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UnbelievableDreams

  • one year ago

Confusing math question

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  1. UnbelievableDreams
    • one year ago
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    csc(3∝+40)=sec(7∝−70)

  2. zzr0ck3r
    • one year ago
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    Do they mean the limit?

  3. jdoe0001
    • one year ago
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    so... what's the question at -> \(\bf csc(3\alpha+40)=sec(7\alpha-70)?\)

  4. UnbelievableDreams
    • one year ago
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    Find the ∝

  5. UnbelievableDreams
    • one year ago
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    @jdoe0001

  6. sdfgsdfgs
    • one year ago
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    3 eqns that can be used to solve the prob. \[\csc A = \frac{ 1 }{ \sin A }\] \[\sec A = \frac{ 1 }{ \cos A }\] and \[\sin (\frac{ \pi }{ 2 } - A) = \cos A\]

  7. anonymous
    • one year ago
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    wow

  8. anonymous
    • one year ago
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    are you supposed to find one answer, or all of them? one will be no problem, all will be almost impossible

  9. UnbelievableDreams
    • one year ago
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    Just one answer, I am confused about tis formula

  10. anonymous
    • one year ago
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    one answer we can start with \[\frac{1}{\cos(3x+40)}=\frac{1}{\sin(7x-70)}\]

  11. anonymous
    • one year ago
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    or \[\cos(3x+40)=\sin(7x-70)\]

  12. anonymous
    • one year ago
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    damn i have that backwards!!

  13. anonymous
    • one year ago
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    \[\sin(3x+40)=\cos(7x-70)\]

  14. UnbelievableDreams
    • one year ago
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    x=12, right?

  15. anonymous
    • one year ago
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    idk i am still working

  16. anonymous
    • one year ago
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    add \(90\) to \(7x-70\) to get \[\cos(7x-70)=\sin(7x+20)\]

  17. anonymous
    • one year ago
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    so one answer will be the solution to \[3x+40=7x+20\]

  18. anonymous
    • one year ago
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    i get \(x=5\)

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spraguer (Moderator)
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