Answer:
- 3
Step-by-step explanation:
[tex]f(x) = \sqrt{x + 1} \\ plug \: x = 0 \\\therefore f(0) = \sqrt{0 + 1} \\\therefore f(0) = \sqrt{1} \\ \therefore f(0) =1 \\ \\ g(x) = 2x - 5 \\ plug \: x = f(0) \\ \therefore g( f(0)) = 2 f(0)- 5 \\ \therefore g( f(0)) = 2 (1) - 5 \\\therefore g( f(0)) = 2 - 5 \\ \\ \huge \red {g( f(0)) = \boxed {- 3} }[/tex]
