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document.write( "A car leaves at 7a.m. on a long trip, traveling at 44kph. A second car leaves at 11 a.m. following the first car at 108kph. At what time does the second car catch the first? \n" );
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Algebra.Com's Answer #437754 by checkley79(3341) ![]() You can put this solution on YOUR website! D=RT \n" ); document.write( "D=44T \n" ); document.write( "D=108(T-4) \n" ); document.write( "44T=108(T-4) \n" ); document.write( "44T=108T-432 \n" ); document.write( "44T-108T=-432 \n" ); document.write( "-64T=-432 \n" ); document.write( "T=-432/-64 \n" ); document.write( "T=6.75 HOURS. \n" ); document.write( "OR 7+6.75=13.75=1 HOUR 45 MIN. THEY WILL MEET. \n" ); document.write( "PROOF: \n" ); document.write( "44*6.75=108(6.75-4) \n" ); document.write( "297=108*2.75 \n" ); document.write( "297=297 \n" ); document.write( " |