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
C=p+R x p
C=30.99 + .05 x 30.99
Order of operations states do the multiplication first
C=30.99 + 1.55
C=32.54
If ABC type question, then sometimes. If its extended, then only when the lines coincide with each other or if one of the lines aren't linear.
Answer:
it's 22 I hope you got it right if not then i'm sorry
The surface area of a sphere is given by
where r represents the radius.
As the radius is given as 60 meters we replace r with 60 in the formula above to get the surface area.
