document.write( "Question 25194: An elevator went from the bottom to the top of a tower at an average speed of 4 m/s, remained at the top for 90 s, and then returned to the bottom at 5 m/s. If the total elapsed time was 4 1/2 min., how high is the tower?
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document.write( "Can you please also give the the equation you used to solve this problem?\r
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document.write( "Thank you! \n" );
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Algebra.Com's Answer #13513 by Paul(988)![]() ![]() ![]() You can put this solution on YOUR website! The elvator went at 4m/s at a distacne of x \n" ); document.write( "The elvator reutrned at 5m/s at a distance of x \n" ); document.write( "Total time in SECONDS. because its meters per second. (4*2+1)/2=4.5*60=270s \n" ); document.write( "270s-90s=180s\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( "4*5 =20 \n" ); document.write( "5*4=20 \n" ); document.write( "Hence the common multiple is 20\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( "Simplfy the expression. \n" ); document.write( "5x+4x=3600 \n" ); document.write( "9x=3600 \n" ); document.write( "x=400\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Hence the tower is 400m tall. \n" ); document.write( "Paul. \n" ); document.write( " |