Let f(x)=Ax−2B2and g(x)=Bx, where B≠0 . If f(g(1))=0 , what is A in terms of B?
f(x) = Ax - 2B^2 g(x) = Bx
g(1) = B(1) = B
So
f( g(1) ) = f(B) = 0
A(B) - 2B^2 = 0
A (B) = 2B^2
A = 2B
Very good! Amazing, CPhill!
It was nothing....really !!!
+4624 
