Answer:
x1 = 2 -![\sqrt{5}](https://tex.z-dn.net/?f=%5Csqrt%7B5%7D)
x2= 2 +![\sqrt{5}](https://tex.z-dn.net/?f=%5Csqrt%7B5%7D)
Step-by-step explanation:
x/4 + 6 = 7 + 1/4x
(x^2 - 1)/4x = 1
x^2 - 1 = 4x
x^2 - 4x -1 = 0
x= (4+
)/2 = 2 -![\sqrt{5}](https://tex.z-dn.net/?f=%5Csqrt%7B5%7D)
x= 2 +![\sqrt{5}](https://tex.z-dn.net/?f=%5Csqrt%7B5%7D)
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
Answer:
5cm
Step-by-step explanation:
volume = πr²h
225π = π × r² × 9
225/9 = r²
r = √225/9 = 15/3 = 5cm
Answer:
12.566 sq inches
Step-by-step explanation: