Placing the function of Machine B before A will be required in other to get a final output of 5
Given an initial input of 6 :
Function of machine A :
- [tex]y = \sqrt{x - 5} [/tex]
Function of machine B :
- [tex]y = {x}^{2} - 6[/tex]
Placing machine B first :
[tex]x = 6[/tex]
[tex]y = {6}^{2} - 6 = 36 - 6 = 30[/tex]
Input to machine A = 30
Put x = 30 in Function of machine A :
[tex]y = \sqrt{30 - 5} = \sqrt{25} = 5[/tex]
Therefore, in other to get a final output of 5 ; place Machine B first then Machine A.
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