[tex]{ \qquad\qquad\huge\underline{{\sf Answer}}} [/tex]
Here we go ~
1. At first Angelica should put the initial value " 6 " in second function ( y = x² - 6 ) amd then the result from second function in first function [tex]{ \sqrt{x - 5}} [/tex] so we will get 5 ~
procedure :
[tex]\qquad \sf \dashrightarrow \: y = {x}^{2} - 6[/tex]
[tex]\qquad \sf \dashrightarrow \: y = (6) {}^{2} - 6[/tex]
[tex]\qquad \sf \dashrightarrow \: y = 36 - 6[/tex]
[tex]\qquad \sf \dashrightarrow \: y = 30[/tex]
not put the result in 1st function ~
[tex]\qquad \sf \dashrightarrow \: y = \sqrt{x - 5} [/tex]
[tex]\qquad \sf \dashrightarrow \: y = \sqrt{30 - 5} [/tex]
[tex]\qquad \sf \dashrightarrow \: y = \sqrt{25} [/tex]
[tex]\qquad \sf \dashrightarrow \: y = 5[/tex]
Hence it's confirmed ~
2. Yes it's possible, if we take the initial value as 6 and put it in first function and then put the result in second function, we will get our required value as -5
procedure :
[tex]\qquad \sf \dashrightarrow \: y = \sqrt{x - 5} [/tex]
[tex]\qquad \sf \dashrightarrow \: y = \sqrt{6 - 5} [/tex]
[tex]\qquad \sf \dashrightarrow \: y = \sqrt{1} [/tex]
[tex]\qquad \sf \dashrightarrow \: y = 1[/tex]
now plug the result in 2nd function ~
[tex]\qquad \sf \dashrightarrow \: y = {x}^{2} - 6[/tex]
[tex]\qquad \sf \dashrightarrow \: y = (1 {)}^{2} - 6[/tex]
[tex]\qquad \sf \dashrightarrow \: y = 1 - 6[/tex]
[tex]\qquad \sf \dashrightarrow \: y = - 5[/tex]
