Answer:
Step-by-step explanation:
c.
[tex]a(x)=x^3-3\\b(x)=2x+1\\c(x)=\frac{3}{x+2} \\d(x)=4-x\\a.\\a(4)=4^3-3=64-3=61\\b.\\\\c(x-5)=\frac{3}{x-5+2} =\frac{3}{x-3} \\c.\\d(-1)=4-(-1)=4+1=5\\b(d(-1))=b(5)=2(5)+1=11\\\\d.\\(coa)(2)=c(a(2))=c(2^3-3)=c(8-3)=c(5)=\frac{3}{5+2} =\frac{3}{7}[/tex]
