Me! two similar triangles are shown on the following coordinate grid: which set of transformations has been preformed on triangle efg to form triangle e’f’g’?

Reflection across y-axis followed by dilation with scale factor 4 and center (0,0).

Step-by-step explanation:

From the given figure it is noticed that the vertices of preimage are E(-2,1), F(-2,0) and G(-3,0). The vertices of image are E'(8,4), F'(8,0) and G'(12,0).

The triangle E'F'G' is enlargement of mirror image of EFG. Therefore reflect the triangle EFG across y-axis. If a point reflects across y-axis, then

The new vertices of the triangle are E(2,1), F(2,0) and G(3,0).

Length of FG is 1 unit and length of F'G' is 4 units. The scale factor is

Length of OF is 2 unit and length of OF' is 8 units.

Rule of dilation with scale factor k and origin as center of dilation is defined as

The vertices of image are E'(8,4), F'(8,0) and G'(12,0).

Since the distance ratio of image and preimage from the origin is same as the scale factor, therefore the center of dilation is origin.

Thus, the triangle EFG can form the triangle E’F’G’ using reflection across y-axis followed by dilation with scale factor 4 and center (0,0).

Notes: In this case dilation with scale factor 4 and center (0,0) followed by reflection across y-axis will give the same results.

step-by-step explanation:

there is a formula i can give you:

it is r^2*n*sin(360/n) divided by 2 where r represents the "radius" and n is the number of sides.

plug them in:

(10sqrt(3))^2*6*sin(360/6) divided by 2

100*3*6*sin(60) divided by 2

900*sin(60)

900*sqrt(3)/2

450*sqrt(3)

approx is 779.4 cm^2