document.write( "Question 175968: Joe drove to work on Monday at 45 mph and arrived 1 minute early. On Tuesday he drove at 40 mph and arrived 1 minute late.\r
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document.write( "a) How far does Joe live from work?
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document.write( "b) What speed does he normally travel?
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document.write( "c) How fast should he go to arrive 5 minutes early? \n" );
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Algebra.Com's Answer #131077 by checkley75(3666)![]() ![]() ![]() You can put this solution on YOUR website! d=rt \n" ); document.write( "45(t-1/60)=40(t+1/60) \n" ); document.write( "45t-45/60=40t+45/60 \n" ); document.write( "45t-40t=90/60 \n" ); document.write( "5t=1.5 \n" ); document.write( "t=1.5/5 \n" ); document.write( "t=3 hours is his normal driving time. \n" ); document.write( "a)45(3-1/60)=135-45/60=135-.75=134.25 miles. \n" ); document.write( "b)(45-40)/2=85/2=42.5 mph. \n" ); document.write( "c) 134.25=r(3-5/60) \n" ); document.write( "134.25=3r-15/60 \n" ); document.write( "134.25+.25=3r \n" ); document.write( "134.50=3r \n" ); document.write( "r=134.5/3 \n" ); document.write( "r=44.8333 mph is his speed in order to arive 5 minutes early.\r \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |