Question 1

Convert the rectangular coordinate (-3V3, -3),to a polar coordinate (r, 0).[ Select ]0[ Select ]

Convert the rectangular coordinate (-3V3, -3),to a polar coordinate (r, 0).[ Select-example-1

Question 2

Step 1: Write out the formula for finding r and Θ

$\begin{gathered} r=\sqrt[]{y^2+x^2} \\ \emptyset=\tan ^(-1)(y)/(x) \\ \text{Where x and y are the given cartesian coordinates and r and }\emptyset\text{ are the equivilent polar coordinates} \\ x=-3\sqrt[]{3} \\ y=-3 \end{gathered}$

Step2: Evaluate r and Θ

$\begin{gathered} r=\text{ }\sqrt[]{(-3)^2}+(-3\sqrt[]{3})^2 \\ =\sqrt[]{9+9(3)} \\ =\sqrt[]{9+27} \\ =\sqrt[]{36}=6 \\ r=6 \end{gathered}$
$\begin{gathered} \varnothing=tan^(-1)(\frac{-3}{-3\sqrt[]{3}}) \\ tan^(-1)(\frac{3}{3\sqrt[]{3}}) \\ By\text{ rationalization, we have} \\ tan^(-1)(\frac{3}{3\sqrt[]{3}}\text{ =}\frac{3}{3\sqrt[]{3}}*\frac{\sqrt[]{3}}{\sqrt[]{3}}=\frac{3\sqrt[]{3}}{9}=\frac{\sqrt[]{3}}{3}) \end{gathered}$
$\varnothing=\tan ^(-1)\frac{\sqrt[]{3}}{3}$

$\varnothing=30^0$

Hence, r=6, Θ =30°

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