document.write( "Question 722710: a candy truck leaves the store at 5:00 am each mornng on its 144 mile route.one day the driver gets a late start and does not leave the store until 5:30 am. to finish his route on time the driver drives 4 miles per hour faster than usual.at what speed does he usually drive ? \n" ); document.write( "
Algebra.Com's Answer #442863 by josgarithmetic(39617)![]() ![]() ![]() You can put this solution on YOUR website! normal speed r in miles per hour. The difference between 5:00am and 5:30am is 1/2 hour.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "____________speed________________time____________________distance \n" ); document.write( "normal_______r___________________ \n" ); document.write( "late________r+4__________________ \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "The normal time being 1/2 hour more than the late time suggests taking the difference between their expressions. Both expressions may be shown as expressions using r, the rate.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "normal time minus late time equals 1/2 hour, \n" ); document.write( " \n" ); document.write( "Multiply left side and right side by \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( "=32 miles per hour --- ANSWER. \n" ); document.write( " |