(HoG)(x) = (2x)2 + 4 simply because HoG(x) is actually H(G(x)). So where ever there was an x in H(x) we substitute our value of G(x). Now the only thing left is to put in the 1. so out answer is (2(1))2 + 4 = 8Only in the 6th grade ; )
5(1+9v) = 5+ 45v
20b+12= 4(5b+3)
And the last one is r=2
Hope this helps!
Permutations are written as nPx where n is the number of total choices possible and x is the number of choices that will be used. This is calculated as nPx = n! / (n-x)!.
Permutations represent the number of ways we can choose x objects from n possibilities where the order of selection matters.
