document.write( "Question 83704: Solve the problem and show the work.
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document.write( "Jim can run 5 miles per hour on level ground on a still day. One windy day, he runs 10 miles with the wind, and in the same amount of time runs 7 miles against the wind. What is the rate of the wind. \n" );
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Algebra.Com's Answer #60181 by ptaylor(2198)![]() ![]() You can put this solution on YOUR website! Let r=rate of wind\r \n" ); document.write( "\n" ); document.write( "Distance(d) equals Rate(r) times Time(t) or d=rt; r=d/t and t=d/r\r \n" ); document.write( "\n" ); document.write( "time it takes to run 10 mi with the wind: \n" ); document.write( "t=10/(5+r)\r \n" ); document.write( "\n" ); document.write( "time it take to run 7 mi against the wind:\r \n" ); document.write( "\n" ); document.write( "t=7/(5-r) \r \n" ); document.write( "\n" ); document.write( "Now we are told that these two times are the same, so:\r \n" ); document.write( "\n" ); document.write( "10/(5+r)=7/(5-r) multiply both sides by(5+r)(5-r) to get rid of fractions. In other words, cross multiply:\r \n" ); document.write( "\n" ); document.write( "10(5-r)=7(5+r) get rid of parens\r \n" ); document.write( "\n" ); document.write( "50-10r=35+7r subtract 7r and also 50 from both sides\r \n" ); document.write( "\n" ); document.write( "50-50-10r-7r=35-50+7r-7r collect like terms\r \n" ); document.write( "\n" ); document.write( "-17r=-15 divide both sides by -17\r \n" ); document.write( "\n" ); document.write( "r=0.882 mph-------------------------------------------rate of wind\r \n" ); document.write( "\n" ); document.write( "ck\r \n" ); document.write( "\n" ); document.write( "10/(5+0.882)=7/(5-0.882)\r \n" ); document.write( "\n" ); document.write( "10/5.882=7/4.118\r \n" ); document.write( "\n" ); document.write( "~1.7=~1.7\r \n" ); document.write( "\n" ); document.write( "Hope this helps-----ptaylor \n" ); document.write( " \n" ); document.write( " |