I really need help finding these angles because I don’t understand

Every time we have a transversal crossing parallel lines we get eight angles. Each is either congruent or supplementary to any of the other seven, there are only two angle measures in play.

Here we have two transversals, so each has two angle measures.   That doesn't count where the two transversals cross; we'll do those last.

On line a we see a 39° angle.  Let's do that first.

39°: ∠13 ∠6 ∠5

It's supplement is 180°-39°=141°

141°: ∠14 ∠12 ∠7 ∠4

Line b has a 74° angle

74°: ∠3 ∠15 ∠18

Supplement:

106°: ∠1 ∠2 ∠16 ∠17

OK, now the middle crossing

∠3 + ∠6 + ∠9 = 180°

74° + 39° + ∠9 = 180°

∠9 = 180° - 74° - 39°

∠9 - ∠11 =  67°

∠8 = ∠10 = 180° - 67° = 113°

let's add {f(x)=x+1}f(x)=x+1 and {g(x)=2x}g(x)=2x together to make a new function.

f(x)+g(x) =(x+1)+(2x)=x+1+2x=3x+1

let's call this new function hh. so we have:

{h(x)}={f(x)}+{g(x)}{=3x+1}h(x)=f(x)+g(x)=3x+1

we can also evaluate combined functions for particular inputs. let's evaluate function hh above for x=2x=2. below are two ways of doing this.

method 1: substitute x=2x=2 into the combined function hh.

h(x)

h(2)

​    

=3x+1

=3(2)+1

=7

​ since h(x)=f(x)+g(x)h(x)=f(x)+g(x), we can also find h(2)h(2) by finding f(2) +g(2)f(2)+g(2).

first, let's find f(2)f(2):

f(x)

f(2)

​    

=x+1

=2+1

=3

​  

now, let's find g(2)g(2):

g(x)

g(2)

​    

=2x

=2⋅2

=4

​  

so f(2)+g(2)=3+4=\greend7f(2)+g(2)=3+4=7.

notice that substituting x =2x=2 directly into function hh and finding f(2) + g(2)f(2)+g(2) gave us the same answer!