document.write( "Question 404549: Charlie is going to rent a truck for one day. There are two companies he can choose from, and they have the following prices. \r
\n" ); document.write( "\n" ); document.write( "Company A charges $127 and allows unlimited mileage.
\n" ); document.write( "Company B has an initial fee of $55.00 and charges an additional $0.90 for every mile driven.
\n" ); document.write( "For what mileages will Company A charge at least as much as Company B? Use m for the number of miles driven, and solve your inequality for m . \r
\n" ); document.write( "\n" ); document.write( "my answer-please check me
\n" ); document.write( "a=127
\n" ); document.write( "b=55+.9m
\n" ); document.write( "127=55+.9m
\n" ); document.write( "72=.9m
\n" ); document.write( "m=80 \r
\n" ); document.write( "\n" ); document.write( "I'm not sure if the answer should be m>=80 or m<=80\r
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Algebra.Com's Answer #285894 by stanbon(75887)\"\" \"About 
You can put this solution on YOUR website!
Charlie is going to rent a truck for one day. There are two companies he can choose from, and they have the following prices.
\n" ); document.write( "Company A charges $127 and allows unlimited mileage.
\n" ); document.write( "Company B has an initial fee of $55.00 and charges an additional $0.90 for every mile driven.
\n" ); document.write( "----
\n" ); document.write( "A(x) = 127x dollars
\n" ); document.write( "B(x) = 55 + 0.9x dollars
\n" ); document.write( "----
\n" ); document.write( "For what mileages will Company A charge \"at least as much\" as Company B? Use m for the number of miles driven, and solve your inequality for m .
\n" ); document.write( "my answer-please check me
\n" ); document.write( "a=127
\n" ); document.write( "b=55+.9m
\n" ); document.write( "Solve: a >= b
\n" ); document.write( "127 >= 55+.9m
\n" ); document.write( "72 >= 0.9m
\n" ); document.write( "m <= 72/0.9
\n" ); document.write( "---
\n" ); document.write( "m <= 80 miles (a >= b when # of miles <= 80)
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\n" ); document.write( "Cheers,
\n" ); document.write( "Stan H.
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