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zaphod

Help in b) iii) only thanks :D

  • one year ago
  • one year ago

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  1. zaphod
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    http://screencast.com/t/Q0eyWYsQ

    • one year ago
  2. zaphod
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    @nbouscal @satellite73 @TuringTest

    • one year ago
  3. zaphod
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    @robtobey

    • one year ago
  4. zaphod
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    @jim_thompson5910 @dpaInc @KingGeorge @AccessDenied @SmoothMath

    • one year ago
  5. KingGeorge
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    I'm pretty sure I have the answer, but I'm not sure yet. What lengths did you get for KM and KN?

    • one year ago
  6. zaphod
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    for KM i got 27 and KN i got 15

    • one year ago
  7. KingGeorge
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    Give me a minute to see if I was correct.

    • one year ago
  8. zaphod
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    okay

    • one year ago
  9. KingGeorge
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    So I have the answer, but I don't think I did it a very good way. Basically, I found the area of KLM using herons formula and the area of LNM using heron's formula. You also know using some alternating angles of the parallel lines, that triangle LNM is twice the area of triangle NPM, so that let's you find the area of NPM, and then you subtract that off from the area of KLM to find the area of the trapezium KLPN

    • one year ago
  10. zaphod
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    what iis herons formula? can u show the woking?

    • one year ago
  11. KingGeorge
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    Heron's formula is a way to find the area of any triangle if you know the side lengths. If you want the general formula, look here: http://en.wikipedia.org/wiki/Heron's_formula Otherwise, I'll just skip the more tedious work and show you how I got the solution.

    • one year ago
  12. KingGeorge
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    (I'm skipping the simplification here, but you should check it on your own) Using that, we get that the area of KLM is \[90\sqrt{2}\]and the area of LNM is \[40\sqrt2\]

    • one year ago
  13. KingGeorge
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    Thus, the area of NPM is \[\frac{40\sqrt2}{2}=20\sqrt2\]Now, we just put all these numbers in the ratio \[\frac{20\sqrt2}{90\sqrt2-20\sqrt2}=\frac{\sqrt2}{\sqrt2}\cdot\frac{20}{70}=\frac{2}{7}\]And there's the solution.

    • one year ago
  14. zaphod
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    but the solution actually is 16/65...according to the markin gsolution...hmm

    • one year ago
  15. KingGeorge
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    It might be a typo since \[\frac{2}{7}=\frac{16}{56}\]So the two possibilities are that there's a typo, or your lengths for KM and KN were incorrect.

    • one year ago
  16. KingGeorge
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    Could you briefly explain to me how you got the length of KN?

    • one year ago
  17. zaphod
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    i used similarity triangle method...like il show the working...first i found KM LK/LN=KM/LM 15/10=KM/18 KM=27 then i found NM= LK/LM=LM/NM 15/10=18/NM NM=12 then KN= KM-NM 27-12= 15cm :)

    • one year ago
  18. KingGeorge
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    Definitely no mistakes there. That means that I'm pretty sure it's a typo in the book.

    • one year ago
  19. zaphod
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    just check if ur answer is wrong, coz it has to be 16/65

    • one year ago
  20. zaphod
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    okay can u help in another one..its algebra?

    • one year ago
  21. zaphod
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    \[\frac{x^{2}}{5} + \frac{5}{x} = \frac{1}{2}x +3\] can u equal them so that one side will have 0

    • one year ago
  22. KingGeorge
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    Sure, give me a few more seconds to finish checking my first solution. I came up with another, far easier way to find the ration.

    • one year ago
  23. KingGeorge
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    I might have found a better solution... It's not working out as well as I had anticipated. As for the algebra question, multiply everything by \(x\). You'll get a cubic where you can easily move everything to one side, and hopefully be able to factor.

    • one year ago
  24. zaphod
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    can u do it, im stuggling like showing a working

    • one year ago
  25. KingGeorge
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    It might even be better to multiply the thing by 10x to get rid of the 5 and the 2 in the denominators as well. That would get you \[2x^3+50=5x^2+30x\]Move everything to one side, \[2x^3-5x^2-30x+50=0\]

    • one year ago
  26. KingGeorge
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    Unfortunately, that doesn't seem to be factoring very easily. And you were right to call me out on my previous solution to the triangle question. I don't think I'm right.

    • one year ago
  27. zaphod
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    well how did u get 2x^3 + 50?

    • one year ago
  28. zaphod
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    any other methods? to do that algebra?

    • one year ago
  29. KingGeorge
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    \[x(\frac{x^{2}}{5} + \frac{5}{x}) = x(\frac{1}{2}x +3)\]Distribute the x\[(\frac{x^{3}}{5} + 5) = (\frac{1}{2}x^2 +3x)\]Now mulitply everything by 10 to get \[(\frac{10x^{3}}{5} + 50) = (\frac{10x^2}{2} +30x)\]Simplify things, \[2x^3 + 50 = 5x^2 +30x\]and move everything to one side.

    • one year ago
  30. zaphod
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    any other easy method, because i can remembr once i did it in an easy way?

    • one year ago
  31. zaphod
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    it is just 1 mark question?

    • one year ago
  32. KingGeorge
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    I'm not sure I understand. Do you just want to set one side equal to 0 and solve for x on the other side?

    • one year ago
  33. zaphod
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    yep?

    • one year ago
  34. zaphod
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    Hallo?

    • one year ago
  35. KingGeorge
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    I'm still here. So if we set one side equal to 0, we want to find solutions to the equations \[\frac{x^2}{5}+\frac{5}{x}=0\]and \[0=\frac{x}{2}+3\] correct?

    • one year ago
  36. zaphod
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    no i mean, both the equations should be simplified and be equal = 0

    • one year ago
  37. KingGeorge
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    Can we come back to that in a minute, I just figured out the easy way to do the triangle problem, and the book's correct. (It really shouldn't have taken me this long).

    • one year ago
  38. zaphod
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    okay

    • one year ago
  39. KingGeorge
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    Since NP is parallel to KL, you can show that triangle NPM is similar to triangle KLM. You know that angle K is the same as angle MNP, by corresponding angles. Also, for the same reason, angle KLM is the same as angle NPM, so this means that the triangles are similar. Now let the area of triangle NPM=A. The ratio between the sidelengths of NPM and KLM is exactly 9/4. Since the formula for area of a triangle is 1/2 bh the area vary as a square, so we square it to get that the area of KLM=81/16 A.

    • one year ago
  40. KingGeorge
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    The area of the trapezium is the area of KLM minus the area of NPM. So that area is given by \[\frac{81A}{16}-A=\frac{65A}{16}\]Now we take A over that, and we get \[\frac{A}{\frac{65A}{16}}=\frac{16A}{65A}=\frac{16}{65}\]is the ratio.

    • one year ago
  41. zaphod
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    how did u get 65/16 in the first place

    • one year ago
  42. KingGeorge
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    \[\frac{81A}{16}-A=\frac{81A}{16}-\frac{16A}{16}=\frac{81A-16A}{16}=\frac{A(81-16)}{16}=\frac{65A}{16}\]

    • one year ago
  43. KingGeorge
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    And I got the 81/16 because the ratio of the side lengths is 9/4, and the ratio of the areas (of the triangles) is the ration of the side lengths squared. So \[\left(\frac{9}{4}\right)^2=\frac{81}{16}\]

    • one year ago
  44. zaphod
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    how did u get 9/4?

    • one year ago
  45. KingGeorge
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    \[\frac{27}{12}=\frac{9}{4}\]I just simplified the ratio between KM and NM

    • one year ago
  46. zaphod
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    oh okay thats the point..

    • one year ago
  47. KingGeorge
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    Yup. I should have seen that earlier :(

    • one year ago
  48. zaphod
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    alright thanks alot man, for ur time...ur a great teacher :)

    • one year ago
  49. KingGeorge
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    thanks for the compliment :)

    • one year ago
  50. KingGeorge
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    I've got to go, so if you still need help with the algebra question, just post it in another question, and hopefully some one will answer it. Good luck!

    • one year ago
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