Can someone answer this? will mark what are the domain and range of the function? explain. f(x)= a. the domain is {x | x ∈ r, x ≠ 0}. the range is {f (x) | f (x) < 0}. b. the domain is {x | x ∈ r, x ≠ 0}. the range is {f (x) | f (x) ∈ r, f(x) ≠ 0}. c. the domain is {x | x ∈ r, x ≠ 0}. the range is {f (x) | f (x) > 0}.

C. The domain is {x | x ∈ R, x ≠ 0}. The range is {f (x) | f (x) > 0}.

Step-by-step explanation:

f(x)= 3/ (5x^2)

the domain is the possible x values

the only values x cannot take is 0  (we cannot divide by 0)

The domain is {x | x ∈ R, x ≠ 0}

the range is the possible y values

y cannot be negative  since x is squared

and cannot be 0   since x^2 cannot equal 0

The range is {f (x) | f (x) > 0}.

∠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.

homozygous. one is homozygous, while two is heterozygous.