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This is about composition of function.
In `(f_0g)(x)` we have to replace x in f(x) terms of x in g(x).
`f(x) = x^2`
`g(x) = x-3`
`(f_0g)(x) = (x-3)^2`
`(g_0g)(x) = (x-3-3) = (x-6)`
Then we can write the answers as;
`(f_0g)(-2) = (-2-3)^2 = 25`
`(g_0g)(7) = (7-6) = 1`
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