The coordinates of the vertices of △jkl△jkl are j(1,4)j(1,4), k(6,4)k(6,4), and l(1,1)l(1,1). the coordinates of the vertices of △j′k′l′△j′k′l′ are j′(0,−4)j′(0,−4), k′(−5,−4)k′(−5,−4), and l′(0,−1)l′(0,−1). what is the sequence of transformations that maps △jkl△jkl to △j′k′l′△j′k′l′ ?

Solution: The ΔJKL reflect along the x-axis, then reflects along the y-axis after that translate 1 unit right to transform ΔJ'K'L'.

Explanation:

The vertices of ΔJKL lies in first quadrant because the coordinated of the vertices are positive. It means the vertices are in the form of (x,y).

If the ΔJKL reflects along the x-axis then the vertices are in the form of (x,-y). Now the  ΔJKL reflects along the y-axis then the vertices are in the form of (-x,-y).

The given coordinates of the the vertices after transformation is J'(0,-4), K'(-5,-4), L'(0,-1). But after reflection along x and y axis the vertices of ΔJKL are J'(-1,-4), K'(-6,-4), L'(-1,-1). Since the y coordinates are same so we have to tralanate horizontally. Since the difference between x coordinates is positive 1, so we have translate the triangle 1 unit right.

Therefore, the ΔJKL reflect along the x-axis, then reflects along the y-axis after that translate 1 unit right to transform ΔJ'K'L'.

how about try it is rotate 90* counterclockwise about the origin and translated 1 unit left that's how you get the right answer

took the test too so

∠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.

step-by-step explanation:

the true statements are:

r = set of real numbers

n = set of natural numbers, 1,2,

w = n u{0} = {0,1,}