document.write( "Question 305147: A cyclist is traveling 85 miles in 5 hours against the wind. He travels 174 miles in 6 hours with the wind. What is the rate of the still air and the rate of the wind. \n" ); document.write( "
Algebra.Com's Answer #218490 by checkley77(12844)![]() ![]() ![]() You can put this solution on YOUR website! D=RT \n" ); document.write( "85=(R-W)*5 AGAINST THE WIND. \n" ); document.write( "85=5R-5W \n" ); document.write( "5R=85+5W \n" ); document.write( "R=(85+5W)/5 \n" ); document.write( "R=17+W \n" ); document.write( "174=(R+W)6 WITH THE WIND. \n" ); document.write( "174=6R+6W \n" ); document.write( "6R=174-6W \n" ); document.write( "R=(174-6W)/6 \n" ); document.write( "R=29-W \n" ); document.write( "17+W=29-W \n" ); document.write( "2W=29-17 \n" ); document.write( "2W=12 \n" ); document.write( "W=12/2 \n" ); document.write( "W=6 MPH. \n" ); document.write( "R=17+6=23 THE SPEED OF THE BIKE IN STILL AIR. \n" ); document.write( "R=29-6=23 \n" ); document.write( "PROOF: \n" ); document.write( "85=(23-6)5 \n" ); document.write( "85=17*5 \n" ); document.write( "85=85 \n" ); document.write( "174=(23+6)*6 \n" ); document.write( "174=29*6 \n" ); document.write( "174=174\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |