document.write( "Question 532320: A cyclist and a jogger start from a town at the same time and head for a destination 19 mi away. The rate of the cyclist is twice the rate of the jogger. The cyclist arrives 1.9 h before the jogger. Find the rate of the cyclist. \n" ); document.write( "
Algebra.Com's Answer #350810 by ankor@dixie-net.com(22740)![]() ![]() You can put this solution on YOUR website! A cyclist and a jogger start from a town at the same time and head for a destination 19 mi away. \n" ); document.write( "The rate of the cyclist is twice the rate of the jogger. The cyclist arrives 1.9 h before the jogger. \n" ); document.write( " Find the rate of the cyclist. \n" ); document.write( ": \n" ); document.write( "let r = the rate of the jogger \n" ); document.write( "then \n" ); document.write( "2r = rate of the cyclist \n" ); document.write( ": \n" ); document.write( "Write a time equation, time = dist/speed \n" ); document.write( ": \n" ); document.write( "jogger time - cyclist time = 1.9 hrs \n" ); document.write( " \n" ); document.write( "multiply by 2r, results: \n" ); document.write( "2(19) - 19 = 2r(1.9) \n" ); document.write( "38 - 19 = 3.8r \n" ); document.write( "19 = 3.8r \n" ); document.write( "r = \n" ); document.write( "r = 5 mph is the rate of the jogger \n" ); document.write( "then \n" ); document.write( "2(5) = 10 mph is the rate of the cyclist \n" ); document.write( ": \n" ); document.write( ": \n" ); document.write( "Check this find the actual time of each \n" ); document.write( "19/5 = 3.8 hrs \n" ); document.write( "19/10= 1.9 hrs \n" ); document.write( "--------------- \n" ); document.write( "time dif: 1.9 hrs \n" ); document.write( " |