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Convert r=12 cos(theta) to rectangular form
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\[x=r\cos(\theta)\\ \frac{x}{r}=\cos(\theta)\]
substitute get \[\frac{r}{12}=\frac{x}{r}\]
making \[r^2=12x\] or \[x^2+y^2=12x\]
I got to x^2+y^2=12x, but wasn't sure if that is rectangular form. Can you confirm that it is?
you could put \[x^2+y^2-12x=0\] or complete the square to make it a circle in standard form
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