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anonymous

  • 4 years ago

Can someone help me solve another inequality?

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  1. anonymous
    • 4 years ago
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  2. anonymous
    • 4 years ago
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    I need to solve the inequality in polar coordinates

  3. anonymous
    • 4 years ago
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    Like I have the answer but I want to understand it

  4. Mertsj
    • 4 years ago
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    That document you posted has nothing to do with an inequality

  5. anonymous
    • 4 years ago
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    hehe it says give the inequalities for r and theta which describ ethe following regions in polar coordinates

  6. nenadmatematika
    • 4 years ago
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    if you look carefuly you will see that angle is between -pi/3 and pi/3 (you get that when you calculate tan of the angle which is sqrt3) and the radius vector (or however you call it in your country) is from zero to 2 (that's what you get when you calculate radius like distance between origin and one of the given points)

  7. nenadmatematika
    • 4 years ago
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    |dw:1328484690164:dw|

  8. anonymous
    • 4 years ago
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    I alos got pi/3 but in my book it is pi/6

  9. Mertsj
    • 4 years ago
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    it is pi/6 because the tangent is 1/sqrt3 or sqrt3/3 and that corresponds to an angle of 30 degrees or pi/6

  10. nenadmatematika
    • 4 years ago
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    ohhhh I'm sorry tan of the angle is (1/sqrt3) which is really pi/6....I'm sleepy sorry :D

  11. Mertsj
    • 4 years ago
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    Tan is sin/cos or y/x

  12. anonymous
    • 4 years ago
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    huh i didnt follow

  13. Mertsj
    • 4 years ago
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    |dw:1328485021278:dw|

  14. nenadmatematika
    • 4 years ago
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    if you look at that point in the first quadrant...its coordinates are (sqrt3, 1)...and tan vertical over horizontal coordinate and you'll get 1/sqrt3....which is a basic value for tan of the angle of 30 degrees of pi/6

  15. anonymous
    • 4 years ago
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    lol oh i did 1/sqrt(3)

  16. anonymous
    • 4 years ago
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    my bad

  17. Mertsj
    • 4 years ago
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    it is 1/sqrt3

  18. nenadmatematika
    • 4 years ago
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    you mean you did sqrt3? :D

  19. anonymous
    • 4 years ago
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    Ok Thank you guys :D You were really clear :DDD

  20. anonymous
    • 4 years ago
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    hehe yes

  21. nenadmatematika
    • 4 years ago
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    you're welcome :D

  22. Mertsj
    • 4 years ago
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    yw

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