document.write( "Question 804535: Hi there and thank you in advance, here is my problem.\r
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document.write( "At noon Vladimir and Estragon depart from the same point. Vladimir walks east at a constant rate of 3mph, While Estragon walks north at a constant rate of 2mph. At what time(to the nearest minute) will they be exactly 20miles apart? \r
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document.write( "So far I have tried 2x + 3y = 20 -> y = 20/3 - 2/3x -> 2x + 3(20/3 - 2/3x) = 20
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document.write( "-> but then I end up with 20 = 20. So I am very confused. Any help would be greatly appreciated.
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Algebra.Com's Answer #484803 by mananth(16946)![]() ![]() You can put this solution on YOUR website! one walks at 3 mph \n" ); document.write( "so at given time he is 3t miles away from starting point\r \n" ); document.write( "\n" ); document.write( "Similarly\r \n" ); document.write( "\n" ); document.write( "the other is 2t miles away\r \n" ); document.write( "\n" ); document.write( "they are walking at right angles to each other\r \n" ); document.write( "\n" ); document.write( "At a given time t they form a right angle with the starting point\r \n" ); document.write( "\n" ); document.write( "Use Pythagoras theorem \n" ); document.write( "(3t)^2+(2t)^2= 20^2\r \n" ); document.write( "\n" ); document.write( "9t^2+4t^2=400\r \n" ); document.write( "\n" ); document.write( "13t^2=400 \n" ); document.write( "t^2=400/13 \n" ); document.write( "t^2= 30.77 \n" ); document.write( "t= 5.55 minutes\r \n" ); document.write( "\n" ); document.write( "say 6 minutes \n" ); document.write( " |