document.write( "Question 1046165: Diana left the coffee shop traveling 6 mph. Then, 3 hours later, Chelsea left traveling the same direction at 12 mph. How long until Chelsea catches up with Diana? \n" ); document.write( "

Algebra.Com's Answer #661662 by josgarithmetic(39627) $\"\"$ $\"About$

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UNKNOWN:
\n" ); document.write( "t, time for Chelsea's travel to catchup
\n" ); document.write( "d, catch-up distance
\n" ); document.write( "KNOWN:
\n" ); document.write( "R=12 mph
\n" ); document.write( "r=6 mph
\n" ); document.write( "h=3, for \"three hours later\"\r
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document.write( "              speed      time        distance\r\n" );
document.write( "Dianna         r          t+h           d\r\n" );
document.write( "Chelsea        R           t            d\r\n" );
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\n" ); document.write( "\n" ); document.write( "The question basically asks for t.
\n" ); document.write( "Travel Rate rule is RT=D to relate RATE, TIME, DISTANCE;
\n" ); document.write( "You will only need to form one equation because d is the same value for Dianna and for Chelsea - the catchup distance. \n" ); document.write( "

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