Question:
1 answer:
A) The vertices of the ABCD square are:
(2,2), (1(4), (3,5), and (4,3).
The reflection over the x-axys keeps the same x-coordinate and changes the y-coordinate ot its negative.
So, the vertices of the square A'B'C'D' are:
(2,-2), (1,-4), (3,-5), and (4,-3).
You can see the figure attached showing the four vertices.
<span>B. Is AB congruent to A'B'? Explain
Answer:
Yes, they are congruent. You can affirm that because reflections are transformations that do not change either sizes or angles, but keep them, so the two squares are congruent.
C. Is the area of ABCD ≈ area of A'B'C'D'? Explain.
Answer:
Yes, the two areas are equal since you have shown that the two figures are congruent.
</span>
(2,2), (1(4), (3,5), and (4,3).
The reflection over the x-axys keeps the same x-coordinate and changes the y-coordinate ot its negative.
So, the vertices of the square A'B'C'D' are:
(2,-2), (1,-4), (3,-5), and (4,-3).
You can see the figure attached showing the four vertices.
<span>B. Is AB congruent to A'B'? Explain
Answer:
Yes, they are congruent. You can affirm that because reflections are transformations that do not change either sizes or angles, but keep them, so the two squares are congruent.
C. Is the area of ABCD ≈ area of A'B'C'D'? Explain.
Answer:
Yes, the two areas are equal since you have shown that the two figures are congruent.
</span>
You might be interested in
Answer:
x = -3/2
Step-by-step explanation:
The removable discontinuity is at the value of x where a denominator factor cancels a numerator factor, and those factors are zero.
That cancellation occurs where 2x+3 = 0, at x=-3/2.
The function p(x) has a removable discontinuity when x = -3/2.
