The similar circles P and Q can be made equal by dilation and translation
- The horizontal distance between the center of circles P and Q is 11.70 units
- The scale factor of dilation from circle P to Q is 2.5
<h3>The horizontal distance between their centers?</h3>
From the figure, we have the centers to be:
P = (-5,4)
Q = (6,8)
The distance is then calculated using:
d = √(x2 - x1)^2 + (y2 - y1)^2
So, we have:
d = √(6 + 5)^2 + (8 - 4)^2
Evaluate the sum
d = √137
Evaluate the root
d = 11.70
Hence, the horizontal distance between the center of circles P and Q is 11.70 units
<h3>The scale factor of dilation from circle P to Q</h3>
We have their radius to be:
P = 2
Q = 5
Divide the radius of Q by P to determine the scale factor (k)
k = Q/P
k = 5/2
k = 2.5
Hence, the scale factor of dilation from circle P to Q is 2.5
Read more about dilation at:
brainly.com/question/3457976
Use the pythagoream theorem A^2+B^2 then square your answer a being your adjacent side and b being your opposite leg.
1 foot = 12 inches. If you multiply 12 by 12 and then add 5, you get 149. 149 is less than 162 so it will be able to fit!
Answer:1 1/20
Step-by-step explanation:
2 5/8 x 2/5
(8x2+5)/8 x 2/5
21/8 x 2/5
(21 x 2) ➗ (8 x 5)
42 ➗ 40
42/40=21/20=1 1/20
