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anonymous
 4 years ago
Can someone help me solve another inequality?
anonymous
 4 years ago
Can someone help me solve another inequality?

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I need to solve the inequality in polar coordinates

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Like I have the answer but I want to understand it

Mertsj
 4 years ago
Best ResponseYou've already chosen the best response.1That document you posted has nothing to do with an inequality

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0hehe it says give the inequalities for r and theta which describ ethe following regions in polar coordinates

nenadmatematika
 4 years ago
Best ResponseYou've already chosen the best response.0if 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
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1328484690164:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I alos got pi/3 but in my book it is pi/6

Mertsj
 4 years ago
Best ResponseYou've already chosen the best response.1it 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
 4 years ago
Best ResponseYou've already chosen the best response.0ohhhh I'm sorry tan of the angle is (1/sqrt3) which is really pi/6....I'm sleepy sorry :D

nenadmatematika
 4 years ago
Best ResponseYou've already chosen the best response.0if 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
 4 years ago
Best ResponseYou've already chosen the best response.0lol oh i did 1/sqrt(3)

nenadmatematika
 4 years ago
Best ResponseYou've already chosen the best response.0you mean you did sqrt3? :D

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Ok Thank you guys :D You were really clear :DDD

nenadmatematika
 4 years ago
Best ResponseYou've already chosen the best response.0you're welcome :D
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