Given: Lines a and b are parallel and line c is a transversal. Prove: Angle2 is supplementary to Angle8 Horizontal and parallel lines a and b are cut by transversal c. On line a where it intersects line c, 4 angles are created. Labeled clockwise, from uppercase left, the angles are: 1, 2, 4, 3. On line b where it intersects line c, 4 angles are created. Labeled clockwise, from uppercase left, the angles are: 5, 6, 8, 7. What is the missing reason in the proof? Statement Reason 1.a || b, is a transv1.given 2.∠6 ≅ ∠22.? 3.m∠6 = m∠23.def. of congruent 4.∠6 is supp. to ∠84.def. of linear pair 5.∠2 is supp. to ∠85.congruent supplements theorem corresponding angles theorem alternate interior angles theorem vertical angles theorem alternate exterior angles theorem

The correct option is;

Corresponding angles theorem

Step-by-step explanation:

Type of lines of lines a and b = Horizontal and parallel lines

The transversal to a and b = Line c

The angles between a and c labelled clockwise from the upper left quarter segment =  1, 2, 4 and 3

The angles between b and c labelled clockwise from the upper left segment =  5, 6, 8 and 7

Therefore, we have;

Statement                                                     Reason

1. a║b, c is a transversal                               Given

2. ∠6 ≅ ∠2                                                    Corresponding angles theorem

3. m∠6 = m∠2                                               Definition of congruent

4. ∠6 is supp. to ∠8                                       Definition of linear pair

5. ∠2 is supp. to ∠8                                       Congruent supplement theorem

Corresponding angles are the angles located in spatially similar or matching corners of two lines that have been crossed by the same transversal. When the two lines having a common transversal are parallel, the corresponding angles will be congruent.                        

geometric series converge the common ratio r has magnitude less than one.

which is equivalent to

for the example we need

|2-x| < 1

-1 < 2-x   and 2-x < 1

x < 3 and x > 1

we can write that

answer: 1 < x < 3

1a. common ratio r=250/25=10 > 1     diverges

1b.   r=50/100=1/2 converges

when these converge they converge to

where a is the first term.

answer: 200

1c. r=-1   so it's not the case that |r| < 1.     diverges

2a. x is the common ratio so we converge when |x|< 1

2b.   r=-10x / -2 = 5x.

we converge when

|5x| < 1

|x| < 1/5

2c.   r=3-2x

-1 < 3-2x and 3-2x < 1

-4 < -2x and 2 < 2x

x < 2 and 1 < x

1 < x < 2

2d. [/tex] r=-48x^3/12 = 4x^3[/tex]

|4x^3| < 1

|x^3| < \frac 1 4

|x| < 4^{-1/3}

just do 5.2 + 2 ( 7.2) and 1/3 x 3 (.9) and youv'e got your equation. 0.9( 9/10) + 7.2 ^2 = 16.4 hope i could !

step-by-step explanation: