document.write( "Question 75243: A car leaves San Francisco for Los Angeles traveling an average of 60mph. At the same time, another car leaves LA for San Francisco traveling at 50mph. If it is 440miles between SanFrancisco and LA, howlong before the 2 cars meet, assuming that each maintains its average speed?
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Algebra.Com's Answer #54024 by ptaylor(2198)\"\" \"About 
You can put this solution on YOUR website!
Distance(d)=rate(r) times time(t)or d=rt; t=d/r; r=d/t\r
\n" ); document.write( "\n" ); document.write( "Car A's average rate of speed=60mph
\n" ); document.write( "Car B's average rate of speed=50mph\r
\n" ); document.write( "\n" ); document.write( "When the distance car A has traveled plus the distance car B has traveled equals 440 mi, then the two cars are together. \r
\n" ); document.write( "\n" ); document.write( "Car A's distance=rt(car A)=60t
\n" ); document.write( "Car B's distance=rt(car B)=50t\r
\n" ); document.write( "\n" ); document.write( "So:\r
\n" ); document.write( "\n" ); document.write( "60t+50t=440
\n" ); document.write( "110t=440 divide both sides by 110\r
\n" ); document.write( "\n" ); document.write( "t=4 hours------------------elapsed time 'till the two cars meet.\r
\n" ); document.write( "\n" ); document.write( "CK
\n" ); document.write( "60*4+50*4=440
\n" ); document.write( "240+200=440
\n" ); document.write( "440=440\r
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\n" ); document.write( "\n" ); document.write( "Hope this helps---ptaylor
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