Answer: AA similarity theorem.
Step-by-step explanation:
Given : AB ∥ DE
Prove: ΔACB ≈ ΔDCE
We are given AB ∥ DE. Because the lines are parallel and segment CB crosses both lines, we can consider segment CB a transversal of the parallel lines. Angles CED and CBA are corresponding angles of transversal CB and are therefore congruent, so ∠CED ≅ ∠CBA.
Also ∠C ≅ ∠C using the reflexive property.
Therefore by AA similarity theorem , ΔACB ≈ ΔDCE
- AA similarity theorem says that if in two triangles the two pairs of corresponding angles are congruent then the triangles are similar .
The value of the expression would be 16.25 or 16 and 1/4, because when you multiply something by 1/2, you’re basically dividing it by 2, and when you put parentheses next to a number (for example 2(4) ), it means that you’re multiplying it, and when more than one values are in the parentheses, they all get multiplied, basically distributive property.
V=s/t
s=300 km = 300000 m
t=1 h and 30 min = 5400 sec
v=300/1,5= 200 km/h
or
v=300000/5400=55,5 m/s
Answer:
51
Step-by-step explanation:
Hope this helps
