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write each polar equation in rectangular form: r=-3sec(Theta) r=costheta+sintheta r=5/3costheta+8sin

Mathematics
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^theta after the last sin
You need to use these equations: \[r^2=x^2+y^2 ; rcos(\theta)=x ; rsin(\theta)=y \]
i did, but like it ended up like x^2+y^2= 25+9x^2r^2-30xr/64y^2 for the last one

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Other answers:

We want the equations just in terms of x and y instead of r and theta.
i don't know what i did wrong
so the last one is: \[r=\frac{5}{3}\cos(\theta)+8\sin(\theta) \]?
no uh (5)/(3costheta+8sintheta)
oh so you meant 5/(3costheta)+8sin(theta)) and not 5/3cos(theta)+8sin(theta)
mhm 5 / (3cos+8sin)
\[r= \frac{5}{3\cos(\theta)+8\sin(\theta)}\] Multiply both sides by that one fraction's denominator
\[r(3\cos(\theta)+8\sin(\theta))=5 \] Distribute then substitute the equations I gave you
3xr+8ry=5 but then you'll need to put r on one side cause you would need to square it. and the answer comes out all weird
well rcos(theta) is x not just cos(theta same thing with the y deal rsin(theta) is y not just sin(theta)
whoops how about the secant one
put it in terms of cos by using an identity
then put it in some form where you can use the equations I gave you
sosoososososo it would be x^3+y^2x=-3 AND uh the other one x^2+y^2=x+y?
for the first one? no...
I haven't looked at the second one... You are making this way too hard it looks like.
first one you did this as first step right?: \[r=\frac{-3}{\cos(\theta)}\]
yep
Notice this only has a cos in it... how do you get cos with the r over there so you can use the equation rcos(theta)=x
multiply both sides with r
no
try again
How do you undo multiplication?
Or division?
oh omg
we have division by cos to undo that you ?
x=-3
yes
thanks
so for that second one...
you actually got it right
but how did you go about it you didn't choose a complicated way right?
yep uh multiplied both side with r and then substituted the stuff in
oh good
do you have any other questions? do you think you got this better?
uh yea i have a few more
for standard forms of polar equation uh how would graph r=2cos3theta
Plug in values of theta and see what the r output is
then remember how you graph using polar coordinates r is the distance from the origin and direction is given by r to from the angle theta is the angle
like how would you do this with a graphing calculator
So for example if I plug in theta=0 \[r=2 \cos (3 \cdot 0)=2\cos(0)=2(1)=2 => \text{ graph the point } (2,0^o)\] |dw:1388269865874:dw|
find 0 degrees and then go to the circle with radius 2
cause wksht says not to use a table
|dw:1388269927627:dw| then plug in more points
oh i know how to graph but not without plugging in points
I haven't looked at a calculator in ages
did you try looking up how to input polar equations into whatever calculator type you have
some calculators require a difference procedure in order to do certain things
http://mathbits.com/MathBits/TISection/PreCalculus/polargraphs.htm
find mode and choose polar or pol or polargc
then just type your equation in
u what's the website
*uh
What ?

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