for #13 letter c. i'm not getting (2-root2)cis(5pi/12)
i keep getting just (root2)cis(5pi/12)
also for #16 letter b. i'm not sure how to get [(x^2+y^2-2x)]^2
if anyone has any idea if they could let me know, thanks.
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i have no idea...sorry
Hey, Can you give me the original problems? I forgot the ditto at school and the link isn't working on schoolwires :(
It could be a mistake on my part...I rushed to complete it before the snow storm.
yeah.
13c) ( -1 + root3i ) / ( 1+i ) write in polar form
16b) r = 1 - 2cos(theta) write in rectangular
thanks...
13c) change to polar ->2cis(2pi/3)/sqrt(2)cis(pi/4)
dividing yields : r1/r2cis(subtract thetas)
gives: (2/sqrt(2))cis(5pi/12)
NOT 2-sqrt(2) ....oops
Final Answer: Sqrt(2)cis(5pi/12)
16b) r=1-2cosT
r=1-2(x/r)
r^2=r-2x (multiplied everything by r)
replace r=+-Sqrt(x^2+y^2)
x^2+y^2 = (+-Sqrt(x^2+y^2))-2x
x^2+y^2+2x = +-Sqrt(x^2+y^2)
Now square both sides
(x^2+y^2+2x)^2 = x^2+y^2
Hope that helps!
ooh alright thanks
Ms. Ricca for #15 Graph r = 6cos2x - 2 ...is that just plugging it into the calculator and sketching it?
Same with #8e The graph of r=2sin2x has 4 petals, true or false
Yep....just plug into calculator
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