What are the vertices of triangle A'B'C' if triangle ABC is dilated by a scale factor of 1/4

A ' (4,0)

B ' (6,3)

C ' (3,4)

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Explanation:

I'm assuming the center of dilation is the origin.

If so, then we multiply the scale factor 1/4 = 0.25 by each of the coordinates of the three points.

Point A(16,0) will move to A ' (4,0) since

Point B(24,12) will move to B ' (6,3) since

We have enough information to see that choice B is the final answer

If you wanted to keep going, then point C(12,16) will move to C ' (3,4) for similar reasoning/steps as the previous two points.

The resulting image triangle A'B'C' will have its side lengths 1/4 of that compared to the side lengths of the preimage triangle ABC.

The scale factor k is such that 0 < k < 1, which indicates the image will be smaller compared to the preimage. If k > 1, then the image would have been larger than the preimage.

2 x^2 -3x+p

sum of roots = 3/2

let other root be x

3+ x = 3/2

x= 3/2 - 3 = -3/2

product of roots = p/2

3× -3/2 = p/2

-9 = p

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you need more information to answer this question

step-by-step explanation: