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
<em>The point where a function (for our problem its the cubic function shown)</em>
<u>and/or</u>
<em>is a zero of the function.</em>
From the graph shown, we can clearly see that it cuts the x-axis at -1 and touches the x-axis at 2.
So the zeros are at -1 and 2.
ANSWER: {-1,2}
Answer:
-15 ft or 15 ft below sea level
Step-by-step explanation:
He started at +8 ft.
Then he went 23 ft down, that is -23 ft.
8 ft + (-23 ft) = 8 ft - 23 ft = -15 ft
Answer: -15 ft
A²+b²=c²
64+b²=196
b²=132
b=11.48912529
Answer:
65 mili= 6.5cm
Step-by-step explanation:
divide the length value by 10