document.write( "Question 162056: Please help!
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document.write( "A plane carries enough fuel for 10 hours of flight at an airspeed of 150 miles per hour. How far can it fly into a 30 mph headwind and still have enough fuel to return to its starting point? (This distance is called the point of no return.)
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Algebra.Com's Answer #119439 by ankor@dixie-net.com(22740)![]() ![]() You can put this solution on YOUR website! A plane carries enough fuel for 10 hours of flight at an airspeed of 150 miles per hour. How far can it fly into a 30 mph headwind and still have enough fuel to return to its starting point? (This distance is called the point of no return.) \n" ); document.write( ": \n" ); document.write( "This assumes that the return will be going with the wind \n" ); document.write( ": \n" ); document.write( "Let d = distance into the wind \n" ); document.write( ": \n" ); document.write( "Speed against the wind: 150 - 30 = 120 mph \n" ); document.write( "Speed with the wind: 150 + 30 = 180 mph \n" ); document.write( ": \n" ); document.write( "Write a time equation: Time = \n" ); document.write( ": \n" ); document.write( "Time into the wind + time with the wind = 10 hrs \n" ); document.write( " \n" ); document.write( ": \n" ); document.write( "Multiply equation by 360 to get rid of the denominators: \n" ); document.write( "360* \n" ); document.write( "Cancel denominators \n" ); document.write( "3d + 2d = 3600 \n" ); document.write( "d = \n" ); document.write( "d = 720 mi to the point of no return \n" ); document.write( ": \n" ); document.write( ": \n" ); document.write( "Check solution by find the times \n" ); document.write( "720/120 = 6 hr \n" ); document.write( "720/180 = 4 hr \n" ); document.write( "-------------- \n" ); document.write( "endurance 10 hrs \n" ); document.write( " \n" ); document.write( " |