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:
8 ≤ x
Step-by-step explanation:
6(2 − 6x) ≤ 4 − 35x
Expand bracket;
12 - 36x ≤ 4 − 35x
Add 36x to both sides to get;
12 ≤ 4 + x
Subtract 4 from both sides to get;
8 ≤ x
Answer:
she will be recivining $24.06
Step-by-step explanation:
4x3.99=15.96
2x4.99=9.98
15.96+9.98=25.94
50-25.94=24.06
Of course with out taxes
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