Find the result of reflecting the figure across the x-axis and then translating it along the vector 〈3, 2〉

[tex]Find the result of reflecting the figure across the x-axis and then translating it along the vect[/tex]

Skip to content# Find the result of reflecting the figure across the x-axis and then translatingit along the vector 〈3, 2〉

Mathematics ##
Comments (8) on “Find the result of reflecting the figure across the x-axis and then translatingit along the vector 〈3, 2〉”

### Leave a Reply Cancel reply

Find the result of reflecting the figure across the x-axis and then translating it along the vector 〈3, 2〉

[tex]Find the result of reflecting the figure across the x-axis and then translating it along the vect[/tex]

A,C,D,F,G,H I just took the test

A, C, D, F, G and H. the only ones that are wrong are B and E

Explanation:

I took the test, trust me, image below.

[tex]Max is trying to prove to his friend that two reflections, one across the x-axis and another across[/tex]

Max is correct

If one reflects a figure across the x-axis, the points of the image can be found using the pattern (x, y) ⇒ (x, –y).

If one reflects a figure across the y-axis, the points of the image can be found using the pattern (x, y) ⇒ (–x, y).

Taking the result from the first reflection (x, –y) and applying the second mapping rule will result in (–x, –y), not (y, x), which reflecting across the line should give.

Step-by-step explanation:

The answer above pretty well explains it.

The net result of the two reflections will be that any figure will retain its orientation (CW or CCW order of vertices). It is equivalent to a rotation by 180°. The single reflection across the line y=x will reverse the orientation (CW ⇔ CCW). They cannot be equivalent.

The student that is correct with statements that will him prove his conjecture are

1. josiah is correct when reflection across the x-axis followed by a reflection across the y-axis will result in a reflection across the line y = x for a pre-image in quadrant ii.

2. if one reflects a figure across the x-axis from quadrant ii, the image will end up in quadrant iii.

3. if one reflects a figure across the y-axis from quadrant iii, the image will end up in quadrant iv.

4. a figure that is reflected from quadrant ii to quadrant iv will be reflected across the line y = x.

the y = x line is a straight line sloping upwards to the right.

60

step-by-step explanation:

every triangle measures a total of 120 degrees. you have to find angle c. we know that isosceles triangles angles are equal. 60 degrees plus 60 degrees gives you 120.

Disjakajshisisisisjsjsjs

The correct option is;

If one reflects a figure across the y-axis, the points of the image can be found using the pattern (x, y) Right-arrow (x, -y).

If one reflects a figure across the y-axis, the points of the image can be found using the pattern (x, y) Right-Arrow (-x, y).

Taking the result from the first reflection (x, -y) and applying the second mapping rule will result in (-x, -y), not (y, x), which reflection across the line y = x should give

Step-by-step explanation:

We have that for reflection across the x-axis, (x, y) → (x, -y)

For reflection across the y-axis, (x, y) → (-x, y)

Therefore, given that the pre-image before the reflection across the y-axis is (x, -y), we have;

For reflection across the y-axis, (x, -y) → (-x, -y)

For reflection across the line, y = x, gives (x, y) → (y, x) which is not the same as (-x, -y)

Did u want to find the value of x?