All three are parts to the triangle congruence theorem. (ASA: angle side angle, AAS: angle angle side, SSA: side side angle). Is there an option D.?
the solution is attached below
the tangential acceleration of car A = -9.8m/s^2
car B = - 13.44 m/s^2
car C = -6.16m/s^2
Answer:
7w+11
Step-by-step explanation:
GEMDAS
You must distribute first, so 3(8p+7r). That will equal 24p +21r. Then you must add all the values together according to like terms, so the variables must match. So simplified it will be 22+24p+30r
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₎
