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Nick2019

  • one year ago

I need help!!!

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  1. Nick2019
    • one year ago
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  2. cherio12
    • one year ago
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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)

  3. cherio12
    • one year ago
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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

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

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

  6. cherio12
    • one year ago
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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

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

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

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

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

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

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

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

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

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

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

  17. cherio12
    • one year ago
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    no problem =D

  18. Nick2019
    • one year ago
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    k

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

  20. cherio12
    • one year ago
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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)

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

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

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

  24. cherio12
    • one year ago
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    yes

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

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

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

  28. Nick2019
    • one year ago
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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

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

  30. Nick2019
    • one year ago
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    yes

  31. cherio12
    • one year ago
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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.

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

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

  34. cherio12
    • one year ago
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    no i got 100

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

  36. Nick2019
    • one year ago
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    ohhh

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

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

  39. cherio12
    • one year ago
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    sure

  40. cherio12
    • one year ago
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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

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

  42. cherio12
    • one year ago
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    21

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

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

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

  46. cherio12
    • one year ago
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    no problem

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

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

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

  50. Nick2019
    • one year ago
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    @cherio12

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

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

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

  54. Nick2019
    • one year ago
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    wait that doesnt work

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

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

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

  58. Nick2019
    • one year ago
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    @cherio12

  59. cherio12
    • one year ago
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    i got 7.937

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

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

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

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

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

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

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

  67. cherio12
    • one year ago
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    yes

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

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

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

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

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

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

  74. Nick2019
    • one year ago
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    so its 15.49

  75. cherio12
    • one year ago
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    yes

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

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

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

  79. Nick2019
    • one year ago
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    @cherio12

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

  81. Nick2019
    • one year ago
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    5.83

  82. Nick2019
    • one year ago
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    wait no its 4

  83. cherio12
    • one year ago
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    yes 4

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

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

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

  87. cherio12
    • one year ago
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    no, no square

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

  89. Nick2019
    • one year ago
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    so 0.8=7

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

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

  92. cherio12
    • one year ago
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    haha, no problem

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

  94. Nick2019
    • one year ago
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    @cherio12

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

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