document.write( "Question 162056: Please help!
\n" ); 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)\"\" \"About 
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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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\n" ); document.write( "This assumes that the return will be going with the wind
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\n" ); document.write( "Let d = distance into the wind
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\n" ); document.write( "Speed against the wind: 150 - 30 = 120 mph
\n" ); document.write( "Speed with the wind: 150 + 30 = 180 mph
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\n" ); document.write( "Write a time equation: Time = \"dist%2Fspeed\"
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\n" ); document.write( "Time into the wind + time with the wind = 10 hrs
\n" ); document.write( "\"d%2F120\" + \"d%2F180\" = 10
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\n" ); document.write( "Multiply equation by 360 to get rid of the denominators:
\n" ); document.write( "360*\"d%2F120\" + 360*\"d%2F180\" = 360(10)
\n" ); document.write( "Cancel denominators
\n" ); document.write( "3d + 2d = 3600
\n" ); document.write( "d = \"3600%2F5\"
\n" ); document.write( "d = 720 mi to the point of no return
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\n" ); document.write( "Check solution by find the times
\n" ); document.write( "720/120 = 6 hr
\n" ); document.write( "720/180 = 4 hr
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\n" ); document.write( "endurance 10 hrs
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