document.write( "Question 1203986: Charles made a business trip of 200.5
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\n" ); document.write( " for the first part of the trip and 59mph
\n" ); document.write( " for the second part. If the trip took 3.5
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Algebra.Com's Answer #839948 by MathLover1(20850) $\"\"$ $\"About$

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\n" ); document.write( "\n" ); document.write( "Let $\"+t+\"$ = time traveled at $\"+53mph\"$
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\n" ); document.write( " $\"+%283.5-t%29\"$ = time traveled at $\"+59mph\"$ \r
\n" ); document.write( "\n" ); document.write( "Total time is given as $\"+3.5h\"$ , therefore\r
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\n" ); document.write( "\n" ); document.write( "Write distance equation; \r
\n" ); document.write( "\n" ); document.write( " $\"+d+=+s%2At\"$
\n" ); document.write( " $\"+53t+%2B+59%283.5-t%29+=200.5\"$
\n" ); document.write( " $\"+206.5+-+6+t+=+200.5\"$
\n" ); document.write( " $\"+206.5+-200.5+=+6+t+\"$
\n" ); document.write( " $\"+6+=+6+t\"$
\n" ); document.write( " $\"+t=1\"$ \r
\n" ); document.write( "\n" ); document.write( "so,\r
\n" ); document.write( "\n" ); document.write( " $\"+t=1\"$ =>Charles traveled $\"+1h+\"$ at $\"+53mph\"$
\n" ); document.write( " $\"+%283.5-t%29=3.5-1=2.5\"$ =>Charles traveled $\"+2.5h\"$ at $\"+59mph\"$ \r
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