document.write( "Question 1099384: A train, traveling at 50 mph., starts 1 and 1/3 hrs. after a slower train, and in 2 hrs., overtakes it. Calculate rate of the slower train.\r
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document.write( "Let x = rate of slower train.\r
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document.write( "rate * time = distance
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document.write( "rate = distance / rate\r
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document.write( "Faster train: 50 * (t - 1.33)= 50t - 66.5\r
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document.write( "Slower: x * t = xt.\r
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document.write( "Not sure how to continue. Non-homework.
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Algebra.Com's Answer #713825 by josgarithmetic(39618)![]() ![]() ![]() You can put this solution on YOUR website! \r\n" ); document.write( " SPEED TIME DISTANCE\r\n" ); document.write( "FAST 50 2 d\r\n" ); document.write( "SLOW r 2+1&1/3 d\r\n" ); document.write( "\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "FAST train data will give the distance. \n" ); document.write( "SLOW train data will allow to find speed r of the slow train.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( "\n" ); document.write( "-\r \n" ); document.write( "\n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( "\n" ); document.write( " |