document.write( "Question 696370:
\n" );
document.write( "when a plane flies into the wind it can travel 3000 km in 6 hours. when it flies with the wind, it goes the same distance in 5 hours. find the rate of the plane in still air. \n" );
document.write( "
Algebra.Com's Answer #429166 by checkley79(3341)![]() ![]() ![]() You can put this solution on YOUR website! D=RT \n" ); document.write( "3000=(R)6 RATE AGAINST THE WIND. \n" ); document.write( "R=3000/6 \n" ); document.write( "R=500 KPH. \n" ); document.write( "3000=(R)5 RATE WITH THE WIND. \n" ); document.write( "R=3000/5 \n" ); document.write( "R=600 KPH. \n" ); document.write( "(600-500)/2 \n" ); document.write( "100/2=50 KPH. IS THE SPEED OF THE WIND. \n" ); document.write( "500-50=450 ANS. FOR THE SPPD OF THE PLANE IN STILL AIR \n" ); document.write( "400+50=450 DITTO. \n" ); document.write( "PROOF: \n" ); document.write( "3000=500*6 \n" ); document.write( "3000=3000 \n" ); document.write( "3000=600*5 \n" ); document.write( "3000=3000\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |