Express the function G in the form f ∘ g ∘ h

G(x) = 3/(7+sqrt of x)^2

hmmm brainliest?

Step-by-step explanation:

Given f (x) = sqrt(x) and g(x) = x – 2, find the domains of ( f o g)(x) and (g o f )(x).

Since  f (x) involves a square root, the inputs have to be non-negative. This means that the domain (the set of x-values) for f (x) is "all x > 0". Then, in (g o f )(x), where I'm plugging x first into f (x) = sqrt(x), the domain is at least restricted to "all x > 0". Let's see what the two compositions look like:

( f o g)(x) =  f (g(x))

   =  f (x – 2)

   = sqrt(         )

   = sqrt(x – 2)

The domain for the square root is all inputs that make "x – 2" non-negative. That is, all x such that x – 2 > 0. Solving this for x, I get that the domain of ( f o g)(x) is "all x > 2".

i think 1: 2

step-by-step explanation:

∠1≅∠2 by the alternate exterior angles theorem.

step-by-step explanation:

given, a ∥ b and ∠1 ≅ ∠3 .we have to prove that e ∥ f

we know that ∠1≅∠3 and that a || b because they are given. we see that by the alternate exterior angles theorem. therefore, ∠2≅∠3 by the transitive property. so, we can conclude that e || f by the converse alternate exterior angles theorem.

we have to fill the missing statement.

transitivity property states that if a = b and b = c, then a = c.

now, given ∠1≅∠3 and by transitivity property ∠2≅∠3 .

hence, to apply transitivity property one angle must be common which is not in result after applying this property which is ∠1.

the only options in which ∠1 is present are ∠1 and ∠2, ∠1 and ∠4

∠1 and ∠4 is not possible ∵ after applying transitivity we didn't get ∠4.

hence, the missing statement is ∠1≅∠2.

so, ∠1≅∠2 by the alternate exterior angles theorem.