Ask your own question, for FREE!
Mathematics 14 Online
OpenStudy (anonymous):

Parallelogram ABCD is reflected over the x-axis. What rule shows the input and output of the reflection, and what is the new coordinate of A'? https://cdn.flvsgl.com/assessment_images/geometry_v15_gs_ca-xml/76546_54075fac/0200_g4_q2.jpg A. (x,y)→(y,-x); A' is at (1, 5) B. (x,y)→(-y,x); A' is at (-1, -5) C. (x,y)→(-x,y); A' is at (5, 1) D. (x,y)→(x,-y); A' is at (-5, -1)

OpenStudy (anonymous):

D

OpenStudy (mynameisnemo):

Remember that people can't open the links and see it.

OpenStudy (anonymous):

lol ok hang on

OpenStudy (mynameisnemo):

You need to screenshot it, the links say forbidden when we click on them. I don't know how @KGabriel99 got that answer without seeing it.

OpenStudy (anonymous):

you sure @KGabriel99 can you see it?

OpenStudy (anonymous):

I can guarantee it. This is VS and I have the notes for it. All x axis reflections are (x,y)->(x,-y)

OpenStudy (mynameisnemo):

@iloveyou;* are you taking these questions off an online class test?

OpenStudy (anonymous):

Is that segment one?

OpenStudy (mynameisnemo):

I have the notes too.

OpenStudy (anonymous):

no its hw

OpenStudy (anonymous):

• Another way to say "reflect across the x-axis" is to say "reflect across the line y=0" since the line created by graphing y=0 is the same as the x-axis. • An image that is a reflection across the x-axis, or across the line y=0, will have the same x-coordinates as the pre-image but opposite y-coordinates. • Therefore, the rule for reflecting an image across the x-axis can be described as (x, y) → (x, −y).

OpenStudy (mynameisnemo):

Yeah he's right .-. I read the question wrong.

OpenStudy (anonymous):

Yup.

OpenStudy (mynameisnemo):

I appologize @KGabriel99

OpenStudy (anonymous):

so D is correct right?

OpenStudy (anonymous):

It's fine. Understandable mistake. @MyNameIsNemo

OpenStudy (mynameisnemo):

Yes D is correct

OpenStudy (anonymous):

thnx guys :D

OpenStudy (anonymous):

anytime

OpenStudy (mynameisnemo):

Np anytime

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!