We can use the Pythagorean theorem, $x^2 + y^2 = r^2$, where $x$ and $y$ are the rectangular coordinates. In this case, $x = 3$ and $y = \sqrt{3}$, so we have $3^2 + (\sqrt{3})^2 = r^2$, which simplifies to $9 + 3 = r^2$. Therefore, $r^2 = 12$ and $r = \sqrt{12} =
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