document.write( "Question 1046368: On a day when the weather bureau announced that the wind was blowing at 5 km/h Scott's radio-controlled airplane flew a measured course in 1 min 48 s with the wind, and in 3 min against the wind. What was the length of the course?What was the rate of the plane in still air?
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document.write( "please help me with this question.
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document.write( "Thank you very much.\r
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Algebra.Com's Answer #661835 by solver91311(24713) You can put this solution on YOUR website! \r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "48 seconds is 48/60ths of a minute which is to say 8/10ths of a minute. Hence the \"with the wind\" leg took 1.8 minutes. Since the wind speed is given in km/hr, we convert the elapsed time to hours. 1.8 divided by 60 is 0.03 hours and 3 divided by 60 is 0.05 hours. Distance equals rate times time. Let \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "But since \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Solve for \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "John \n" ); document.write( " \n" ); document.write( "My calculator said it, I believe it, that settles it \n" ); document.write( " ![]() \n" ); document.write( " |