document.write( "Question 247408: Donna drove half the distance of a trip at 40 mi/h. At what speed would she have to drive for the rest of the distance so that the average speed for the entire trip would be 45 mi/h? \n" ); document.write( "
Algebra.Com's Answer #180512 by ankor@dixie-net.com(22740)\"\" \"About 
You can put this solution on YOUR website!
half the distance of a trip at 40 mi/h.
\n" ); document.write( " At what speed would she have to drive for the rest of the distance so that the average speed for the entire trip would be 45 mi/h?
\n" ); document.write( ":
\n" ); document.write( "Let s = speed required on the 2nd half to average 45 mph for the trip
\n" ); document.write( "Let d = total distance of the trip
\n" ); document.write( ":
\n" ); document.write( "Write time equation; time = dist/speed
\n" ); document.write( "\".5d%2F40\" + \".5d%2Fs\" = \"d%2F45\"
\n" ); document.write( "multiply equation by 360s to get rid of the denominator:
\n" ); document.write( "9s(.5d) + 360(.5d) = 8s(d)
\n" ); document.write( ":
\n" ); document.write( "4.5sd + 180d = 8sd
\n" ); document.write( ":
\n" ); document.write( "Divide thru by d
\n" ); document.write( "4.5s + 180 = 8s
\n" ); document.write( ":
\n" ); document.write( "180 = 8s - 4.5s
\n" ); document.write( ":
\n" ); document.write( "180 = 3.5s
\n" ); document.write( "s = \"180%2F3.5\"
\n" ); document.write( "s = 51.43 mph
\n" ); document.write( ":
\n" ); document.write( "
\n" ); document.write( "
\n" );