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anonymous

  • 5 years ago

i need someone to walk me step by step on a Pythagorean theorem i am having trouble with it and just a little bit of extra help

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  1. anonymous
    • 5 years ago
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  2. anonymous
    • 5 years ago
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    Zed

  3. anonymous
    • 5 years ago
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    hey tinkerbell :)

  4. anonymous
    • 5 years ago
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    help

  5. anonymous
    • 5 years ago
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    sure, just reading the file

  6. anonymous
    • 5 years ago
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    ok

  7. watchmath
    • 5 years ago
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    hello satellite :)

  8. anonymous
    • 5 years ago
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    brb zed im going to get a drink

  9. anonymous
    • 5 years ago
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    man that is confusing. pick two numbers, m and n, one even one odd. lets say i pick m = 4, n 1 then take \[m^2-n^2=4^2-1^2=15\] also \[2mn=2\times 4\times 1=8\] and finally \[m^2+n^2=4^2+1^2=17\] those three number, \[15,8,17\] will be a pythagorean triple. that is \[15^2+8^2=17^2\]

  10. anonymous
    • 5 years ago
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    watchmath!! long time no see. i just mentioned you the other day

  11. anonymous
    • 5 years ago
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    saying i missed your interesting questions/puzzles

  12. watchmath
    • 5 years ago
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    yes, you are a legend now here :). Good job!

  13. anonymous
    • 5 years ago
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    legend smegend. when i get to 10,000 medals i am going to have a whiskey and call it a day

  14. anonymous
    • 5 years ago
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    hope to see you more regularly.

  15. watchmath
    • 5 years ago
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    :). I will think some interesting problem for you

  16. anonymous
    • 5 years ago
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    ok im here sorry had to grab a bottle of water i have been sick but im back now

  17. anonymous
    • 5 years ago
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    ok, but right now i am puzzling over how to make truth tables interesting because they are boring me to death. if you have any good puzzles or exercises involving them, let me know.

  18. anonymous
    • 5 years ago
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    @tinkerbell, i wrote out a method for finding triples for you above. the worksheet you sent is rather vague (method on line) etc

  19. anonymous
    • 5 years ago
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    i have been down ill with the flu really bad

  20. anonymous
    • 5 years ago
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    i can walk you through another one if you like

  21. anonymous
    • 5 years ago
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    tinkerbell do you need to use different methods each time?

  22. anonymous
    • 5 years ago
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    It can be proved as the 2 dimensional case of the Parseval equality I think! Functional Analysis is goood. If not, then just remember this: The square on the hypotenuse is equal to the sum of the squares on the other two sides!

  23. anonymous
    • 5 years ago
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    this is the only one i see , a^2+b^2 = c^2.

  24. anonymous
    • 5 years ago
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    maybe it is not clear what you are looking for. not any three numbers a, b and c, but three whole numbers a, b and c with \[a^2+b^2=c^2\]

  25. anonymous
    • 5 years ago
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    thats what i see

  26. anonymous
    • 5 years ago
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    did you open the link on the worksheet?

  27. anonymous
    • 5 years ago
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    my assignment

  28. anonymous
    • 5 years ago
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    this link http://en.wikipedia.org/wiki/Formulas_for_generating_Pythagorean_triples

  29. anonymous
    • 5 years ago
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    that confused me

  30. anonymous
    • 5 years ago
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    i see. it looks like your worksheet is asking for 5 different triples. i wrote one above. would you like to do another one?

  31. anonymous
    • 5 years ago
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    can you rewrite it and explain as you go

  32. anonymous
    • 5 years ago
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    ok we can try the first one If a is odd, then b = a2/2 − 1/2 and c = b + 1 If a is even, then b = a2/4 − 1 and c = b + 2 pick an odd number

  33. anonymous
    • 5 years ago
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    7

  34. anonymous
    • 5 years ago
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    ok then to find "b" we compute \[b=\frac{7^2}{2}-\frac{1}{2}\] what do you get?

  35. anonymous
    • 5 years ago
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    there any other way we can do this with out fractions

  36. anonymous
    • 5 years ago
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    will 7^2 =49 so it would be 49/2-1/2

  37. anonymous
    • 5 years ago
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    this is a fraction but don't fret \[\frac{7^2}{2}-\frac{1}{2}=\frac{49-1}{2}=\frac{48}{2}=24\]

  38. anonymous
    • 5 years ago
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    a nice whole number so \[a=7,b=24\] and now \[c=b+1=24+1=25\] and that is your "triple" \[7,24,25\]

  39. anonymous
    • 5 years ago
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    and you are supposed to check that \[7^2+24^2=25^2\] which you can do with a calculator

  40. anonymous
    • 5 years ago
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    slow down

  41. anonymous
    • 5 years ago
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    let me know if you have any questions about what i wrote

  42. anonymous
    • 5 years ago
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    no i have no questions so far let me type it into the template real qick

  43. anonymous
    • 5 years ago
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    ok that gives me 625

  44. anonymous
    • 5 years ago
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    good so we have one. now we can try another one

  45. anonymous
    • 5 years ago
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    yes please

  46. anonymous
    • 5 years ago
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    Ready when you are

  47. anonymous
    • 5 years ago
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    ok now we try a different method

  48. anonymous
    • 5 years ago
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    pick two numbers m and n, where one is even and one is odd. make them not too big

  49. anonymous
    • 5 years ago
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    88 and 13

  50. anonymous
    • 5 years ago
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    whoa nice and small!

  51. anonymous
    • 5 years ago
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    you did not say that

  52. anonymous
    • 5 years ago
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    we want to make this easy, not hard. pick small numbers

  53. anonymous
    • 5 years ago
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    11 and 17

  54. anonymous
    • 5 years ago
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    ok i will pick them, one even and the other odd. i pick 5 and 2 now we compute \[5^2-2^2=25-4=21\]

  55. anonymous
    • 5 years ago
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    then i take \[2\times 5\times 2=20\] and finally \[5^2+2^2=25+4=29\] and the three numbers \[21,20,29\] are also a triple and you can check that \[21^2+20^2=29^2\]

  56. anonymous
    • 5 years ago
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    ok

  57. anonymous
    • 5 years ago
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    now we have two. do you have to use a different method for each one?

  58. anonymous
    • 5 years ago
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    yes i would like if it is possible

  59. anonymous
    • 5 years ago
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    i will le let you know when im done entering it in the template

  60. anonymous
    • 5 years ago
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    well maybe we can do one more, but if i were you i would use the last method again with two different numbers

  61. anonymous
    • 5 years ago
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    ok hang on im 3^2-8^2=9-64=55

  62. anonymous
    • 5 years ago
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    am i right so far

  63. anonymous
    • 5 years ago
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    you are on the right track, but you should make it \[8^2-3^2=64-9=55\] so you don't get negative numbers

  64. anonymous
    • 5 years ago
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    so you picked 8 and 3,and now you have 55 next two numbers will be \[2\times 8\times 3\] and \[8^2+3^2\]

  65. anonymous
    • 5 years ago
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    2x 8x 3=48

  66. anonymous
    • 5 years ago
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    8^2 + 3^2 is to check right

  67. anonymous
    • 5 years ago
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    no that is the third number, the long side

  68. anonymous
    • 5 years ago
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    73

  69. anonymous
    • 5 years ago
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    the three numbers are \[55,48,73\] and the check is \[55^2+48^2=73^2\]

  70. anonymous
    • 5 years ago
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    ok be right with you

  71. anonymous
    • 5 years ago
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    unfortunately i have to run. but you have 3 so far, and it looks like you know what you are doing, so i think you should be in good shape

  72. anonymous
    • 5 years ago
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    i just have to rememer to put the bigger number in front

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