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Mathematics 8 Online
OpenStudy (anonymous):

Convert the rectangular coordinates to polar coordinates with r > 0 and 0 < θ < 2π. (-9, 9)

OpenStudy (anonymous):

You can use x= R cos t and y = R sin t

OpenStudy (amistre64):

r = 9sqrt(2)

OpenStudy (amistre64):

and its some form of 45 degree angle

OpenStudy (amistre64):

q2

OpenStudy (anonymous):

Find R by: R= sqrt(x^2 + y ^2)

OpenStudy (amistre64):

angle = 135

OpenStudy (anonymous):

I know how to use the formula, but I can't use decimals as the answer.

OpenStudy (anonymous):

I have to put the answer in an x and y format.

OpenStudy (amistre64):

(9sqrt(2), 135) or the radian (3pi/4)

OpenStudy (anonymous):

how do u get this answer?

OpenStudy (amistre64):

been going over spherical and cylindrical coordinates today lol

OpenStudy (amistre64):

well; i notice that both ax and y are the same; the only angle with that is a 45 degree; the hyp of a 45 degree is sqrt(2)

OpenStudy (amistre64):

r = 9 times bigger than sqrt(2) = 9sqrt(2)

OpenStudy (anonymous):

What about this one? (-√21, -√7)

OpenStudy (anonymous):

oh ok

OpenStudy (amistre64):

and the x=- and y=+ tells me its in the Q2

OpenStudy (amistre64):

r = sqrt(28) the x=- and y=- is in the Q3

OpenStudy (anonymous):

sqrt28?

OpenStudy (amistre64):

tan^1(sqrt(7)/sqrt(21)) is the angle

OpenStudy (amistre64):

sqrt(21+7) = sqrt(28) right?

OpenStudy (anonymous):

yeah, where to u get the 7 from?

OpenStudy (anonymous):

or why use the 7

OpenStudy (amistre64):

sqrt(7)^2 = 7

OpenStudy (amistre64):

7*4 = 28 so...2sqrt(7) = r

OpenStudy (amistre64):

tan^-1(sqrt(7)/sqrt(21)) + pi is the angle

OpenStudy (amistre64):

we got 30 +pi :)

OpenStudy (anonymous):

so the y would be 7pi/6

OpenStudy (amistre64):

yes :)

OpenStudy (amistre64):

the angle = 7pi/6

OpenStudy (anonymous):

(-1, 0)

OpenStudy (anonymous):

would it be (sqrt1,0)

OpenStudy (amistre64):

thats facing left.... r = 1; theta = pi

OpenStudy (anonymous):

why does theta = 0

OpenStudy (anonymous):

i mean pi

OpenStudy (amistre64):

it 1 from the origin; so r=1 and its 180 or pi from the pos x axis

OpenStudy (amistre64):

theta doesnt equal 0, the y coord = 0

OpenStudy (anonymous):

the angle = pi

OpenStudy (amistre64):

if we turn by pi and move out by 1 we come to -1,0

OpenStudy (anonymous):

it's (1,pi)

OpenStudy (amistre64):

yes :) 1,pi

OpenStudy (anonymous):

ok, a little change in direction.

OpenStudy (anonymous):

Convert the equation to polar form. 5x=5y

OpenStudy (amistre64):

thats the same as y = x which is r = t cos(pi/4)

OpenStudy (amistre64):

or rather: (r,pi/4) i think

OpenStudy (amistre64):

could be +-r, pi/4

OpenStudy (amistre64):

whats our options for answers?

OpenStudy (anonymous):

pi/4 is wrong, i tried thaat

OpenStudy (amistre64):

y=x is a 45 degree line.... cant be another angle :)

OpenStudy (amistre64):

x = r cos(pi/4) y = r sin(pi/4)

OpenStudy (anonymous):

i'm so confused right now :(

OpenStudy (amistre64):

maybe 5r, but thats rather redundant

OpenStudy (amistre64):

5x = 5y right? 5r cos(pi/4) = 5r sin(pi/4)

OpenStudy (amistre64):

the 5rs can be removed and you get; cos(pi/4)=sin(pi/4)

OpenStudy (anonymous):

is it 5r/sqrt2

OpenStudy (amistre64):

dont know; what formats you got to choose from?

OpenStudy (anonymous):

i have no idea. what formats are there?

OpenStudy (amistre64):

if its multiple choice; then they give you options to choose from....any options to narrow this down with?

OpenStudy (anonymous):

no, no multiple choice.

OpenStudy (anonymous):

here's another example: x^2+y^2=16

OpenStudy (amistre64):

r^2 = 16

OpenStudy (anonymous):

that would be 4. I tried that but it says it's wrong.

OpenStudy (amistre64):

then i got no idea what your program wants as an answer :)

OpenStudy (anonymous):

i got it i had to put it in as r=4

OpenStudy (anonymous):

y=2

OpenStudy (amistre64):

:)

OpenStudy (anonymous):

would it be sqrt2

OpenStudy (amistre64):

y=2 r sin(t) = 2

OpenStudy (amistre64):

r = 2/sin(t) ?

OpenStudy (anonymous):

I don't get it.

OpenStudy (anonymous):

the question is Convert the equation to polar form. y=2

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