt tri with 45 degrees is sqrt(2); this goes 8 times longer... so: 0 <= radius <= 8sqrt(2)

  21. amistre64
    • 5 years ago
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    pi/4 <= theta <= sqrt(8y), just gotta figure out if that sqrt(8y) need to be altered...

  22. anonymous
    • 5 years ago
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    hello anthony sir,,,how are you ??????you are excellent.

  23. amistre64
    • 5 years ago
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    howdy sath :) im fine thnx

  24. anonymous
    • 5 years ago
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    i wish i were there with you!

  25. amistre64
    • 5 years ago
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    you just bored? or bad times?

  26. anonymous
    • 5 years ago
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    nah...but am great of yours.wanna learn something from you.

  27. anonymous
    • 5 years ago
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    atleast way to study.

  28. anonymous
    • 5 years ago
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    im trying to calculate it and I got 256pi/3...but it turns out to be wrong :(

  29. amistre64
    • 5 years ago
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    {SS} f(x,y) dA <=> {SS} f(rcos, rsin)r dr.dt

  30. anonymous
    • 5 years ago
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    so i have to integrate this right int(int(r^2, r = 0 .. 8*sqrt(2)), theta = (1/4)*Pi .. (1/2)*Pi)

  31. amistre64
    • 5 years ago
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    i think the upper limit for the theta is spose to correspond to the x = sqrt(8y) function somehow

  32. amistre64
    • 5 years ago
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    x = r cos(t) y = r sin(r) right r cos(t) = sqrt(8y) r cos(t) = sqrt(8 rsin(t))

  33. amistre64
    • 5 years ago
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    r^2 cos^2(t) = 8r sin(t) solve for 't' i think

  34. amistre64
    • 5 years ago
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    or solve for r :)

  35. amistre64
    • 5 years ago
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    finding 't' sounds more reasonable to me; unless we can convert around to make it simpler

  36. anonymous
    • 5 years ago
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    this is sooo difficult.. :(

  37. amistre64
    • 5 years ago
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    well the good news is that the equation sqrt(x^2 +y^2) is just the unit sphere :) and should be easy to convert to polars

  38. anonymous
    • 5 years ago
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    yeah..as I said earlier after I used those boundaries (theetha bound as pi/4to pi/2 and r bound 0 to \[8\sqrt{2}\] the answer I got was 256pi/3 but its wrong

  39. amistre64
    • 5 years ago
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    r^2 cos^2(t) = 8r sin(t) ; /r r cos^2(t) = 8 sin(t) ; indentity r (1-sin^2(t)) = 8 sin(t) r - r sin^2(t) = 8 sin(t)...... it gonna be wrong because you sweep 90 degrees each time thru. You have to only sweep out the are from p/4 to the curve x=sqrt(8y)

  40. amistre64
    • 5 years ago
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    you are ending up taking the area 1/16th of a sphere if you do pi/4 to pi/2. and we dont want the whole area...

  41. amistre64
    • 5 years ago
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    does that make sense?

  42. anonymous
    • 5 years ago
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    so is the answer pi/16?

  43. amistre64
    • 5 years ago
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    i got no clue what the answer is yet :) I could check it with the x and y stuff, but the polar version I aint got ...

  44. anonymous
    • 5 years ago
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    yeah please...

  45. amistre64
    • 5 years ago
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    hold on...

  46. anonymous
    • 5 years ago
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    sure

  47. amistre64
    • 5 years ago
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    well, i had to seek a higher source lol

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  48. amistre64
    • 5 years ago
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    http://www.wolframalpha.com/widgets/gallery/view.jsp?id=f5f3cbf14f4f5d6d2085bf2d0fb76e8a

  49. anonymous
    • 5 years ago
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    but still it turns out to be incorrect..

  50. amistre64
    • 5 years ago
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    is it looking for an exact? or an approximation?

  51. anonymous
    • 5 years ago
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    I think we should keep the answers in terms of pi..

  52. amistre64
    • 5 years ago
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    i aint having no luck at it....

  53. anonymous
    • 5 years ago
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    its fine then..thanks a lot :)

  54. anonymous
    • 5 years ago
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    i have the answer exactly if you want. 1024/45+(1024/45)*sqrt(2)

  55. anonymous
    • 5 years ago
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    and i have the correct integral. I used maple to solve though. I can try to do it by hand if you want.

  56. anonymous
    • 5 years ago
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    \[\int\limits_{0}^{\pi/4}\int\limits_{0}^{\frac{8\sin(\theta)}{\cos^{2}(\theta)}}r^{2}drd \theta\]

  57. anonymous
    • 5 years ago
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    can you read that?

  58. anonymous
    • 5 years ago
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    ya i can read it..

  59. anonymous
    • 5 years ago
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    omg its crct....pls tell me hw u do it

  60. anonymous
    • 5 years ago
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    i have it down to this: \[8^3/6*\int\limits_{0}^{\pi/4}x*\sqrt{1+x}dx\]

  61. anonymous
    • 5 years ago
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    ok did you draw the region?

  62. anonymous
    • 5 years ago
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    wait nevermind i made a mistake that simplified integral is wrong

  63. anonymous
    • 5 years ago
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    but the first one is right

  64. anonymous
    • 5 years ago
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    amistre64 sent me the pic http://assets.openstudy.com/updates/attachments/4de1249dc9a28b0b38e067c2-amistre64-1306602087960-likethis.jpg

  65. anonymous
    • 5 years ago
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    the picture is a little strange the curve should be y=(x^2)/8, but it doesnt make much difference

  66. anonymous
    • 5 years ago
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    so we know at the top limit: x=sqrt(8y) x^2=8y r^2cos^2(theta)=8*r*sin(theta) r=8sin(theta)/(cos^2(theta)) thats where the upper limit came from

  67. anonymous
    • 5 years ago
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    and as a lower limit r=0

  68. anonymous
    • 5 years ago
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    then the maximum r is 8sqrt(2) so we want to find corresponding theta 8sqrt(2)=8sin(theta)/cos^2(theta) sqrt(2)=sin(theta)/cos^2(theta) at Pi/4 sin(theta)=sqrt(2)/2 and cos^2(theta)=1/2 so Pi/4 is our solution. theta from 0 to Pi/4

  69. anonymous
    • 5 years ago
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    so you have that integral i sent you, and you need to solve that

  70. anonymous
    • 5 years ago
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    the first integration is standard, but the second is a bit messy.

  71. anonymous
    • 5 years ago
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    here is the correct final integral: \[8^{3}/2\int\limits_{0}^{1}x \sqrt{1+x}dx\]

  72. anonymous
    • 5 years ago
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    sorry should be 8^3/6

  73. anonymous
    • 5 years ago
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    you can do the final integral by parts

  74. anonymous
    • 5 years ago
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    if you want to know how I got to the integral above let me know

  75. anonymous
    • 5 years ago
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    oh can u please tell me how u say that the maximum r is 8sqrt2

  76. anonymous
    • 5 years ago
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    sure r is in the shaded region and the maximum is at the intersection of the two curves. there is a 45 45 90 right triangle with legs of size 8 so the hypotenuse is 8sqrt(2)

  77. anonymous
    • 5 years ago
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    sqrt(8^2+8^2)=sqrt(128)=sqrt(64*2)=8sqrt(2)

  78. anonymous
    • 5 years ago
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    does the rest of the initial set up make sense?

  79. anonymous
    • 5 years ago
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    Need few minutes more..Im looking at your solution

  80. anonymous
    • 5 years ago
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    when you do the first integration and plug in limits you get: \[\int\limits_{0}^{\pi/4}8^{3}/3*(\sin^{3}(\theta)/\cos^{6}(\theta)d \theta\]

  81. anonymous
    • 5 years ago
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    or: \[8^{3}/3\int\limits_{0}^{\pi/4}\tan^{3}(\theta)\sec^{3}(\theta)d \theta\]

  82. anonymous
    • 5 years ago
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    use substitution and chose tan^2(theta)=x 2tan(theta)sec^2(theta)=dx 1+tan^2(theta)=1+x sec^2(theta)=1+x sec(theta)=sqrt(1+x) putting that all together and changing the limits of integration: \[8^{3}/6\int\limits_{0}^{1}x \sqrt{1+x} dx\]

  83. anonymous
    • 5 years ago
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    now by parts choose u=x dv=(1+x)^1/2 du=1 v=2((1+x)^(3/2))//3 and then you have a final simple power rule integral to solve

  84. anonymous
    • 5 years ago
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    \[\int\limits_{0}^{\pi/4}\int\limits_{0}^{8\sqrt{2}}r^2 drd \Theta\]

  85. anonymous
    • 5 years ago
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    so what I have to do is solve this right

  86. anonymous
    • 5 years ago
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    sorry i didnt mean the upper limit of integration on r was 8sqrt(2), I meant that was the r that let you find the limits of theta. the upper limit on r=8sin(theta)/cos^2(theta)

  87. anonymous
    • 5 years ago
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    \[\int\limits_{0}^{\frac{8\sin(\theta)}{\cos^{2}(\theta)}}r^{2}dr\] thats the inner integral

  88. anonymous
    • 5 years ago
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    whole thing you need to solve: \[\int\limits\limits_{0}^{\pi/4}\int\limits\limits_{0}^{\frac{8\sin(\theta)}{\cos^{2}(\theta)}}r^{2}drd \theta\]

  89. anonymous
    • 5 years ago
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    theta from 0 to Pi/4 you can see if you draw the picture correctly

  90. anonymous
    • 5 years ago
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    oh i see.but like isnt solving the inner integral difficult?

  91. anonymous
    • 5 years ago
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    no its not bad. Did you see my posts above, I wrote out the steps

  92. anonymous
    • 5 years ago
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    let me check..I missed it i guess :) thanks a tons btw

  93. anonymous
    • 5 years ago
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    you're welcome :) i want to be a math professor so this is fun for me.

  94. anonymous
    • 5 years ago
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    btw hw dd u get 8^3/3 for r

  95. anonymous
    • 5 years ago
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    thats from integrating r^2 int(r^2)=r^3/3. then plug in the upper and lower limits for r the constant term is 8^3/3

  96. anonymous
    • 5 years ago
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    YEAH BUT THE upper and lower limits for r is 0 to sin(theeta)/cos^2(theta) right? that meas we have to substitue those two right?

  97. anonymous
    • 5 years ago
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    0 to 8*sin(theta)/cos^2(theta).

  98. anonymous
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    there's an 8 in front of the second limit

  99. anonymous
    • 5 years ago
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    so you get 8^3/3* sin^3(theta)/cos^6(theta)

  100. anonymous
    • 5 years ago
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    then from there I posted most of the rest of the work

  101. anonymous
    • 5 years ago
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    if you scroll up a bunch I wrote the work to find that upper limit of 8sin(theta)/cos^2(theta)

  102. anonymous
    • 5 years ago
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    oh i geuss amistre's pic did make a difference he got the wrong bounds on theta

  103. anonymous
    • 5 years ago
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    yeah I see..BTW if u dont mind me asking..are u a student or what? :)

  104. anonymous
    • 5 years ago
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    Im a graduate student in math. Ill have a masters in a year and then probly go on to phd. How about you?

  105. anonymous
    • 5 years ago
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    i have 3 years towards an engineering degree and a physics minor too, but I decided to do math

  106. anonymous
    • 5 years ago
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    Engineering and Ohio State, math at Cleveland State. I am looking at University of Chicago, Michigan, Illinois, OSU, and some others for phd. Where are you studying?

  107. anonymous
    • 5 years ago
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    EE is a lot of differential equations have you taken that yet?

  108. anonymous
    • 5 years ago
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    engineering @ Ohio State I mean*

  109. anonymous
    • 5 years ago
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    I'm Ryan. What nationality is your name? I've never seen it before

  110. anonymous
    • 5 years ago
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    I've heard of Sri Lanka just not the name. Nice to meet you too:) I'm looking through your school's math courses and it looks like you have a good math program there.

  111. anonymous
    • 5 years ago
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    oh yeah..are college runs by quarter system and the pressure is a waaay tooo much to bear..:)

  112. anonymous
    • 5 years ago
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    ohio state was quarters I didn't really like it

  113. anonymous
    • 5 years ago
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    yeah...the speed is hard to bear..LOL..anyways I am new to this study room. Is this like fb I mean like can we add friends and stuff?

  114. anonymous
    • 5 years ago
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    I've only been using it for a couple weeks I'm not really sure. I was trying to figure out if you could do that and have private chats, etc. but I don't think you can.

  115. anonymous
    • 5 years ago
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    yeah it would have been great if they had that option..anyways I like to be friends with you..:) u have been really helpful for me..

  116. anonymous
    • 5 years ago
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    the picture should come up. Smiley face drawn in paint. same as my icon here

  117. anonymous
    • 5 years ago
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    k, sounds good. Ill talk to you later

  118. anonymous
    • 5 years ago
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    sure thnx and see ya

  119. anonymous
    • 5 years ago
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    hmm im on facebook now and it's not coming up

  120. anonymous
    • 5 years ago
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    shud come up any second//

  121. anonymous
    • 5 years ago
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    k

  122. anonymous
    • 5 years ago
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    allright got it. cya!

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