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math456

  • 2 years ago

Can anyone help me???

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  1. math456
    • 2 years ago
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  2. brattygirl0962
    • 2 years ago
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    with

  3. math456
    • 2 years ago
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    attached file!

  4. brattygirl0962
    • 2 years ago
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    idk about that because i havent figerd that out my self i am new lol

  5. math456
    • 2 years ago
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    lol its okay..

  6. math456
    • 2 years ago
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    welcome to open study!!

  7. brattygirl0962
    • 2 years ago
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    i would i would not say no but im sorry i have no clue how to do anything rite now i would but i do not know how rite now....

  8. math456
    • 2 years ago
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    its okay, but thanks for putting up the effort!!

  9. brattygirl0962
    • 2 years ago
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    your welcome any time

  10. math456
    • 2 years ago
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    :)

  11. Romero
    • 2 years ago
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    Pythagoras!!!

  12. Romero
    • 2 years ago
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    \[x^2 + y^2 = z^2\]

  13. math456
    • 2 years ago
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    but we dont know the values..

  14. Romero
    • 2 years ago
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    Only for the second one lol sorry.

  15. Romero
    • 2 years ago
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    \[x^2 + 7^2 = a^2\]

  16. Romero
    • 2 years ago
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    solve for x

  17. math456
    • 2 years ago
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    sq. root a-7

  18. Romero
    • 2 years ago
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    \[\pm \sqrt{a^2 - 7^2}\]sq. root a^2-7^2

  19. Romero
    • 2 years ago
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    For the other ones you have to use cos sin and tan

  20. math456
    • 2 years ago
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    ohh okay, so this is all we have to do?

  21. Romero
    • 2 years ago
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    I did b

  22. math456
    • 2 years ago
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    how about angles?

  23. math456
    • 2 years ago
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    like for a) is it x=+/- sq. root a-43 degress

  24. Romero
    • 2 years ago
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    That's what I mean. You have to use cos sin tan SIN COS TAN SOH CAH TOA O=opposite side A=adjacent side H=Hypotnus

  25. Romero
    • 2 years ago
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    Go learn how to use cos sin and tan.

  26. brattygirl0962
    • 2 years ago
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    what do u neeed help with the same thing

  27. math456
    • 2 years ago
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    i know abt them already.. but not sure how to use it here..

  28. Romero
    • 2 years ago
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    If you know it already then you would know how to use them exactly. Even with my information that I have you about cos sin and tan GO back to that section and review it.

  29. math456
    • 2 years ago
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    is it like sin(43)=x/a-->a sin(43)=x?

  30. math456
    • 2 years ago
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    thanks for part b!!

  31. Romero
    • 2 years ago
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    Cos(43)=a/x Adjacent side is a over hypotnus which is x CAH

  32. Romero
    • 2 years ago
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    c) use tan d) use sin

  33. math456
    • 2 years ago
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    thank you..

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