anonymous
  • anonymous
Can someone help me solve another inequality?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
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anonymous
  • anonymous
I need to solve the inequality in polar coordinates
anonymous
  • anonymous
Like I have the answer but I want to understand it

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Mertsj
  • Mertsj
That document you posted has nothing to do with an inequality
anonymous
  • anonymous
hehe it says give the inequalities for r and theta which describ ethe following regions in polar coordinates
nenadmatematika
  • nenadmatematika
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)
nenadmatematika
  • nenadmatematika
|dw:1328484690164:dw|
anonymous
  • anonymous
I alos got pi/3 but in my book it is pi/6
Mertsj
  • Mertsj
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
nenadmatematika
  • nenadmatematika
ohhhh I'm sorry tan of the angle is (1/sqrt3) which is really pi/6....I'm sleepy sorry :D
Mertsj
  • Mertsj
Tan is sin/cos or y/x
anonymous
  • anonymous
huh i didnt follow
Mertsj
  • Mertsj
|dw:1328485021278:dw|
nenadmatematika
  • nenadmatematika
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
anonymous
  • anonymous
lol oh i did 1/sqrt(3)
anonymous
  • anonymous
my bad
Mertsj
  • Mertsj
it is 1/sqrt3
nenadmatematika
  • nenadmatematika
you mean you did sqrt3? :D
anonymous
  • anonymous
Ok Thank you guys :D You were really clear :DDD
anonymous
  • anonymous
hehe yes
nenadmatematika
  • nenadmatematika
you're welcome :D
Mertsj
  • Mertsj
yw

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