Two similar triangles are shown on the coordinate grid: Which set of transformations has been performed on triangle ABC to form triangle A′B′C′?

dilation by a scale factor of 4 followed by reflection about the x-axis
dilation by a scale factor of 2 followed by reflection about the x-axis
dilation by a scale factor of 4 followed by reflection about the y-axis
dilation by a scale factor of 2 followed by reflection about the y-axis

The set of transformations has been performed on triangle ABC is dilation by a scale factor of 2 followed by reflection about the x-axis

Step-by-step explanation:

Let us write the vertices of Δ ABC

∵ Vertex A = (-2, -1)

∵ Vertex B = (0, 0)

∵ Vertex C = (1, -3)

Let us Write the vertices of Δ A'B'C'

∵ Vertex A' = (-4, 2)

∵ Vertex B' = (0, 0)

∵ Vertex C' = (2, 6)

Let us find the scale factor k between the x-coordinates of the points and their images.

∵ -4 = k (-2) ⇒ Divide both side by -2

∵ =

∴ 2 = k

∴ The scale factor of a dilation is 2

∵ The signs of the y-coordinates of the images are the opposite of

    the signs of the y-coordinates of the points

∴ The triangle is reflected about the x-axis

∴ The set of transformations has been performed on triangle ABC

   is dilation by a scale factor of 2 followed by reflection about the x-axis

quit

step-by-step explanation:

x = ±sqrt(10)

step-by-step explanation:

5 - 2x² = -15

subtract 5 from each side

5-5 - 2x² = -15-5

-2x^2   = -20

divide by -2 on each side

-2x^2   /-2= -20/-2

x^2 =10

take the square root on each side

sqrt(x^2 )=±sqrt(10)

x = ±sqrt(10)

check:

5 - 2x² = -15

+sqrt(10)

5 - 2(sqrt(10))² = -15

5 -2(10) = -15

5 -20 = -15

-15 = -15

-sqrt(10)

5 - 2(-sqrt(10))² = -15

5 -2(10) = -15

5 -20 = -15

-15 = -15