document.write( "Question 1110002: At 09 00 car A starts its journey and traveling at 70 km/h at 10 30.Car B started from the same place and traveled steadily on the same road. Car B took 3 1/2 h to catch up with car A. \r
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Algebra.Com's Answer #725001 by josgarithmetic(39616) $\"\"$ $\"About$

You can put this solution on YOUR website!
Last part, is a little unclear.
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\n" ); document.write( "Time quantity from 9:00 to 10:30 MIGHT be 1 and 1/2 hours.
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document.write( "CAR           SPEED      TIME     DISTANCE(catch-up)\r\n" );
document.write( "A              70        x+3.5      d\r\n" );
document.write( "B              r          3.5       d\r\n" );
document.write( "Difference                1.5\r\n" );
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\n" ); document.write( "\n" ); document.write( "x should be 1.5 hours.\r
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document.write( "CAR           SPEED      TIME     DISTANCE(catch-up)\r\n" );
document.write( "A              70          5        d\r\n" );
document.write( "B              r          3.5       d\r\n" );
document.write( "Difference                1.5\r\n" );
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\n" ); document.write( "You can find d.
\n" ); document.write( " $\"70%2A5=d\"$ \r
\n" ); document.write( "\n" ); document.write( " $\"highlight%28d=350%29\"$ \r
\n" ); document.write( "\n" ); document.write( "and you can then find r:
\n" ); document.write( " $\"350%2F3.5=r\"$
\n" ); document.write( " $\"highlight%28r=100%29\"$ ------speed for car B. \n" ); document.write( "

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