Intermediate Algebra: Connecting Concepts through Application

a. $C=150+5(t)$ for $t\gt 100$ b. 1650 dollars c. 270 T-shirts d. 5.50 dollars
a. C=cost, t= # of T-shirts $C=150+5(t)$ for $t\gt 100$ b. t=300 $C=150+5(300)$ $C=150+1500$ $C=1650$ c. C=1500 $1500=150+5(t)$ $1500-150=5(t)$ $1350=5(t)$ $1350\div5=t$ $t=270$ d. P=0, R=revenue, C=1650, x=Price of each t-shirt $P=R-C$ $0=300(x)-1650$ $0+1650=300(x)$ $1650=300(x)$ $1650\div300=x$ $x=5.5$