document.write( "Question 712445: two planes leaves the dallas- fort worth airport at the same time. one travels east at 550 mphs, and the other travels west at 500 mph. Assuming no wind, how long will it take for the planes to be 2100 mi apart? \n" ); document.write( "
Algebra.Com's Answer #437947 by algebrahouse.com(1659) You can put this solution on YOUR website! distance = rate x time\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "one plane \n" ); document.write( "rate = 550 \n" ); document.write( "time = t \n" ); document.write( "distance = 550t\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "other plane \n" ); document.write( "rate = 500 \n" ); document.write( "time = t \n" ); document.write( "distance = 500t\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "When will their combined distances equal 2100\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "550t + 500t = 2100 {added distances and set equal to 2100} \n" ); document.write( "1050t = 2100 {combined like terms} \n" ); document.write( "t = 2 {divided each side by 1050}\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "In 2 hours they will be 2100 mi. apart \n" ); document.write( " For more help from me, visit: www.algebrahouse.com \n" ); document.write( " |