Let #f(x)=x^2+5# and #g(x)= (x+5)/x#. What is #(g*f)(-3)#?

Question

A) 1/4
B) 14/19
C) 19/14
D) 49/9

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Answer 1 · 2018-07-30T10:54:39

#19/14to(C)#

#(g@f)(x)=g(f(x))#

#=(x^2+5+5)/(x^2+5)=(x^2+10)/(x^2+5)#

#(g@f)(-3)=((-3)^2+10)/((-3)^2+5)=19/14to(C)#