document.write( "Question 1071256: Stacy and Heather are 20 miles apart and walk towards each other along the same route. Stacy walks at constant rate that is 1 mile per hour fast than heather's constant rate of 5 miles/hour. If Heather starts her journey 24 minutes after Stacy, how far from the original destination has Heather walked when the two meet?.
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Algebra.Com's Answer #686191 by josmiceli(19441)\"\" \"About 
You can put this solution on YOUR website!
How far does Stacy get in \"+24+\" min ?
\n" ); document.write( "\"+d%5B1%5D+=+6%2A%28+24%2F60+%29+\"
\n" ); document.write( "\"+d%5B1%5D+=+24%2F10+\"
\n" ); document.write( "\"+d%5B1%5D+=+2.4+\" mi
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\n" ); document.write( "After \"+24+\" min, they will be
\n" ); document.write( "\"+20+-+2.4+=+17.6+\" mi apart,
\n" ); document.write( "and you start a stopwatch to
\n" ); document.write( "measure time \"+t+\" in hrs untill they meet
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\n" ); document.write( "Let \"+d+\" = distance in miles Heather has
\n" ); document.write( "to walk utill they meet
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\n" ); document.write( "Heather's equation:
\n" ); document.write( "(1) \"+d+=+5t+\"
\n" ); document.write( "Stacy's equation:
\n" ); document.write( "(2) \"+17.6+-+d+=+6t+\"
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\n" ); document.write( "Plug (1) into (2)
\n" ); document.write( "(2) \"+17.6+-+5t+=+6t+\"
\n" ); document.write( "(2) \"+11t+=+17.6+\"
\n" ); document.write( "(2) \"+t+=+1.6+\" hrs
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\n" ); document.write( "In \"+1.6+\" hrs, Heather has walked:
\n" ); document.write( "\"+d+=+5%2A1.6+\"
\n" ); document.write( "\"+d+=+8+\"
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\n" ); document.write( "Heather walked 8 miles from her original
\n" ); document.write( "destination until they meet
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\n" ); document.write( "check:
\n" ); document.write( "Stacy must have walked \"+17.6+-+8+=+9.6+\" mi
\n" ); document.write( "\"+9.6+=+6%2A1.6+\"
\n" ); document.write( "\"+9.6+=+9.6+\"
\n" ); document.write( "So Stacy walked a total of \"+2.4+%2B+9.6+=+12+\" mi
\n" ); document.write( "and added to Heather's distance,
\n" ); document.write( "\"+12+%2B+8+=+20+\" mi
\n" ); document.write( "OK\r
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