document.write( "Question 381977: Bill and Roger want to race. Bill runs an average of 12 mph and Roger runs an average of 10 mph. To 'even things up' Bill lets Roger have a head start of .15 hour. How long before Bill catches up?\r
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document.write( "this is what I did:
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document.write( "12(x)=10(.15)
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document.write( "x=.125
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document.write( "or 7.5 minutes\r
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document.write( "Is that correct?
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document.write( "I also came up with 45 minutes... \n" );
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Algebra.Com's Answer #270845 by mananth(16949) You can put this solution on YOUR website! Bill runs an average of 12 mph \n" ); document.write( " Roger runs an average of 10 mph. \n" ); document.write( "head start of .15 hour. for roger \n" ); document.write( ".. \n" ); document.write( "make up speed for Bill is 12-10 = 2mph. \n" ); document.write( ".. \n" ); document.write( "Roger has head start of 0.15 hour \n" ); document.write( "so distance = 0.15 *10= 1.5 miles\r \n" ); document.write( "\n" ); document.write( ".. \n" ); document.write( "t = d/r \n" ); document.write( "1.5/2 \n" ); document.write( "=0.75 hours OR 45 minutes to catch up \n" ); document.write( "... \n" ); document.write( "m.ananth@hotmail.ca \n" ); document.write( " |