document.write( "Question 978778: A man flies a small airplane from Fargo to Bismarck, North Dakota --- a distance of 180 miles. Because he is flying into a head wind, the trip takes him 2 hours. On the way back, the wind is still blowing at the same speed, so the return trip takes only 1 hour 12 minutes. What is his speed in still air, and how fast is the wind blowing?\r
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document.write( "Your answer is
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document.write( "his speed equals
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document.write( "the wind speed equals \n" );
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Algebra.Com's Answer #600159 by richwmiller(17219) You can put this solution on YOUR website! r*t=d \n" ); document.write( "with the wind \n" ); document.write( "(r+w)*1.2=180 \n" ); document.write( "(r+w)=180/1.2 \n" ); document.write( "r+w=150.0 \n" ); document.write( "against the wind \n" ); document.write( "(r-w)*2=180 \n" ); document.write( "(r-w)=180/2\r \n" ); document.write( "\n" ); document.write( "r-w=90 \n" ); document.write( "r+w=150.0 \n" ); document.write( "add \n" ); document.write( "r+r=150.0+90 \n" ); document.write( "2r=240.0 \n" ); document.write( "r=120.0 mph rate\r \n" ); document.write( "\n" ); document.write( "120-w=90 \n" ); document.write( "w=120-90 \n" ); document.write( "w=30 mph wind \n" ); document.write( "check \n" ); document.write( "(120+30)*1.2=180 \n" ); document.write( "150*1.2=180 \n" ); document.write( "180.0=180 \n" ); document.write( "ok \n" ); document.write( "(r-w)*2=180 \n" ); document.write( "(120-30)*2=180 \n" ); document.write( "90*2=180 \n" ); document.write( "180=180 \n" ); document.write( "ok \n" ); document.write( "codedrtdtt \n" ); document.write( " \n" ); document.write( " |