The distance from the center of dilation, P, to.the image vertice S' is; 6 units.
<h3>What is the distance from the center of dilation, P, to the image S'?</h3>
It follows from the task content that the center of dilation of the triangle QRS is point P and the length of segment PS in the pre-image is; 8 units.
Hence, since the dilation factor as given in the task content is; three-fourths, it therefore follows that the distance of point P to S' in the image is; (3/4) × 8 = 6units.
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The answer is 0.008a^3 because you would do 0.2 to the power of 3
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