document.write( "Question 206985: A truck can be rented from Basic Rental for $50 per day plus $0.20 per mile. Roadrunners charges $20 per day plus $0.50 per mile to rent the same truck. How many miles must be driven in a day to make the rental cost for Basic Rental a better deal than that for Roadrunners?\r
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Algebra.Com's Answer #156546 by MathTherapy(10552)![]() ![]() You can put this solution on YOUR website! Since Basic Rental charges $50, plus $0.20 per mile, and since Roadrunners \r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "charges $20, plus $0.50, per mile, then in order for Basic Rental to have a \r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "better rental deal than Roadrunners, we'll have: \r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "50 + .2(m) < 20 + .5(m) (using m for amount of miles)\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "50 + .2m < 20 + .5m\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "-.3m < - 30\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Therefore, in order for Basic Rental to have a better deal than Roadrunners, \r \n" ); document.write( "\n" ); document.write( "the renter has to rent for MORE THAN \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Check:\r \n" ); document.write( "\n" ); document.write( "Since m > 100, we'll use value of 101 \r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "50 + .2(101) < 20 + .5(101)\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "50 + 20.20 < 20 + 50.50\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "70.20 < 70.50 \n" ); document.write( " \n" ); document.write( " |