Best answer will get brainliest! Use the figure below to answer the question that follows:

Intersecting triangles ACE and BDF. They intersect at points G, H, I, and J.

What must be given to prove that ΔBIJ ~ ΔBDF?

Select one:

a. segment GH is parallel to segment JI

b. segment JI is parallel to segment FD

c. ∠BGH ≅∠BJI

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6: 4= 3/2

18: 8= 9/4

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asnwer is (g o h o f)(x) = 2x - 6/x + 7

step-by-step explanation:

given,

f(x) = x - 3/x

g(x) = x + 3

h(x) = 2x + 1

we need to find : (g o h o f)(x)

we can write (g o h o f)(x) = g[ h{ f(x) } ]

so first we find h{f(x)}

h{f(x)} = f(x) + 3 = (x - 3/x) + 3

h{f(x)} = x - 3/x + 3

now we find g[ h{ f(x) } ].

g[ h{ f(x) } ] = g(x - 3/x + 3)       {plug h{f(x)} = x - 3/x + 3 in g[h{f(x)}] }

g(x - 3/x + 3) = 2 (x - 3/x + 3) + 1

                    = 2x - 6/x + 6 + 1

                    = 2x - 6/x + 7

hence we get (g o h o f)(x) = 2x - 6/x + 7

all of them are elements inside set of real numbers since none of them appears to have complex number form.

hope this .

r3t40