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:
Step-by-step explanation:
Move the decimal to the left once on both of them and time the numbers together
Answer:
A) -1.7 x 10^-4 D)-0.00017
Step-by-step explanation:
Since its negative in the base B and C wont work. The negative in the exponent means the decimal moves left which exclude E.
Answer: -72a^3 +108a^2
Collect like terms
(2a)(6a)(2a-8a+9)= (2a)(6a)(-6a+9)=12a^2(-6a+9)
= -72a^3 +108a^2
Answer:
Step-by-step explanation:
you need more information bu the stem-leaf is meant to split the number
