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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Answer:
-45˚
Step-by-step explanation:
Because the quadrant is 90˚ and 90 ÷ 2 = 45 and it in in the negative area. Therefore he angle is -45˚
Answer:
B and C both equal 0.060 after evaluation.