Iliana rotated trapezoid WXYZ 180° about the origin. On a coordinate plane, trapezoid W X Y Z has points (negative 4, 2), (negative 3, 4), (negative 1, 4), (0, 2).

What is the correct set of image points for trapezoid W’X’Y’Z’?

W’(4, –2), X’(3, –4), Y’(1, –4), Z’(0, –2)
W’(4, 2), X’(3, 4), Y’(1, 4), Z’(0, 2)
W’(–2, –4), X’(–4, –3), Y’(–4, –1), Z’(–2, 0)
W’(2, 4), X’(4, 3), Y’(4, 1), Z’(2, 0)

The answer is A. W’(4, –2), X’(3, –4), Y’(1, –4), Z’(0, –2)

Step-by-step explanation:

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}