Max is trying to prove to his friend that two reflections, one across the x-axis and another across the y-axis, will not result in a reflection across the line y = x for a pre-image in quadrant II. His friend Josiah is trying to prove that a 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. Which student is correct, and which statements below will help him prove his conjecture? Check all that apply. Max is correct.
Josiah is correct.
If one reflects a figure across the x-axis from quadrant II, the image will end up in quadrant III.
If one reflects a figure across the y-axis from quadrant III, the image will end up in quadrant IV.
A figure that is reflected from quadrant II to quadrant IV will be reflected across the line y = x.
If one reflects a figure across the x-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 reflecting across the line y = x should give.

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)

something that characterises a right triangle is the fact that we may use pythagoras' theorem to find any one of its sides given that we have the length of the other two sides. here, we have the length of the hypotenuse, √(87) yd, and one other side, √(23) yd, thus we can easily find the length of the remaining side using pythagoras' theorem as defined below:

c^2 = a^2 + b^2, where c is the length of the hypotenuse, and a and b are the lengths of the other two sides.

thus, given that c = √(87) and a = √(23), we may substitute these into the formula we defined above to get:

c^2 = a^2 + b^2

(√87)^2 = (√23)^2 + b^2 (substitute c = √87 and a = √23 into the equation)

87 = 23 + b^2 (evaluate (√87)^2 = 87 and (√23)^2 = 23 - remember that the square root of a number, squared will equal that very same number)

64 = b^2 (subtract 23 from both sides)

√64 = b (take the square root of both sides)

8 = b (evaluate √64 = 8)

therefor, the length of the unknown leg in the right triangle is 8 yd (first answer).