document.write( "Question 130565: A patrol plane flies 220 miles per hour in still air. It carries fuel for 4 hours of safe flying. If it takes off on patrol against a wind of 20 miles per hour, how far can it fly and return safely? \n" ); document.write( "
Algebra.Com's Answer #95366 by ptaylor(2198)![]() ![]() You can put this solution on YOUR website! \n" ); document.write( "Distance(d)=Rate(r) times Time(t) or d=rt; r=d/t and t=d/r\r \n" ); document.write( "\n" ); document.write( "rate of the plane against the wind=220-20=200 mph \n" ); document.write( "rate of the plane with the wind=220+20=240 mph \n" ); document.write( "distance flown with the wind = distance flown against the wind \n" ); document.write( "Time flying against the wind=d/200 \n" ); document.write( "Time flying with the wind=d/240\r \n" ); document.write( "\n" ); document.write( "And we are told that these two times add up to 4 hours, so:\r \n" ); document.write( "\n" ); document.write( "(d/200)+(d/240)=4 multiply each term by 4 \n" ); document.write( "d/50 + d/60=16 \n" ); document.write( "6d/300 + 5d/300=16 \n" ); document.write( "6d+5d=4800 \n" ); document.write( "11d=4800 \n" ); document.write( "d=436.36 mi---distance the plane can fly out and return safely (actually if I was the pilot I would shoot for about 420 mi out)\r \n" ); document.write( "\n" ); document.write( "CK \n" ); document.write( "436.36/200+436.36/240=4 \n" ); document.write( "3.99999999999999999999---~~4\r \n" ); document.write( "\n" ); document.write( "Hope this helps---ptaylor \n" ); document.write( " |