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Convert the polar equation to rectangular form and identify the graph. r= 7/ (2cosΘ+5sinΘ)

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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

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Other answers:

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?
\[r=\frac{7}{2\cos(\theta)+5 \sin(\theta)} => r(2\cos(\theta)+5 \sin(\theta))=7\] Distributive property!
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
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
putting an equation in terms of just x and y ( or just x or just y) is called a cartesian equation if that was the directions to just right as a cartesian equation then you are done
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
yep. :) that is right you wrote the poloar equation as a cartesian equation if that is all the directions said to do then you are done.
do you stop there
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?
you could subtract both sides by 3x giving you \[x^2+y^2-3x=3r \] then square both sides and replace the r^2 with x^2+y^2
why square both sides?
I just prefer my answer without radicals and me squaring both sides will allow me to write my answer without radicals
so what would be final answer then
what if you with what I was saying take my equation and then square both sides and then replace the r^2 with x^2+y^2 you will have your final answer
yeah but when you square the whole left side you get x^4+y^4-9x^2 = 3x^2+y^2?
\[(x^2+y^2-3x)^2=(3r)^2 \] the us what you get when you square both sides your equation is not equivalent to that you squared each term sorta but not really it is whatever you do to one side of equation you can do to the other and the equation still holds
\[\text {recall law of exponents} (3r)^2=3^2r^2=9r^2=9(x^2+y^2) \]
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
You can multiply it out if you want. Just do it correctly, but the if the directions just say write as a rectangular (aka cartesian equation) than you are done. If the left hand side was already in terms of x and y. r^2 was the only thing that needed to be converted.

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