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anonymous

  • 5 years ago

Suppose that you take out an unsubsidized Stafford loan on September 1 before your junior year for $4500 and plan to begin paying it back on December 1 after graduation and grace period 27 months later. The interest rate is 6.8%. How much of what you will owe will be interest?

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  1. anonymous
    • 5 years ago
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    i believe its $661.76

  2. myininaya
    • 5 years ago
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    cantorset can we use induction on your problem? do you know what that is?.

  3. anonymous
    • 5 years ago
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    yes we can

  4. anonymous
    • 5 years ago
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    how did you find me, lol

  5. myininaya
    • 5 years ago
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    i went to the homepage and you can see what everyone is doing

  6. anonymous
    • 5 years ago
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    the chat program doesnt work for me

  7. anonymous
    • 5 years ago
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    ok lets do this one for all n>=5 , n^3 < 3^n

  8. myininaya
    • 5 years ago
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    ok

  9. myininaya
    • 5 years ago
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    for n=5, we have 5^3=125<243=3^5. Sold the inequality holds for n=5

  10. anonymous
    • 5 years ago
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    right, basis case

  11. myininaya
    • 5 years ago
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    now we should assume it is true for integers k>=5 Then we show it is true for k+1 the induction step is the hardest so let me think about it

  12. myininaya
    • 5 years ago
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    it seems that you are going over logs so i'm pretty sure we may have to involved that somehow

  13. anonymous
    • 5 years ago
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    hmmm, they may be unrelated though. here is what i got ,

  14. anonymous
    • 5 years ago
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    that was a different question

  15. anonymous
    • 5 years ago
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    is there a way to post without having to click my mouse on post?

  16. myininaya
    • 5 years ago
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    no i guess not

  17. anonymous
    • 5 years ago
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    show that (k+1)^3 < 3^(k+1)

  18. anonymous
    • 5 years ago
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    k^3 + 3k^2 + 3k + 1 < 3^(k+1)

  19. myininaya
    • 5 years ago
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    omg i got it

  20. myininaya
    • 5 years ago
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    i got so excited i clicked out sorry. so we have (k+1)^3<K^3+1^3=k^3+1<3^k+1

  21. anonymous
    • 5 years ago
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    k ^3 + 3k^2 + 3k + 1 < k^3 + 3k^2 + 3k + 3k = k^3 + 3k^2 + 6k < k^3 + k^3 + k^3 = 3k^3 < 3*3^k = 3^(k+1)

  22. anonymous
    • 5 years ago
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    (k+1)^3 = k ^3 + 3k^2 + 3k + 1 < k^3 + 3k^2 + 3k + 3k = k^3 + 3k^2 + 6k < k^3 + k^3 + k^3 = 3k^3 < 3*3^k = 3^(k+1)

  23. myininaya
    • 5 years ago
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    i got excited ignore what i wrote

  24. anonymous
    • 5 years ago
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    (k+1)^3<K^3+1^3=k^3+1<3^k+1 this is false

  25. anonymous
    • 5 years ago
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    my proof isnt elegant, but i think it works, might have to do some side induction arguments for the intermediate steps

  26. myininaya
    • 5 years ago
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    you are right i got excited

  27. anonymous
    • 5 years ago
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    also you better use parenthesses :) more often

  28. anonymous
    • 5 years ago
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    the log proof is A LOT easier

  29. myininaya
    • 5 years ago
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    yep

  30. anonymous
    • 5 years ago
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    what did you get? log_3 n < n if n < 3^n , assuming we raise both sides to 3

  31. anonymous
    • 5 years ago
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    or you can start, if we can show that n < 3^n , for n>=1, then taking the log base 3 of both sides we have our result. what was your logic?

  32. myininaya
    • 5 years ago
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    ok lets skip the case for n=1 that is easy to do and we know it holds. so assume for integer k, log3(k)<k (3is base) now we want to show log3(k+1)<k+1 so we have ...

  33. anonymous
    • 5 years ago
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    so that proof goes, 1 < 3^n, assume k < 3^k then add 1 to both sides of that inequality we have k + 1 < 3^k + 1 < 3^k + 3 = 3^(k+1)

  34. myininaya
    • 5 years ago
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    yea maybe we should base 3 on both sides to get rid of log

  35. anonymous
    • 5 years ago
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    you mean exponentiate?

  36. anonymous
    • 5 years ago
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    you cant get rid of log with log :)

  37. anonymous
    • 5 years ago
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    can you help me with surface area of integral

  38. myininaya
    • 5 years ago
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    surface area of a solid?

  39. anonymous
    • 5 years ago
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    surface area of a torus

  40. anonymous
    • 5 years ago
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    so you cant finish the proof with log, no problemo

  41. myininaya
    • 5 years ago
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    one thing at a time

  42. anonymous
    • 5 years ago
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    how do you find the question i posted earlier

  43. anonymous
    • 5 years ago
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    this website is so confusing

  44. myininaya
    • 5 years ago
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    are you given the equation for the torus or is it a parametric curve/

  45. anonymous
    • 5 years ago
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    Calculus question: surface area of a torus. it is generated by circle with radius r , off center. Suppose the circle is centered at (x - R)^2 + y^2 = r^2, and R> r. so the distance from (0,0) to the center of the circle is (R,0) which has radius r. Ok the integral i got is int 2pi y sqrt [ 1+ (dy/dx)^2] = int 2pi sqrt ( r^2 - ( x-R)^2) [ 1 + (-2(x-R)/(2sqrt( r^2 - ( x-R)^2) ^2 ]

  46. anonymous
    • 5 years ago
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    we are given a circle , and its going to generate the torus by revolving about the y axis

  47. anonymous
    • 5 years ago
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    the final answer should be 4pi^2 R*r

  48. myininaya
    • 5 years ago
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    so that last thing you have the square cancels the sqrt

  49. myininaya
    • 5 years ago
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    ok im just going to do it on paper and scan it and you can open it give a few min

  50. anonymous
    • 5 years ago
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    right, good :) ok here is what i got

  51. anonymous
    • 5 years ago
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    integral 2pi y dS, remember ds can be sqrt 1 + dx/dy ^2 or sqrt 1 + dy/dx ^2

  52. myininaya
    • 5 years ago
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    we need another pi in there somewhere

  53. myininaya
    • 5 years ago
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    since are answer has pi^2

  54. myininaya
    • 5 years ago
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    this is what i got using the integral you found we can get times another pi it would be correct

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  55. myininaya
    • 5 years ago
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    don't look at the name of the file. i know we are not finding volume

  56. anonymous
    • 5 years ago
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    right, its surface area, here lets use twiddla

  57. anonymous
    • 5 years ago
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    http://www.twiddla.com/521818

  58. anonymous
    • 5 years ago
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    if you dont mind, we can draw on whiteboard

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