document.write( "Question 1204929: Convert the given Cartesian coordinates to polar coordinates with
\n" ); document.write( "r > 0, 0 ≤ 𝜃 ≤ 2𝜋. Remember to consider the quadrant in which the given point is located.
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Algebra.Com's Answer #841478 by MathLover1(20850)\"\" \"About 
You can put this solution on YOUR website!
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\n" ); document.write( "\n" ); document.write( "given point:\r
\n" ); document.write( "\n" ); document.write( "(\"2\", \"2\")\r
\n" ); document.write( "\n" ); document.write( "convert to polar coordinates with\r
\n" ); document.write( "\n" ); document.write( "\"r+%3E+0\", \"0+%3C=+theta+%3C=+2pi\"\r
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\n" ); document.write( "\n" ); document.write( "find \"r\" using formula:\r
\n" ); document.write( "\n" ); document.write( "\"r=sqrt%28x%5E2%2By%5E2%29\"\r
\n" ); document.write( "\n" ); document.write( "\"r=sqrt%282%5E2%2B2%5E2%29\"\r
\n" ); document.write( "\n" ); document.write( "\"r=sqrt%282%2A2%5E2%29\"\r
\n" ); document.write( "\n" ); document.write( "\"r=2sqrt%282%29\"\r
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\n" ); document.write( "\n" ); document.write( "\"theta=tan%5E-1%28y%2Fx%29\"\r
\n" ); document.write( "\n" ); document.write( "\"theta=tan%5E-1%282%2F2%29\"\r
\n" ); document.write( "\n" ); document.write( "\"theta=tan%5E-1%281%29\"\r
\n" ); document.write( "\n" ); document.write( "\"theta=pi%2F4\"\r
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\n" ); document.write( "\n" ); document.write( "since given point is in \"Q+\"\"I\", solution is \"theta=pi%2F4\" \r
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\n" ); document.write( "\n" ); document.write( "so, polar coordinates are (\"r\", \"theta\")=(\"2sqrt%282%29\", \"pi%2F4\")\r
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