document.write( "Question 136647This question is from textbook Algebra Structure and Method
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document.write( ": Jacqui commutes 30 mi to her job each day. She finds that if she drives 10 mi/h faster, it takes her 6 min less to get to work. Find her new speed. \n" );
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Algebra.Com's Answer #100068 by ankor@dixie-net.com(22740)![]() ![]() You can put this solution on YOUR website! Jacqui commutes 30 mi to her job each day. She finds that if she drives 10 mi/h faster, it takes her 6 min less to get to work. Find her new speed \n" ); document.write( ": \n" ); document.write( "Let s = new speed \n" ); document.write( "then \n" ); document.write( "(s-10) = original speed \n" ); document.write( ": \n" ); document.write( "Convert 6 min to hrs: 6/60 = .1 hrs \n" ); document.write( ": \n" ); document.write( "Write a time equation: Time = dist/speed \n" ); document.write( ": \n" ); document.write( "Fast trip + 6 min = slow trip \n" ); document.write( " \n" ); document.write( "Multiply equation by s(s-10) to get rid of the denominators: \n" ); document.write( "s(s-10)* \n" ); document.write( "Results \n" ); document.write( "30(s-10) + .1s^2 - 1s = 30s \n" ); document.write( ": \n" ); document.write( "30s - 300 + .1s^2 - s - 30s = 0 \n" ); document.write( ": \n" ); document.write( ".1s^2 - s - 300 = 0; a quadratic equation \n" ); document.write( ": \n" ); document.write( "s^2 - 10s - 3000 = 0; mult by 10 to get rid of the decimal \n" ); document.write( "Factors to: \n" ); document.write( "(s-60)(s+50) = 0 \n" ); document.write( ": \n" ); document.write( "S = 60 mph is the new speed \n" ); document.write( ": \n" ); document.write( ": \n" ); document.write( "Check solution by finding the times of each trip: \n" ); document.write( "30/50 = .6 hrs or 36 min \n" ); document.write( "30/60 = .5 hrs or 30 min \n" ); document.write( "------------------------ \n" ); document.write( "difference time = 6 min\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " = \n" ); document.write( " |