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
The slope would be -1 since it’s the coefficient of -x
9514 1404 393
Answer:
(a) positive and increasing
Step-by-step explanation:
Point c is on the x-axis, and point d is above it. The curve between points c and d is positive and increasing.
Hello from MrBillDoesMath!
Answer:
f(2-k) = -2 + 5k
Discussion:
If f(x) = 8 - 5x
, then
f(2-k) = 8 - 5(2-k) =>
f(2-k) = 8 - 10 + 5k =>
f(2-k) = -2 + 5k
Regards,
MrB
