document.write( "Question 692583: David is 1/4 mile from home bicycling at 21 mph when his brother takes off on his bike from home trying to catch him. How fast must David’s brother go to catch David a mile from home? \n" ); document.write( "
Algebra.Com's Answer #427245 by ankor@dixie-net.com(22740)![]() ![]() You can put this solution on YOUR website! David is 1/4 mile from home bicycling at 21 mph, when his brother takes off on his bike from home trying to catch him. \n" ); document.write( " How fast must David’s brother go to catch David a mile from home? \n" ); document.write( ": \n" ); document.write( "Brother has to ride 1 mile while David rides 3/4 of a mile (use .75 mi) \n" ); document.write( ": \n" ); document.write( "Let s = brothers required speed, to accomplish this \n" ); document.write( ": \n" ); document.write( "Write a time equation, time = dist/speed \n" ); document.write( ": \n" ); document.write( " \n" ); document.write( "cross multiply \n" ); document.write( ".75s = 1 * 21 \n" ); document.write( "s = \n" ); document.write( "s = 28 mph \n" ); document.write( "; \n" ); document.write( ": \n" ); document.write( "Find the times, should be equal \n" ); document.write( "1/28 = .0357 hrs \n" ); document.write( ".75/21 = .0357 \n" ); document.write( " |