PLEASE HELP ASAP!! Complete the question below

Step-by-step explanation:

you mean questions ...

17.

angle 6 is also 30° (because for 2 crossing lines the angles on top are mirrored below).

18.

angle BHG is 90 degrees. so, angle 6 is the remainder of 115 degrees. and angle 3 is the same as angle 6 :

angle 3 = 115 - 90 = 25°

19.

angle 1 is also 27°.

GD and BH are perpendicular lines, just as AE and CF.

GD and BH are a rotated version (around H) of AE and CF. the amount of rotation is defined by angle 6. and every "spoke" of the wheel turns by the same degrees, so angle 1 is the same as angle 6.

20.

angle CHG is also 133°

again, mirrored angles in both sides of crossing lines.

21.

90° = angle 2 + angle 3 = angle 2 + 32

58° = angle 2

angle 2 is the complementary angle to angle 3 (together they are 90 degrees), because BH and CG are perpendicular.

12.

due to the mentioned mirroring of angles in big sides of crossing lines, the opposing angles must be equal.

therefore

8x + 7 = 9x - 4

11 = x

and

5y = 7y - 34

34 = 2y

17 = y

13.

the same principle as 12.

4x = 6x - 26

26 = 2x

13 = x

and

7y - 12 = 6y + 8

y = 20

14.

and the same principle again.

10x - 4 = 6(x + 2) = 6x + 12

4x = 16

x = 4

and

16y = 18y - 18

18 = 2y

9 = y

42.

angle 1 is equivalent to angle 3.

but also angle 1 is equivalent to angle 2.

therefore, in the same way, angle 3 is equivalent to angle 4.

43.

QRS and PSR are supplementary (together they are 180°).

QRS and QRL are also supplementary.

so, QRL and PSR are equivalent.

2.2°  

step-by-step explanation:

let θ = the angle of elevation.

tanθ =77/2020

tanθ = 0.038 12

      θ = tan⁻¹0.038 12

      θ = 2.2°

1^3 + 2^3 + 3^3 +4^3 = (1+2+3+4)^2

step-by-step explanation:

i am only able to answer the first part. i am not sure about the second part. sorry