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anonymous
 one year ago
PLEASE Determine two pairs of polar coordinates for the point (5, 5) with 0° ≤ θ < 360°.
(5 square root of 2, 315°), (5 square root of 2, 135°)
(5 square root of 2, 45°), (5 square root of 2, 225°)
(5 square root of 2, 135°), (5 square root of 2, 315°)
(5 square root of 2, 225°), (5 square root of 2, 45°)
anonymous
 one year ago
PLEASE Determine two pairs of polar coordinates for the point (5, 5) with 0° ≤ θ < 360°. (5 square root of 2, 315°), (5 square root of 2, 135°) (5 square root of 2, 45°), (5 square root of 2, 225°) (5 square root of 2, 135°), (5 square root of 2, 315°) (5 square root of 2, 225°), (5 square root of 2, 45°)

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cwrw238
 one year ago
Best ResponseYou've already chosen the best response.0Hint the point is in the 4th quadrant

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1440031939497:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'm not sure what you mean. I have never worked with this kind of problem before..

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1440032084293:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok when converting to polar form use following equations for given point (x,y) \[r = \sqrt{x^2 + y^2}\] \[\tan \theta = \frac{y}{x}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Okay so polar coordinates are the opposite coordinates?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so \[r=\sqrt{5^2+(5^2)}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0therefore r= \[\sqrt{50}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0now there are multiple ways of writing a given point in polar form different angles can be used as long as tan = y/x r can be neg, which reflects the point 180 degrees

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0correct, simplify the radical \[\sqrt{50} = 5 \sqrt{2}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Alright, and I got \[\tan \theta=5/5\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0What am I to do with this information?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok good so we have 1 angle (45 or pi/4) now use my circle drawings above to see how to get an equivalent angle

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I dont understand..I can see the equivalent angle

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I just dont know how to get it

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0here is a general form showing all 4 possible points for a given (x,y) \[(r,\theta)\] \[(r, \theta +2\pi)\] \[(r, \theta + \pi)\] \[(r, \theta  \pi)\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1440033135732:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So what would be my theta and my r?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you already calculated those ..... r = sqrt(50) , theta = pi/4

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh, okay. I would have the following possible answers then: A. \[5\sqrt{2},\left(\begin{matrix}\pi \\ 4\end{matrix}\right)\] B. \[5\sqrt{2},\left(\begin{matrix}\pi \\ 4\end{matrix}\right)+2pi\] C. \[5\sqrt{2},\left(\begin{matrix}\pi \\ 4\end{matrix}\right)+pi\] D. \[5\sqrt{2},\left(\begin{matrix}\pi \\ 4\end{matrix}\right)pi\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0what do you mean by 'find 2 match the given pairs'

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0figure it out...gotta go good luck!
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