Determine two pairs of polar coordinates for the point (3, -3) with 0° ≤ θ < 360°. choices: (3 , 315°), (-3 , 135°) (3 , 225°), (-3 , 45°) (3 , 45°), (-3 , 225°) (3 , 135°), (-3 , 315°)
Is the given point (3, -3) in polar form? or cartesian form?
polar
ok so I'm guessing that (3, -3) means r = 3 and theta = -3 radians right?
or I guess theta = -3 degrees not exactly sure
it would be degrees
ok well -3 degrees is coterminal to 360 - 3 = 357 degrees
wait but thats not on there.. lol
hmm let me think for a sec
okay haha
something seems to be missing a symbol maybe can you post a screenshot of the entire problem please?
i have no idea how to screen shot lol
above the insert key is the print screen key
on your keyboard
but i copy and pasted it word for word and i checked and no symbols are missing
hit that button then open up MS paint paste it in save it, post it
oh my gosh i feel like an idiot.. there is something missing
im sorry let me fix it
that's ok, and the site is going verrry slow for me lol, so fair warning
hopefully you figured out how to take a screenshot
\[(3 \sqrt{2,} 315°), (-3\sqrt{2} , 135°)\] \[ (3\sqrt{2} , 225°), (-3\sqrt{2} , 45°) \] \[ (3\sqrt{2} , 45°), (-3 \sqrt{2}, 225°)\] \[ (3\sqrt{2} , 135°), (-3 , 315°)\]
oh ok, those are the answer choices?
oh the last one is \[ (3 \sqrt{2}, 135°), (-3\sqrt{2} , 315°)\]
yes, my bad
alright, that's ok and thx
does that make more sense?
if the given point is in cartesian form, then r = sqrt(x^2 + y^2) r = sqrt(3^2 + (-3)^2) r = sqrt(9 + 9) r = sqrt(18) r = 3*sqrt(2)
the angle is theta = arctan(y/x) theta = arctan(-3/3) theta = arctan(-1) theta = -45 which is coterminal to 360 + (-45) = 315 degrees
now if you add 180 degrees to this, you basically make r negative (because you're going in reverse now) so 315 + 180 = 495 which is coterminal to 495 - 360 = 135 degrees
oh so it's the first!
so yep, it's A
thank you so much i really appreciate the help!
yw
oh wait i lied.. lol im sorry! hold on let me fix the choices!
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