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anonymous

  • one year ago

Which combination of integers can be used to generate the Pythagorean triple 7,24,25

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  1. misty1212
    • one year ago
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    HI!!

  2. misty1212
    • one year ago
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    the question is a little unclear, but maybe it is this: all triples can be found by taking whole number \(m,n\) and computing \[m^2-n^2, 2mn, m^2+n^2\] is that what you have to find, \(m\) and \(\)?

  3. Elsa213
    • one year ago
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    Welcome to Openstudy! :D

  4. misty1212
    • one year ago
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    @Elsa213 thanks dear, welcome to you as well!

  5. Elsa213
    • one year ago
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    Thank you cx

  6. anonymous
    • one year ago
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    its apex so the answers would be like A.x=4,y=3 B.X=1, y+3 C.x=3,y=2 D.x=2 y=2

  7. misty1212
    • one year ago
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    what does "it is apex" mean?

  8. anonymous
    • one year ago
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    and thanks for the warm welcome and its an online class

  9. misty1212
    • one year ago
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    ok it looks like they used \(x\) and \(y\) whereas i used \(m\) and \(n\) but it is the same thing

  10. anonymous
    • one year ago
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    what would be the answer?

  11. misty1212
    • one year ago
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    you have \[x^2-y^2=7\] that is the same as \[(x-y)(x+y)=7\] there is only one way to factor \(7\) as \(1\times 7\) so \[x-y=1\\ x+y=7\] and using that you can solve for \(x\) and \(y\)

  12. anonymous
    • one year ago
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    so i basically plug in and solve

  13. misty1212
    • one year ago
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    you can solve that in your head: two numbers that add to 7 and are one apart

  14. anonymous
    • one year ago
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    basicaly A

  15. misty1212
    • one year ago
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    basically A?

  16. anonymous
    • one year ago
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    the answer?

  17. misty1212
    • one year ago
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    ooh i see A is \(x=4,y=3\) yes, that is right

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