Polygon abcde has the vertices a(2, 8), b(4, 12), c(10, 12), d(8, 8), and e(6, 6). polygon mnopq has the vertices m(-2, 8), n(-4, 12), o(-10, 12), p(-8, 8), and q(-6, 6). a transformation or sequence of transformations that can be performed on polygon abcde to show that it is congruent to polygon mnopq is a . if polygon mnopq is translated 3 units right and 5 units down, it will coincide with a congruent polygon, vwxyz, with its vertices at .

A transformation or sequence of transformations that can be performed on polygon ABCDE to show that it is congruent to polygon MNOPQ is a:

Reflection across the y-axis

Step-by-step explanation:

Please, see the attached graph.

The polygon ABCDE is on the right side, first (I) quadrant.

The polygon MNOPQ is on the left side, second (II) quadrant.

We can see that the polygon MNOPQ is a reflection of the polygon ABCDE across the y-axis (only the abscissas change their signs from positive to negative) .

last part is

v(1, 3)

w(-1, 7)

x(-7, 7)

y(-5, 3)

z(-3, 1)

what he has is correct he just forgot about the last part

(1, –1), (16, –4), (36, –6)

step-by-step explanation:

given that the function is

this can also be written as

we have to graph this function and check which list of points lie on the graph

see attachment for graph:

we find that the points lying are (1,-1) (16,-4) and (36,-6)

since square root is there we find that x can only be posiitve but y can be since square root both positive and negative.

answer: what is the question you're trying to ask?

step-by-step explanation: