document.write( "Question 1194658: Janet leaves home at 10:15a jogging 8mph.
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document.write( "Sue leaves the same home at 10:35am biking 18mph.
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document.write( "What time do they meet and at what mile?\r
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document.write( "solved for
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document.write( "their ratio of speed S4:J9 and distance of J9:S4
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document.write( "and at 10:25 Janet would have gone 2 2/3 miles. \n" );
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Algebra.Com's Answer #826897 by josgarithmetic(39617)![]() ![]() ![]() You can put this solution on YOUR website! Sue leaves 20 minutes or one-third hour after Janet. In that time, the distance between the two is \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "If Janet catches up in x hours, then \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "For number of minutes, \n" ); document.write( "and you can compute the time on the clock.\r \n" ); document.write( "\n" ); document.write( "-- \n" ); document.write( "-- \n" ); document.write( "--\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "What time met? \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "10:51AM\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "What distance when met? \n" ); document.write( "18 miles per hour for 16 minutes \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " |