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userneedinghelp

I need help with the problem photographed at http://pbckt.com/pK.uB4qrd Thanks. Note tis not spam.

  • one year ago
  • one year ago

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  1. userneedinghelp
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    @AriPotta

    • one year ago
  2. AriPotta
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    does that box say 225cm^3?

    • one year ago
  3. userneedinghelp
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    yes ma'am

    • one year ago
  4. across
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    Use the Pythagorean theorem.

    • one year ago
  5. AriPotta
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    mmk. so we know it's a square and it has an area of 225. how would we find the side length?

    • one year ago
  6. userneedinghelp
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    a2+b2=c2?

    • one year ago
  7. userneedinghelp
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    o side length is the square root of the area

    • one year ago
  8. across
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    Yes, now identify \(a\), \(b\) and \(c\) and solve for the unknown variable.

    • one year ago
  9. userneedinghelp
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    . ill solve n tell u what i got in five mins s, tnx

    • one year ago
  10. userneedinghelp
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    will you wait five mins plz?

    • one year ago
  11. userneedinghelp
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    u find the sq root of 17 and 225?

    • one year ago
  12. AriPotta
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    just find the square root of 225. 17 is the side length of that one square

    • one year ago
  13. userneedinghelp
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    oh. alright,.

    • one year ago
  14. userneedinghelp
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    tis 15

    • one year ago
  15. userneedinghelp
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    the sq root of 17 is about 8.5

    • one year ago
  16. AriPotta
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    you don't need the square root of 17

    • one year ago
  17. userneedinghelp
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    according to yahoo answers

    • one year ago
  18. userneedinghelp
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    oh mk

    • one year ago
  19. userneedinghelp
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    so 15+ 17 = c2?

    • one year ago
  20. AriPotta
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    a^2 + 15^2 = 17^2

    • one year ago
  21. userneedinghelp
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    wait. 17 is c. sorry

    • one year ago
  22. userneedinghelp
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    yea mk. let me solve one second.

    • one year ago
  23. userneedinghelp
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    i got 8

    • one year ago
  24. AriPotta
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    me too

    • one year ago
  25. userneedinghelp
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    cm^2

    • one year ago
  26. AriPotta
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    no, just 8cm

    • one year ago
  27. userneedinghelp
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    oh mk. one sec:)

    • one year ago
  28. userneedinghelp
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    thar we go:D

    • one year ago
  29. userneedinghelp
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    thank u

    • one year ago
  30. userneedinghelp
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    ill upload number 2 next. one sec

    • one year ago
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