Which type of transformation maps abc to a'b'c?

Transformation Shifted by 4 units right and 2 units down .

Step-by-step explanation:

Given  : Graph .

To find : Which type of transformation maps ABC to A'B'C.

Solution : We have given graph with  transformation maps ABC to A'B'C.

Coordinates of A ( -3 , 2) , B( -2 ,4) , C( -1 ,3) .

Coordinates of A' (1 , 0) , B'( 2 ,2) , C'( 3 ,1) .

We can see from given graph triangle ABC is shifted right ( x axis) by 4 units and move down by 2 units (y axis ).

Then

A ( -3 + 4 , 2 -2) →→A' (1 , 0)

B( -2+4 ,4 -2)  →→  B'( 2 ,2)

C( -1+ 4 ,3 -2)  →→ C'( 3 ,1) .

Therefore, Transformation Shifted by 4 units right and 2 units down .

∠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.