document.write( "Question 784848: please help me solve:\r
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document.write( "Kevin drove 320 miles to a resort. He return trip took 20 min longer because his speed returning was 4 mi/h slower than going \n" );
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Algebra.Com's Answer #477294 by ankor@dixie-net.com(22740)![]() ![]() You can put this solution on YOUR website! Kevin drove 320 miles to a resort. \n" ); document.write( " He return trip took 20 min longer because his speed returning was 4 mi/h slower than going \n" ); document.write( ": \n" ); document.write( "Let (s+4) = speed of the outbound trip \n" ); document.write( "then \n" ); document.write( "s = speed of the return trip \n" ); document.write( ": \n" ); document.write( "Change 20 min to 1/3 hr \n" ); document.write( ": \n" ); document.write( "Write a time equation; time = dist/speed \n" ); document.write( ": \n" ); document.write( "return time - outbound time = 20 min \n" ); document.write( " \n" ); document.write( "multiply by 3s(s+4) to clear the denominators, resulting in:\ \n" ); document.write( "3(s+4)(320) - 3s(320) = s(s+4) \n" ); document.write( "960s + 3840 - 960s = s^2 + 4s \n" ); document.write( ": \n" ); document.write( "Arrange as a quadratic equation on the right \n" ); document.write( "0 = s^2 + 4s - 3840 \n" ); document.write( ": \n" ); document.write( "You can use the quadratic formula here but this will factor to \n" ); document.write( "(s+64)(s-60) = 0 \n" ); document.write( "the positive solution \n" ); document.write( "s = 60 mph on the return trip \n" ); document.write( "and obviously 64 mph on the outbound trip \n" ); document.write( ": \n" ); document.write( ": \n" ); document.write( "Confirm this by finding the actual times \n" ); document.write( "320/60 = 5.33 \n" ); document.write( "320/64 = 5.00 \n" ); document.write( "--------------- \n" ); document.write( "dif: .33 hrs which is 20 min\r \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |