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anonymous

  • one year ago

Determine two pairs of polar coordinates for the point (2, -2) with 0° ≤ θ < 360°.

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  1. MTALHAHASSAN2
    • one year ago
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    |dw:1441922948541:dw|

  2. anonymous
    • one year ago
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    would it be (2 square root of 2, 135°), (-2 square root of 2, 315°)?

  3. MTALHAHASSAN2
    • one year ago
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    wait what are you trying to find??

  4. anonymous
    • one year ago
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    2 pairs of polar coordinates

  5. MTALHAHASSAN2
    • one year ago
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    oh ok

  6. MTALHAHASSAN2
    • one year ago
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    give me a sec

  7. MTALHAHASSAN2
    • one year ago
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    @chachi123

  8. MTALHAHASSAN2
    • one year ago
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    @nanaruiz123 do you still need help??

  9. anonymous
    • one year ago
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    ya i dont really get how to do this lol

  10. MTALHAHASSAN2
    • one year ago
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    @dtan5457 can you plz help

  11. MTALHAHASSAN2
    • one year ago
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    @Nnesha

  12. MTALHAHASSAN2
    • one year ago
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    @ASAAD123

  13. MTALHAHASSAN2
    • one year ago
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    @nanaruiz123 this is the hard question thou

  14. anonymous
    • one year ago
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    ya its pre-calc

  15. anonymous
    • one year ago
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    converting rectangular (x,y) to polar (r,θ) : \[r=\sqrt{x^2+y^2}\] \[\theta=\tan^{-1} (\frac{ y }{ x})\] \[r=\sqrt{2^2+(-2)^2}\]= \[2\sqrt{2}\] \[\theta=\tan^{-1} (\frac{ -2 }{ 2})=315\] \[(2\sqrt{2},315)\] the other pair of polar coordinate with negative radius>>(-r,θ+(2k+1)*180) let k=-1 so the other pair \[(-2\sqrt{2},135)\]

  16. anonymous
    • one year ago
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    oh my god thank you so much @ASAAD123

  17. anonymous
    • one year ago
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    welcome ^_^ @nanaruiz123

  18. MTALHAHASSAN2
    • one year ago
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    woo nice job @ASAAD123

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