Find a polar equation for the curve represented by the given Cartesian equation: a) 5x+2y=5 Write the answer in the form r=f(t) where t stands for θ. Find r b) x2−y2=3 Write the answer in the form r2=f(t) where t stands for θ. Find r^2

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Find a polar equation for the curve represented by the given Cartesian equation: a) 5x+2y=5 Write the answer in the form r=f(t) where t stands for θ. Find r b) x2−y2=3 Write the answer in the form r2=f(t) where t stands for θ. Find r^2

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For polar coordinates we have \[x=rcos(\theta)\] \[y = rsin(\theta)\]\[r = \sqrt{x^2+y^2}\]
\[5x+2y=5 \implies 5(rcos(\theta)+2rsin(\theta))=5\] now mess around with it, and see what you get :-)
i don't know how to solve further can you help?

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Each term contains an r, try factoring! :)
cos()+1/2sin() =r but wasn't right
I don't understand what you did... where are the 5's? \(\large\rm 5r\cos\theta+2r\sin\theta=5\) Factor out an r from each term, \(\large\rm r(5\cos\theta+2\sin\theta)=5\) Then just divide both sides by that big bracketed thing, ya?
cos(t)+5/2sin(t) ?
still wasn't right
?? 0_o
\[\large\rm r(stuff)=5\qquad\to\qquad r=\frac{5}{stuff}\]
The division should be simple, I think maybe you're over complicating it :O
Zeps right, you're just one step away from completing it :P
(5/5cos(t))+(5/2sint) is what i put but it wasn't right
\[r = \frac{ 5 }{ (5 \cos \theta+2\sin \theta) }\] let theta = t
you can't split it into two fractions like that nick.
\[\large\rm \frac{a}{b+c}\ne\frac{a}{b}+\frac{a}{c}\]
oh okay thanks

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