The question: a parallelogram has vertices at (0,0), (0,6), (4,4), and (4, -2). which transformation maps the parallelogram to itself?

C

Step-by-step explanation:

Consider all options.

A. A reflection across the line x=2 gives us the parallelogram A'B'C'D' (see first attached diagram) with vertices A'(4,0), B'(4,6), C'(0,4) and D'(0,-2). As you can see this is not the same parallelogram as parallelogram ABCD.

B. A reflection across the line y=2 gives us the parallelogram A''B''C''D'' (see first attached diagram) with vertices A''(0,4), B''(0,-2), C''(4,0) and D''(4,6). As you can see this is not the same parallelogram as parallelogram ABCD.

C. A rotation of 180° about the point (2,2) gives us the same parallelogram ABCD (see second attached diagram), because the image of A is C, the image of B is D, the image of C is A and the image of D is B.

D. A rotation of 180° about the point (0,0) gives us the  parallelogram A'''B'''C'''D''' (see second attached diagram) that differs from parallelogram ABCD.

see explanation

step-by-step explanation:

since the sequence is geometric there is a common ratio r between consecutive terms, that is

r = 4 ÷ - 2 = - 2

r = - 8 ÷ 4 = - 2

r = 16 ÷ 8 = - 2

the next term - 13 must be incorrect as it should be - 32 for the sequence to be geometric.

the first 4 terms are therefore geometric with common ratio r = - 2

answer: idk

step-by-step explanation: