Triangle FGH is graphed on the coordinate plane below. On a coordinate plane, triangle F G H has points (negative 6, negative 8), (negative 2, negative 3), (negative 4, negative 2).

The figure is rotated 180° using the origin as the center of rotation. How do the coordinates of the vertices of the preimage compare to the coordinates of the vertices of the image?
(x, y) right-arrow (y, x)
(x, y) right-arrow (negative y, negative x)
(x, y) right-arrow (negative x, negative y)
(x, y) right-arrow (x, negative y)

(x, y) right-arrow (negative x, negative y) ⇒ (x, y) → (-x, -y)

Step-by-step explanation:

Let us revise the rotation about the origin

If point (x, y) rotated about the origin by angle 90° counterclockwise (or 270° clockwise), then its image is (-y, x) . The rule is (x, y) → (-y, x)If point (x, y) rotated about the origin by angle 180° counterclockwise ( or 180° clockwise), then its image is (-x, -y) . The rule is (x, y) → (-x, -y)If point (x, y) rotated about the origin by angle 270° counterclockwise (90° clockwise), then its image is (y, -x) . The rule is (x, y) → (y, -x)

Let us solve the question

∵ The vertices of triangle FGH are (-6, -8), (-2, -3), (-4, -2)

∵ The figure is rotated 180° using the origin as the center of rotation

→ By using the 2nd rule above

∴ The vertices of its image are (6, 8), (2, 3), (4, 2)

∴ The rule is (x, y) → (-x, -y)

The coordinates of the vertices of the preimage compare to the coordinates of the vertices of the image do (x, y) right-arrow (negative x, negative y) ⇒ (x, y) → (-x, -y)

1

Step-by-step explanation:

According to law of indices

log a b = 1/ log b a

hence log c (2) = 1/log 2 (c)

log 2 (c)⋅log c (2)

=  log 2 (c)⋅ 1/log 2 (c)

= log 2 (c)/log 2 (c)

= 1

Hence the required answer is 1