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anonymous

  • 5 years ago

Find the rectangular coordinates for the point whose polar coordinates are given. (6√2, 5π/3)

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  1. anonymous
    • 5 years ago
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    \[x=rcos(\theta) , y=rsin(\theta)\]

  2. anonymous
    • 5 years ago
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    \[ \sin(5\pi/3) = \sin(\pi/3) = (\frac{\sqrt{3}}{2})\]

  3. anonymous
    • 5 years ago
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    For (6sqrt2))cos(5pi/3) I get 4.242640607. Do I round that up or what else?

  4. anonymous
    • 5 years ago
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    what form do i put it in

  5. anonymous
    • 5 years ago
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    \[\cos(5\pi/3) = - \cos(\pi/3) = -\frac{1}{2}\]

  6. anonymous
    • 5 years ago
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    you can get exact values!,

  7. anonymous
    • 5 years ago
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    they want exact values

  8. anonymous
    • 5 years ago
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    do i not put the 6sqrt2 in the front

  9. anonymous
    • 5 years ago
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    yeah, exact values

  10. anonymous
    • 5 years ago
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    no, I was just finding the angles

  11. anonymous
    • 5 years ago
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    the answers u gave are wrong.

  12. anonymous
    • 5 years ago
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    is 1/2 the x value? and sqrt3/2 y

  13. anonymous
    • 5 years ago
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    \[ x =-6\sqrt{2} \frac{1}{2} , y= 6\sqrt{2}\frac{\sqrt{3}}{2}\]

  14. anonymous
    • 5 years ago
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    lol plz dont tell u you put them in :(

  15. anonymous
    • 5 years ago
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    ^ thats the FINAL answer

  16. anonymous
    • 5 years ago
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    i did.

  17. anonymous
    • 5 years ago
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    well simplify it a bit

  18. anonymous
    • 5 years ago
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    x = -3sqrt(2) , y= 3sqrt(6)

  19. anonymous
    • 5 years ago
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    the answer for x seems very obscure

  20. anonymous
    • 5 years ago
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    the other one's r wrong to

  21. anonymous
    • 5 years ago
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    no their not

  22. anonymous
    • 5 years ago
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    that's what the my hw program says. i hate it :P

  23. amistre64
    • 5 years ago
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    x = 3sqrt(2) y = -3sqrt(6)

  24. anonymous
    • 5 years ago
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    no,

  25. amistre64
    • 5 years ago
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    5[pi/3 is in the 4th quadrant; +x, -y

  26. anonymous
    • 5 years ago
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    wait lols , maybe

  27. amistre64
    • 5 years ago
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    360-60 :)

  28. anonymous
    • 5 years ago
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    yeh I was thinking 2pi/3 :|

  29. anonymous
    • 5 years ago
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    amistre to the rescue: Thank you for all ur help eng :)

  30. amistre64
    • 5 years ago
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    elec did al the hard stuff; got rid of the errors I would have done lol

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