document.write( "Question 764193: angela leaves from jocelyn's house on her bicycle traveling at 12 mph. ten minutes later jocelyn leaves her house on her bicycle traveling at 15 mph to catch up with angela. how long in minutes does it take jocelyn to reach angela \n" ); document.write( "
Algebra.Com's Answer #465302 by ankor@dixie-net.com(22740)![]() ![]() You can put this solution on YOUR website! angela leaves from jocelyn's house on her bicycle traveling at 12 mph. \n" ); document.write( " ten minutes later jocelyn leaves her house on her bicycle traveling at \n" ); document.write( " 15 mph to catch up with angela. \n" ); document.write( "how long in minutes does it take jocelyn to reach angela. \n" ); document.write( ": \n" ); document.write( "We're using mph so change 10 min to \n" ); document.write( ": \n" ); document.write( "let t = time for J to catch A \n" ); document.write( "then \n" ); document.write( "(t+ \n" ); document.write( ": \n" ); document.write( "When this happens they will have traveled the same distance \n" ); document.write( "Write a distance equation; dist = speed * time \n" ); document.write( ": \n" ); document.write( "J's dist = A's dist \n" ); document.write( "15t = 12(t+ \n" ); document.write( "15t = 12t + 2 \n" ); document.write( "15t - 12t = 2 \n" ); document.write( "3t = 2 \n" ); document.write( "t = 2/3 hrs which is: \n" ); document.write( ": \n" ); document.write( ": \n" ); document.write( "Check this by finding the actual dist of each, should be equal \n" ); document.write( "A's travel time is 50 min (5/6 hr) \n" ); document.write( "12 * 5/6 = 10 mi is A's dist \n" ); document.write( "15 * 4/6 = 10 mi also \n" ); document.write( " \n" ); document.write( " |