document.write( "Question 106367: Becky and Rita are 648 miles apart. Becky walks toward Rita at the rate of 2.25 m/s and Rita runs towards Becky. What is Rita's average speed if she reaches Becky in 1.6 minutes? \n" ); document.write( "
Algebra.Com's Answer #77375 by stanbon(75887)\"\" \"About 
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Becky and Rita are 648 miles apart. Becky walks toward Rita at the rate of 2.25 m/s and Rita runs towards Becky. What is Rita's average speed if she reaches Becky in 1.6 minutes?
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\n" ); document.write( "Comment: I don't know if your 2.25 m/s is meters/second or 2.25 miles/second
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\n" ); document.write( "Becky DATA:
\n" ); document.write( "Rate = 2.25 miles/sec ; distance = x miles ; time = d/r = x/2.25 seconds
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\n" ); document.write( "Rita DATA:
\n" ); document.write( "Time = 96 seconds ; distance = 648-x miles ; rate = d/t = (648-x)/96 miles/sec.
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\n" ); document.write( "EQUATION:
\n" ); document.write( "Becky time = Rita time
\n" ); document.write( "x/2.25 = 96
\n" ); document.write( "x = 216 miles (Becky's distance)
\n" ); document.write( "Rita's speed = (648-216)/96 = 432/96 = 4.5 miles/second
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\n" ); document.write( "Cheers,
\n" ); document.write( "Stan H.\r
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