Mike and molly are having an argument about domain. if f(x) = sq root of x-1 and g(x) = x^2, what is the domain of g(f(x))\)? mike says that since g(f(x)) = g(sq root x-1) = (sq root x-1)^2 = x-1, the domain of g(f(x)) is all real numbers. molly says that he is wrong since f(x) has a restricted domain. who is correct here? mike or molly? if molly is correct, what is the domain of g(f(x)

125 yards

step-by-step explanation:

i take 75 plus 50.

hello,

as we don't know if g(x) is continue,

answer b

f(x)=2x²-8x+1=2(x²-4x+4)+1-8=2(x-2)²-7

vertex is (2,-7) ( a minimum since coefficient of x² is > 0)

step-by-step explanation:

let pq = qr = rs = x, and xz = y and zu = z.

if the horizontal lines are parallel, then corresponding segments are in proportion:

substitute:

therefore 3 = y = z.

xu = y + z → xu = 3 + 3 = 6