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:
Option D.
Step-by-step explanation:
The given statement is

x-4 is difference between x and 4.
|x-4| distance from x because distance can not be negative.
Inequality sign "<" means less than.
represent all values of x whose distance from 4 is less than 0.5.
Therefore, the correct option is D.
Answer:
3rd quadrant
Step-by-step explanation:
- First we just convert the angle from radians to degrees
- Now that's too big, all this means is if we start rotating from the positive y-axis in a circle we will cross the starting point 2 times, 2 full circles;
- Now in which quadrant it 210 degrees?
- 0 degrees to 90 degrees is 1st quadrant
- 90 degrees to 180 degrees is 2nd quadrant
- 180 degrees to 270 degrees is 3rd quadrant
- 270 degrees to 360 degrees is 4th quadrant
- So our answer is the 3rd quadrant.
Answer:
(2, 3)
Step-by-step explanation:
The point with integer coordinates nearest the solution point seems to be (2, 3).
Answer:
Step-by-step explanation:
So we would have to multiply the "2x - 8" by 5 each resulting in 10x - 40 + 15. Then we subtract which results in 10x - 25 = - 15. We add 25 to 25 and 15 resulting in 10x = 10. We divide each by 10 which results in x = 1.