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: 16m^4−25
Step-by-step explanation:
(4m^2−5)(4m^2+5)
=(4m^2+−5)(4m^2+5)
=(4m^2)(4m^2)+(4m^2)(5)+(−5)(4m^2)+(−5)(5)
=16m^4+20m^2−20m^2−25
=16m^4−25
Answer: 216/343 or 0.6297
Step-by-step explanation:
Y=1/3x+1 is the equation for ur question
