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:
A = 76.85
B =65.28
Step-by-step explanation:
(30/100)A = 10 + (20/100)B
0.3A - 0.2B = 10 ...... equation (i)
(30/100)B + 35 = (20/100)A
0.2A - 0.3B = 35 ........ equation (ii)
From equation (i)
0.3A = 10 - 0.2B
A = (10 - 0.2b) / 0.3
A = 33.33 - 0.67B ........equation (iii)
Put equation (iii) into equation (ii)
0.2(33.33 - 0.67B) - 0.3B = 35
6.67 - 0.134B - 0.3B = 35
0.434B = 35 - 6.67
B = 28.33 / 0.434
B = 65. 275 = 65.28
Put B = 65.28 into equation (i)
0.3A - 0.2B = 10
0.3A - 0.2(65.28) = 10
0.3A - 13.056 = 10
0.3A = 10 + 13.056
A = 23.056/ 0.3
A = 76.85
1/6(12z-18)=2z-3
12/6z-18/6 =2z-3
2z-3=2z-3
there are infinite solutions for this equation or all real numbers
