Atriangle has vertices at b(−3, 0), c(2, −1), d(−1, 2). which transformation would produce an image with vertices b″(−2, 1), c″(3, 2), d″(0, −1)? (x, y) → (x, −y), (x, y) → (x + 1, y + 1) (x, y) → (−x, y), (x, y) → (x + 1, y + 1) (x, y) → (x, −y), (x, y) → (x + 2, y + 2) (x, y) → (−x, y), (x, y) → (x + 2, y + 2)

A. (x, y) → (x,-y), (x, y) → (x + 1, y + 1)

Step-by-step explanation:

Plot points B, C and D on the coordinate plane (blue points in attached diagram).

1 transformation is reflection across the x-axis with the rule

(x,y)→(x,-y)

and it maps these points to

These image points are marked in green in attached diagram.

2 transformation is translation 1 unit to the right and 1 unit up with the rule

(x,y)→(x+1,y+1)

and it maps previous image points to

x > 1.3

step by step:

8.5−0.9x> 9.8−1.9x

step 1: simplify both sides of the inequality.

−0.9x+8.5> −1.9x+9.8

step 2: add 1.9x to both sides.

−0.9x+8.5+1.9x> −1.9x+9.8+1.9x

1x+8.5> 9.8

step 3: subtract 8.5 from both sides.

1x+8.5−8.5> 9.8−8.5

1x> 1.3

step 4: divide both sides by 1.

1x/1 > 1.3/1

b.my answer is b.

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