document.write( "Question 941053: Leah leaves work at 17:54, halfway home she meets her daughter who left home at 17:46. Knowing that Leah walks 1.5x faster than her daughter and that they are travelling in a straight line, at what time do they meet? \n" ); document.write( "
Algebra.Com's Answer #573600 by josmiceli(19441)\"\" \"About 
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Let \"+d+\" = the distance in feet from work to home
\n" ); document.write( "Find Leah's head start:
\n" ); document.write( "17:54 minus 17:46 = 8 min
\n" ); document.write( "Let \"+s+\" = the daughter's walking speed in ft/min
\n" ); document.write( "\"+1.5s+\" = Leah's walking speed
\n" ); document.write( "\"+d%5B1%5D+=+1.5s%2A8+\"
\n" ); document.write( "\"+d%5B1%5D+=+12s+\" ft
\n" ); document.write( "-----------------------
\n" ); document.write( "Start a stop watch when daughter leaves
\n" ); document.write( "add their speeds to find when they meet
\n" ); document.write( "\"+d+-+12s+=+%28+s+%2B+1.5s+%29%2At+\"
\n" ); document.write( "\"+t+=+%28+d+-+12s+%29+%2F+%282.5s%29+\"
\n" ); document.write( "\"+t+=+d%2F%282.5s%29+-+4.8+\" minutes
\n" ); document.write( "---------------------
\n" ); document.write( "They meet at 17:46 + d/(2.5s) - 4.8\r
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