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
900,000+80,000+500+7 is 980,507
Answer:
Step-by-step explanation:
Given that you have eight cards. Five are green and three are yellow. The five green cards are numbered 1, 2, 3, 4, and 5. The three yellow cards are numbered 1, 2, and 3. The cards are well shuffled. You randomly draw one card.
G = card drawn is green
Y = card drawn is yellow
E = card drawn is even-numbered
List:
Sample space = {G1, G2, G3, G4, G5, Y1, Y2, Y3}
2) P(G) = 5/8
3) P(G/E) = P(GE)/P(E)
GE = {G2, G4}
Hence P(G/E) = 2/5
4) GE = {G2, G4}
P(GE) = 2/8 = 1/4
5) P(G or E) = P(G)+P(E)-P(GE)
= 5/8 + 3/8-2/8 = 3/5
6) No there is common element as G2 and G4
Cannot be mutually exclusive
Answer:
attach a picture
Step-by-step explanation:
