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|dw:1393722700861:dw|

y = r sin(theta)
x = r cos(theta)
x^2 + y^2 = r^2
plug and chug

yea so how would i plug it in

cuz i did this last week and i kinda forgot sorry

Try getting your cos and sin with r over there by multiplying both sides by 2cos(theta)+5sin(theta)

okay. i did that

So you have:
\[2rcos(\theta)+5rsin(\theta)=7 ?\]

why would you put r w/ each one of the terms isnt there just 1?

oh ok i gotchu

you should have it from here. Use the equations sour gave you.

thannk you(:

what if the question's like r= -3sinΘ ?

Try multiplying r on both sides
recall rsin(theta)=y
and r^2=x^2+y^2

hmm ok...

so r^2= -3rsinΘ?

right so now you are able to replace rsin(theta) with y and r^2 with x^2+y^2

okay

then youd have?

the square root of that right?

just do exactly what I said replace r sin(theta) with y
replace r^2 with x^2+y^2

i did tht

you are done
what else do you want?

is tht my final answer

oh ok

write*

right, sorry. i missed a problem on my homework it was r= 2sinΘ-4cosΘ

multiply both sides by r

do you square both sides 1st

thts what i emant

replace r^2 with x^2+y^2
replace r sin(theta) with y
replace r cos(theta) with x

ok i did that: x^2+y^2= 2y-4x

do you stop there

ok

aand i missed r= 3+3cosΘ

so i multiplied both sides by r 1st

and got x^2+y^2=3r +3x but dont i need to get rid of the r?

why square both sides?

so what would be final answer then

yeah but when you square the whole left side you get x^4+y^4-9x^2 = 3x^2+y^2?

\[\text {recall law of exponents} (3r)^2=3^2r^2=9r^2=9(x^2+y^2) \]

so..

im not following are you saying keep left side just squared in parentheses?

It is easier than multiplying it out, don't you think?

yes. so only change right side. like substitue