Mathematics, 29.06.2019 21:00 crtlq

Which transformations could be performed to show that △abc is similar to △a"b"c"? a reflection over the x-axis, then a dilation by a scale factor of 3 b reflection over the x-axis, c then a dilation by a scale factor of a 180° rotation about the origin, d then a dilation by a scale factor of 3 a 180° rotation about the origin, then a dilation by a scale factor of

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The similarity of triangles ABC and A'B'C' is performed by rotating one triangle about the origin at $180^{\circ}$ . Hence, option (D) is correct.

As per the given problem, the triangle is ABC is an original geometry. And triangle A'B'C' is an reference geometry .

As per the concept of similarity of triangles using the scale factor, which says that, "In order for two figures to be similar, they must have congruent corresponding angle measures and proportional sides". If two triangles ABC and A'B'C' are similar then we take point C of triangle ABC to find the transformation performed form C to C'. The scale factor of 3 says that Coordinates of C of triangle ABC are (0, 3) and the coordinates of C' of triangle A'B'C' are (0, -1).

This shows that C is rotated 180 degrees about origin to get the new coordinates as (0, -3) and then new coordinates were dilated by 1/3 which forms C' as (0, -1)

Thus, we can conclude that the similarity of triangles ABC and A'B'C' is performed by rotating one triangle about the origin at $180^{\circ}$ . Hence, option (D) is correct.

Learn more about scale factor and dilation here:

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Answer from: starajaclark1429

Option D.

Step-by-step explanation:

If two triangles ΔABC and ΔA'B'C' are similar then we take point C of ΔABC to find the transformation performed form C to C'.

Coordinates of C are (0, 3) and the coordinates of C' are (0, -1).

This shows that C is rotated 180° about origin to get the new coordinates as (0, -3) and then new coordinates were dilated by 1/3 which forms C' as (0, -1)

Therefore option D is the correct option.

Answer from: Quest

X^2 - 4x +(4/2^2) + y^2 + 8y + (8/2^2) = -4 + (4/2^2) + (8/2^2)(x + 2)^2 + (y + 4)^2 = -4 + 4 + 16your center is (-2, -4) and the radius is the sqrt of -4 + 4 + 16 which is 4

Answer from: Quest

4.6875 pounds

step-by-step explanation:

Iam also having trouble with this one too pls belp

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