document.write( "Question 174188This question is from textbook algebra 1
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Algebra.Com's Answer #129126 by jim_thompson5910(35256)\"\" \"About 
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\n" ); document.write( "\n" ); document.write( "\"d=sqrt%28%28x%5B1%5D-x%5B2%5D%29%5E2%2B%28y%5B1%5D-y%5B2%5D%29%5E2%29\" Start with the distance formula.\r
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\n" ); document.write( "\n" ); document.write( "\"d=sqrt%28%283-4%29%5E2%2B%284-5%29%5E2%29\" Plug in \"x%5B1%5D=3\", \"x%5B2%5D=4\", \"y%5B1%5D=4\", and \"y%5B2%5D=5\".\r
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\n" ); document.write( "\n" ); document.write( "\"d=sqrt%28%28-1%29%5E2%2B%284-5%29%5E2%29\" Subtract \"4\" from \"3\" to get \"-1\".\r
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\n" ); document.write( "\n" ); document.write( "\"d=sqrt%28%28-1%29%5E2%2B%28-1%29%5E2%29\" Subtract \"5\" from \"4\" to get \"-1\".\r
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\n" ); document.write( "\n" ); document.write( "\"d=sqrt%281%2B%28-1%29%5E2%29\" Square \"-1\" to get \"1\".\r
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\n" ); document.write( "\n" ); document.write( "\"d=sqrt%281%2B1%29\" Square \"-1\" to get \"1\".\r
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\n" ); document.write( "\n" ); document.write( "\"d=sqrt%282%29\" Add \"1\" to \"1\" to get \"2\".\r
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\n" ); document.write( "\n" ); document.write( "So our answer is \"d=sqrt%282%29\" \r
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\n" ); document.write( "\n" ); document.write( "Which approximates to \"d=1.414\" \r
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\n" ); document.write( "\n" ); document.write( "So the distance between the two points is approximately 1.414 units. \r
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