Answer:
The sequence of transformations that maps ΔABC to ΔA'B'C' is the reflection across the <u>line y = x</u> and a translation <u>10 units right and 4 units up</u>, equivalent to T₍₁₀, ₄₎
Step-by-step explanation:
For a reflection across the line y = -x, we have, (x, y) → (y, x)
Therefore, the point of the preimage A(-6, 2) before the reflection, becomes the point A''(2, -6) after the reflection across the line y = -x
The translation from the point A''(2, -6) to the point A'(12, -2) is T(10, 4)
Given that rotation and translation transformations are rigid transformations, the transformations that maps point A to A' will also map points B and C to points B' and C'
Therefore, a sequence of transformation maps ΔABC to ΔA'B'C'. The sequence of transformations that maps ΔABC to ΔA'B'C' is the reflection across the line y = x and a translation 10 units right and 4 units up, which is T₍₁₀, ₄₎
add 2q to both sides.
subtract 5 to both sides.
divide 5 to both sides.
the answer is:
If the diagonals of a quadrilateral are perpendicular, then it is a rhombus. False, diagonals don't have to be congruent or bisect each other. The diagonals of a rectangle bisect its angles. A kite with all consecutive angles congruent must be a square.
Answer:
(7*100) can be written as 7*10^2
4 * / 100 i don’t know
if it was 4 * 100 then it’s 4*10^2
if it was divided then it’s 4*10^-2
8* 1/1000 = 8/1000 = 8*10^-3
add together to get 1.100008 * 10^3 (if it was 4*100)
or 7.00048 * 10^2 ( if it was 4/100)
