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zaphod Group Title

Help in b) iii) only thanks :D

  • 2 years ago
  • 2 years ago

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

    • 2 years ago
  2. zaphod Group Title
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    @nbouscal @satellite73 @TuringTest

    • 2 years ago
  3. zaphod Group Title
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    @robtobey

    • 2 years ago
  4. zaphod Group Title
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    @jim_thompson5910 @dpaInc @KingGeorge @AccessDenied @SmoothMath

    • 2 years ago
  5. KingGeorge Group Title
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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?

    • 2 years ago
  6. zaphod Group Title
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    for KM i got 27 and KN i got 15

    • 2 years ago
  7. KingGeorge Group Title
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    Give me a minute to see if I was correct.

    • 2 years ago
  8. zaphod Group Title
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    okay

    • 2 years ago
  9. KingGeorge Group Title
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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

    • 2 years ago
  10. zaphod Group Title
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    what iis herons formula? can u show the woking?

    • 2 years ago
  11. KingGeorge Group Title
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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.

    • 2 years ago
  12. KingGeorge Group Title
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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 \[40\sqrt2\]

    • 2 years ago
  13. KingGeorge Group Title
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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.

    • 2 years ago
  14. zaphod Group Title
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    but the solution actually is 16/65...according to the markin gsolution...hmm

    • 2 years ago
  15. KingGeorge Group Title
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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.

    • 2 years ago
  16. KingGeorge Group Title
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    Could you briefly explain to me how you got the length of KN?

    • 2 years ago
  17. zaphod Group Title
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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 :)

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

    • 2 years ago
  19. zaphod Group Title
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    just check if ur answer is wrong, coz it has to be 16/65

    • 2 years ago
  20. zaphod Group Title
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    okay can u help in another one..its algebra?

    • 2 years ago
  21. zaphod Group Title
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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

    • 2 years ago
  22. KingGeorge Group Title
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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.

    • 2 years ago
  23. KingGeorge Group Title
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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.

    • 2 years ago
  24. zaphod Group Title
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    can u do it, im stuggling like showing a working

    • 2 years ago
  25. KingGeorge Group Title
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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, \[2x^3-5x^2-30x+50=0\]

    • 2 years ago
  26. KingGeorge Group Title
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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.

    • 2 years ago
  27. zaphod Group Title
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    well how did u get 2x^3 + 50?

    • 2 years ago
  28. zaphod Group Title
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    any other methods? to do that algebra?

    • 2 years ago
  29. KingGeorge Group Title
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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.

    • 2 years ago
  30. zaphod Group Title
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    any other easy method, because i can remembr once i did it in an easy way?

    • 2 years ago
  31. zaphod Group Title
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    it is just 1 mark question?

    • 2 years ago
  32. KingGeorge Group Title
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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?

    • 2 years ago
  33. zaphod Group Title
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    yep?

    • 2 years ago
  34. zaphod Group Title
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    Hallo?

    • 2 years ago
  35. KingGeorge Group Title
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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?

    • 2 years ago
  36. zaphod Group Title
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    no i mean, both the equations should be simplified and be equal = 0

    • 2 years ago
  37. KingGeorge Group Title
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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).

    • 2 years ago
  38. zaphod Group Title
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    okay

    • 2 years ago
  39. KingGeorge Group Title
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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.

    • 2 years ago
  40. KingGeorge Group Title
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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.

    • 2 years ago
  41. zaphod Group Title
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    how did u get 65/16 in the first place

    • 2 years ago
  42. KingGeorge Group Title
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    \[\frac{81A}{16}-A=\frac{81A}{16}-\frac{16A}{16}=\frac{81A-16A}{16}=\frac{A(81-16)}{16}=\frac{65A}{16}\]

    • 2 years ago
  43. KingGeorge Group Title
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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 \[\left(\frac{9}{4}\right)^2=\frac{81}{16}\]

    • 2 years ago
  44. zaphod Group Title
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    how did u get 9/4?

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

    • 2 years ago
  46. zaphod Group Title
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    oh okay thats the point..

    • 2 years ago
  47. KingGeorge Group Title
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    Yup. I should have seen that earlier :(

    • 2 years ago
  48. zaphod Group Title
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    alright thanks alot man, for ur time...ur a great teacher :)

    • 2 years ago
  49. KingGeorge Group Title
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    thanks for the compliment :)

    • 2 years ago
  50. KingGeorge Group Title
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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!

    • 2 years ago
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