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
Lentgh * width = area so, 3.5*9 right?
The equation y = x^2 has axis of symmetry of x = 0
The answer to the question is y = x^2 + 2 which is y = x^2 moved vertically up to 2 units
Answer:
units.
Step-by-step explanation:
Let x be the width of rectangle.
We have been given that the length of garden is 2 units more than 1.5 times it’s width. So length of the rectangle will be: 
.
To find the length of total fencing we need to figure out perimeter of rectangle with width x and length .
Since we know that perimeter of a rectangle is two times the sum of its length and width.
Upon substituting length and width of garden in above formula we will get,
Upon using distributive property we will get,
Therefore, the length of required fencing will be units.
