## Intermediate Algebra: Connecting Concepts through Application

a. $C=23,250+145(x)$ b. 37, 750 dollars c. This is impossible... their costs can never be lower than their fixed cost of 23,250 dollars. d. 377.50 dollars
a. C=cost per month, x=number of sets produced $C=23,250+145(x)$ b. x=100 $C=23,250+145(100)$ $C=23,250+14500$ $C=37,750$ c. C=20,000 $20,000=23,250+145(x)$ $20,000-23,250=145(x)$ $-3,250=145(x)$ $-3,250\div145=x$ $x=-22.41$ This is impossible... there costs can never be lower than their fixed cost of 23,250. d. p=price per set, R=revenue=100(p), C=37,750 C needs to equal R to break even. $37,750=100(p)$ $37,750\div100=p$ $p=377.50$