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Nick2019 Group Title

I need help!!!

  • one year ago
  • one year ago

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  1. Nick2019 Group Title
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    • one year ago
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  2. cherio12 Group Title
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    okay well the theorem is \[z^2=x^2+y^2\] and you know everything has to be an integer (whole number)

    • one year ago
  3. cherio12 Group Title
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    you can just try different values. For instance for the first one i might try z=51 and x=24. so 51^2=24^2+y^2. Solving i get y=45. So this works

    • one year ago
  4. Nick2019 Group Title
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    can you try 19?

    • one year ago
  5. Nick2019 Group Title
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    @cherio12 can you try #19 plz

    • one year ago
  6. cherio12 Group Title
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    you just have to play with the numbers. I first tried \[z^2=25^2+20^\] but i got z= 32.0156..so this didn't work next i tried \[25^2=20^2+y^2\] y=15 (this is an integer) so it checks out

    • one year ago
  7. cherio12 Group Title
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    does this make sense?

    • one year ago
  8. Nick2019 Group Title
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    thats the answer in the book so i should be able to learn off your 2 answers

    • one year ago
  9. Nick2019 Group Title
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    wait so on 18# i got 2601=576+y2 then do i get y by itself? @cherio12

    • one year ago
  10. cherio12 Group Title
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    yes. you should go 2601-576=y^2. then take the squareroot of both sides to just get y

    • one year ago
  11. Nick2019 Group Title
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    ok how do you know were to put the numbers on the equation?

    • one year ago
  12. cherio12 Group Title
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    what do you mean?

    • one year ago
  13. Nick2019 Group Title
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    wait oh i see you said move them around and it has to be a whole number

    • one year ago
  14. cherio12 Group Title
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    yup, its always (leg)^2+(leg)^2=(hypotenuse)^2

    • one year ago
  15. Nick2019 Group Title
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    thanks man! helped alot

    • one year ago
  16. cherio12 Group Title
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    its commonly written as x^2+y^2=z^2

    • one year ago
  17. cherio12 Group Title
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    no problem =D

    • one year ago
  18. Nick2019 Group Title
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    k

    • one year ago
  19. Nick2019 Group Title
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    wait so on # 20 i got 91.82592222ect

    • one year ago
  20. cherio12 Group Title
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    okay so that means you picked the wrong combo of hypotenuse and legs. try another combo (by the way the hypotenuse will always be the largest of the 3 numbers)

    • one year ago
  21. Nick2019 Group Title
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    so set it up as x2= 96 + 28?

    • one year ago
  22. cherio12 Group Title
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    i would try \[96^2=x^2+28^2\]

    • one year ago
  23. Nick2019 Group Title
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    do get x by its self?

    • one year ago
  24. cherio12 Group Title
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    yes

    • one year ago
  25. cherio12 Group Title
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    |dw:1359577137858:dw|

    • one year ago
  26. Nick2019 Group Title
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    but i still will get that 91.etc right?

    • one year ago
  27. cherio12 Group Title
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    i was just showing you the triangle to help show what i meant by x, y and z

    • one year ago
  28. Nick2019 Group Title
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    i know im saying when i get x by its self and take 9216 and square it i get that 91.00202 number

    • one year ago
  29. cherio12 Group Title
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    o you mean for the 96, 28 problem?

    • one year ago
  30. Nick2019 Group Title
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    yes

    • one year ago
  31. cherio12 Group Title
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    \[96^2=x^2+28^2\] \[9216=x^2+784\] \[9216-784=x^2\] \[x=\sqrt{8432}\] x=91.8259 so this way didn't work.

    • one year ago
  32. cherio12 Group Title
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    i would then try \[z^2=96^2+28^2\]

    • one year ago
  33. Nick2019 Group Title
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    but you would still get 91.8259 right?

    • one year ago
  34. cherio12 Group Title
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    no i got 100

    • one year ago
  35. cherio12 Group Title
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    you add 96^2 and 28^2 in this example instead of subtracting them (like we tried before)

    • one year ago
  36. Nick2019 Group Title
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    ohhh

    • one year ago
  37. Nick2019 Group Title
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    now i see thakns agian lol

    • one year ago
  38. Nick2019 Group Title
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    hey can you help @cherio12 with # 23

    • one year ago
  39. cherio12 Group Title
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    sure

    • one year ago
  40. cherio12 Group Title
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    okay so i first tried z^2=72^2+75^2 But i got z=103.966 so that's not right then i tried 75^2=x^2+72^2

    • one year ago
  41. Nick2019 Group Title
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    ok i did the first but didnt try that one what did you get?

    • one year ago
  42. cherio12 Group Title
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    21

    • one year ago
  43. Nick2019 Group Title
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    can you show how you did that?

    • one year ago
  44. cherio12 Group Title
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    75^2=x^2+72^2 solve for x

    • one year ago
  45. Nick2019 Group Title
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    ok thanks i got 21

    • one year ago
  46. cherio12 Group Title
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    no problem

    • one year ago
  47. Nick2019 Group Title
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    hey @cherio12 for #24 you you do the same?

    • one year ago
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  48. cherio12 Group Title
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    yes, your two legs are 6 and 3

    • one year ago
  49. Nick2019 Group Title
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    so would it be 6(2)=3(2)+x2

    • one year ago
  50. Nick2019 Group Title
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    @cherio12

    • one year ago
  51. cherio12 Group Title
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    by(2) do you mean you are squaring it?

    • one year ago
  52. Nick2019 Group Title
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    no 6 to the power 2 like 6*6

    • one year ago
  53. cherio12 Group Title
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    that means squaring, but yes that is correct

    • one year ago
  54. Nick2019 Group Title
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    wait that doesnt work

    • one year ago
  55. Nick2019 Group Title
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    i tried both ways and got not a whole number

    • one year ago
  56. cherio12 Group Title
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    I don't believe you are supossed to get an integer for this problem

    • one year ago
  57. Nick2019 Group Title
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    so would 5.19 work?

    • one year ago
  58. Nick2019 Group Title
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    @cherio12

    • one year ago
  59. cherio12 Group Title
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    i got 7.937

    • one year ago
  60. cherio12 Group Title
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    6^2+3^2=x^2

    • one year ago
  61. Nick2019 Group Title
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    ya that was my second way

    • one year ago
  62. Nick2019 Group Title
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    so would you set up # 25 and # 26 the same way?

    • one year ago
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  63. cherio12 Group Title
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    yes for 25, but for 26 you will use a ratio

    • one year ago
  64. cherio12 Group Title
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    i gotta go eat..be back later

    • one year ago
  65. Nick2019 Group Title
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    ok when you get back can you explain ratio?

    • one year ago
  66. Nick2019 Group Title
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    @cherio12 you back?

    • one year ago
  67. cherio12 Group Title
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    yes

    • one year ago
  68. cherio12 Group Title
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    okay well for starters find the other leg on the 5,3 triangle

    • one year ago
  69. Nick2019 Group Title
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    wait on the # 25 its just 11 and x so is it 121 then sqare root answer is 11?

    • one year ago
  70. cherio12 Group Title
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    no, there are two 11 sides

    • one year ago
  71. Nick2019 Group Title
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    oh i see cuase of that line thingy

    • one year ago
  72. Nick2019 Group Title
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    so the answer is 0

    • one year ago
  73. cherio12 Group Title
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    no for 25, it is 11^1+11^2=x^2

    • one year ago
  74. Nick2019 Group Title
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    so its 15.49

    • one year ago
  75. cherio12 Group Title
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    yes

    • one year ago
  76. cherio12 Group Title
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    does that make sense why?

    • one year ago
  77. Nick2019 Group Title
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    yes i just didnt try that way and remember there not whole numbers

    • one year ago
  78. Nick2019 Group Title
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    so how do you do #26 is it 2 problems in 1

    • one year ago
  79. Nick2019 Group Title
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    @cherio12

    • one year ago
  80. cherio12 Group Title
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    find the other leg in the 5,3 triangle first

    • one year ago
  81. Nick2019 Group Title
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    5.83

    • one year ago
  82. Nick2019 Group Title
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    wait no its 4

    • one year ago
  83. cherio12 Group Title
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    yes 4

    • one year ago
  84. Nick2019 Group Title
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    now do the other triangel?

    • one year ago
  85. cherio12 Group Title
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    now we can use like triangle ratios 4/5=7/x

    • one year ago
  86. Nick2019 Group Title
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    so 4 sqare + 5 square = 7sqare?

    • one year ago
  87. cherio12 Group Title
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    no, no square

    • one year ago
  88. cherio12 Group Title
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    just ratio, just division

    • one year ago
  89. Nick2019 Group Title
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    so 0.8=7

    • one year ago
  90. cherio12 Group Title
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    4/5=7/x .8=7/x .8x=7

    • one year ago
  91. Nick2019 Group Title
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    oh ok wow being sick for one day of notes can rweally heart you

    • one year ago
  92. cherio12 Group Title
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    haha, no problem

    • one year ago
  93. Nick2019 Group Title
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    wait # 28 its 2x and 2x+4 and the length of the hypot is 4x-4

    • one year ago
  94. Nick2019 Group Title
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    @cherio12

    • one year ago
  95. cherio12 Group Title
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    yes, just apply the Pythagorean theorem

    • one year ago
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