Which scenario is best represented by a linear function? A
The area of a square, A, where the length of each side of the square is represented by
S (A= s?)
The perimeter of a square, P, where the length of each side of the square is
B represented by s
(P = 4s)
The surface area of a cube, A, where the length of each side of the cube is
C represented by s
(A = 6s )
The volume of a cube, V, where the length of each side of the cube is represented by
s(V = $)​

Step-by-step explanation:

since we can see what linear function is

the second condition is the best fit

therefore

our answer is b

∠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.