I will mark brainliest How is quadrilateral ABCD related to quadrilateral A'B'C'D'?
'
A
A'B'C'D' is the 90 degree clockwise rotation of ABCD with the origin as its center.
B
A'B'C'D' is the 90 degree counter-clockwise rotation of ABCD about point D.
C
A'B'C'D' is the reflection of ABCD across the line y = x.
D
A'B'C'D' is the reflection of ABCD across the line y = -x.

D

Step-by-step explanation:

I'm pretty sure it's D because if you graph the line y=-x its a perfect reflection.

intensity of the vector is v= √37 ≈ 6.08 units and make angle ∡α ≈ 9.46° with east direction

step-by-step explanation:

required vector is consists of the two componentsvx= 2+4=6 units   and vy= 1 unit and vx ⊥ vywe will use pythagorian theorem to find intensity of the vector vv∧2 = vx∧2 + vy∧2 => v = √vx∧2 + vy∧2 = √6∧2 + 1∧2 = √36+1 = √37 ≈ 6.08 unitsthe angle ∡α between vector and east direction we wil find with tanαtanα = 1/6 => α = arc tanα = arc 1/6 => α ≈ 9.46°

hope this

i think 1: 2

step-by-step explanation: