Aline segment has endpoints at (–4, –6) and (–6, 4). which reflection will produce an image with endpoints at (4, –6) and (6, 4)?

Reflection over y axis will produce an image with endpoints at (4, –6) and (6, 4).

Step-by-step explanation:

Given  : A line segment has endpoints at (–4, –6) and (–6, 4).

To find : Which reflection will produce an image with endpoints at (4, –6) and (6, 4).

Solution : We have given

Parent points :  (–4, –6) and (–6, 4).

Transformed function : (4, –6) and (6, 4).

(–4, –6) →→ (4, –6).

(–6, 4) →→ (6, 4).

By the reflection over y axis : ( x ,y) →→ ( -x , y). it reflect the only x coordinates.

Here (–4, –6) →→ (-(-)4, –6)→→(4, –6).

(–6, 4) →→ (-(-)6, 4)→→(6, 4).

Therefore,reflection over y axis will produce an image with endpoints at (4, –6) and (6, 4).

x ≥ 18 is your answer

step-by-step explanation:

let "a number" = x

(2/3)x + 17 ≥ 29

isolate the variable (x). do the opposite of pemdas (parenthesis, exponent (& root), multiplication, division, addition, subtraction).

first, subtract 17 from both sides.

(2/3)x + 17 (-17) ≥ 29 (-17)

(2/3)x ≥ 29 - 17

(2/3)x ≥ 12

next, multiply 3/2 to both sides to isolate the x

(3/2)(2/3)x ≥ 12(3/2)

x ≥ 12(3/2)

simplify. multiply, then divide

x ≥ (12 * 3)/2

x ≥ (36)/2

x ≥ 18

x ≥ 18 is your answer

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