document.write( "Question 722710: a candy truck leaves the store at 5:00 am each mornng on its 144 mile route.one day the driver gets a late start and does not leave the store until 5:30 am. to finish his route on time the driver drives 4 miles per hour faster than usual.at what speed does he usually drive ? \n" ); document.write( "
Algebra.Com's Answer #442863 by josgarithmetic(39617)\"\" \"About 
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normal speed r in miles per hour. The difference between 5:00am and 5:30am is 1/2 hour.\r
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\n" ); document.write( "\n" ); document.write( "____________speed________________time____________________distance
\n" ); document.write( "normal_______r___________________\"t=144%2Fr\"_______________________144
\n" ); document.write( "late________r+4__________________\"t-1%2F2=144%2F%28r%2B4%29\"___________________144\r
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\n" ); document.write( "\n" ); document.write( "The normal time being 1/2 hour more than the late time suggests taking the difference between their expressions. Both expressions may be shown as expressions using r, the rate.\r
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\n" ); document.write( "\n" ); document.write( "normal time minus late time equals 1/2 hour,
\n" ); document.write( "\"highlight%28144%2Fr-144%2F%28r%2B4%29=1%2F2%29\".
\n" ); document.write( "Multiply left side and right side by \"2r%28r%2B4%29\" to clear the fractions.
\n" ); document.write( "\"288%28r%2B4%29-288r=r%28r%2B4%29\"
\n" ); document.write( "\"288r%2B1152-288r=r%5E2%2B4r\"
\n" ); document.write( "\"1152=r%5E2%2B4r\"
\n" ); document.write( "\"highlight%28r%5E2%2B4r-1152=0%29\", .\r
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\n" ); document.write( "\n" ); document.write( "\"r=%28-4%2Bsqrt%2816-4%2A%28-1152%29%29%29%2F2=%2868-4%29%2F2\"
\n" ); document.write( "=32 miles per hour --- ANSWER.
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