document.write( "Question 974726: Sandy left Burbank driving 79 miles per hour. Mary, in order to catch up, drove 88 miles per hour. Mary caught up in 8 hours. How long was Sandy driving before Mary caught up?\r
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Algebra.Com's Answer #596577 by josgarithmetic(39620)![]() ![]() ![]() You can put this solution on YOUR website! Sandy left first and Mary left later; but traveled less time than Sandy. Think about that \n" ); document.write( "a while,... and then assign t as the travel time for Sandy BEFORE Mary began her travel.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Each Sandy and Mary travel the same distance, d, upon catch-up.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "______________________speed____________time__________distance \n" ); document.write( "Sandy_________________79_______________t+8____________d \n" ); document.write( "Mary__________________88________________8_____________d\r \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( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "- \n" ); document.write( "1 HOUR 46 SECONDS \n" ); document.write( " \n" ); document.write( " |