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vovikov84 [41]
3 years ago
13

Parallelogram efgh has been reflected across the x axis and then rotated 180 degrees around the origin. which of the following t

ransformations would return the parallelogram to it's original position?
a. reflection across the line y=x
b.reflection across the x-axis
c.reflection across the x-axis and then reflection across the y-axis
d.reflection across the y-axis <--- (this one??)

Mathematics
2 answers:
Brilliant_brown [7]3 years ago
6 0
You are correct, it would be a reflection across the y-axis. I am great at visualizing but for those who are not (Not saying you) all they have to do is cut out a parallelogram and mimic the movements.

nikklg [1K]3 years ago
5 0

Answer:

Correct answer is D

Step-by-step explanation:

Parallelogram EFGH has been reflected across the x-axis to form the parallelogram ABCD. The reflection across the x-axis has a rule:

(x,y)→(x,-y).

Then parallelogram ABCD has been rotated 180° around the origin to form the parallelogram IJKL. This transformation has a rule:

(x,y)→(-x,-y).

Both these transformation together have a rule:

(x,y)→(x,-y)→(-x,y)

i.e.

(x,y)→(-x,y).

If you have to return the parallelogram to it's original position, then point (-x,y) should turn in point (x,y). So you have to apply the rule:

(x,y)→(-x,y)

that is reflection across the y-axis (see attached diagram for illustration).

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4 0
3 years ago
QHome Spring 2020
raketka [301]

Answer:

(D)Equilateral Triangle

Step-by-step explanation:

Given a circle centre M; and

The measure of major arc JL = 300 degrees

The triangle formed by radii ML and MJ and chord JL is Triangle JLM.

Since ML=MJ (radii of a circle), the base angles are equal.

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\angle MLJ= \angle MJL\\\angle LMJ =60^\circ\\$Therefore:\\\angle LMJ+2\angle MLJ=180^\circ\\60^\circ+2\angle MLJ=180^\circ\\2\angle MLJ=180^\circ-60^\circ\\2\angle MLJ=120^\circ\\\angle MLJ=60^\circ

We can see that all the angles of triangle JLM are 60 degrees, therefore Triangle JLM is an Equilateral Triangle.

8 0
3 years ago
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