document.write( "Question 887023: Please help me! \r
\n" ); document.write( "\n" ); document.write( "Starting at home, Jessica travelled uphill to the hardware store for 75 minutes at just 4 mph. She then travelled back home along the same path downhill at a speed of 12 mph.\r
\n" ); document.write( "\n" ); document.write( "What is her average speed for the entire trip from home to the hardware store and back?
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Algebra.Com's Answer #536321 by josgarithmetic(39617)\"\" \"About 
You can put this solution on YOUR website!
Same distance d both ways.
\n" ); document.write( "RT=D rate time distance.\r
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\n" ); document.write( "\n" ); document.write( "_______________rate________time_________distance
\n" ); document.write( "UP_____________4___________\"75%2F60\"__________d
\n" ); document.write( "DOWN__________12___________t______________d\r
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\n" ); document.write( "\n" ); document.write( "Simplify the UP time hours and compute d:\r
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\n" ); document.write( "\n" ); document.write( "_______________rate________time____________distance
\n" ); document.write( "UP_____________4___________\"1%261%2F4\"__________d=4(5/4)=5
\n" ); document.write( "DOWN__________12___________t_________________d=5\r
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\n" ); document.write( "\n" ); document.write( "The time going DOWN can now easily be computed as \"t=5%2F12\"\r
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\n" ); document.write( "\n" ); document.write( "Total distance is BOTH WAYS, up and down \"5%2B5=10\" miles.
\n" ); document.write( "Total time in hours is \"5%2F4%2B5%2F12=15%2F12%2B5%2F12=20%2F12=5%2F3\" hours.\r
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\n" ); document.write( "\n" ); document.write( "Average Speed, \"10%2F%285%2F3%29=%2810%2A3%29%2F5=highlight%286%29\", miles per hour.
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