The answer is True.
Explanation:
Let a = major axis
Let b = minor axis
Let c = focal length.
Consider the right focus, located a distance c from the center of the ellipse (at the origin).
From the right focus to the right point on the major axis is equal to a-c. This is the minimum distance.
From the right focus to the left point on the major axis is equal to a+c. This is the maximum distance.
Answer:
C. √2 - 1
Step-by-step explanation:
If we draw a square from the center of the large circle to the center of one of the small circles, we can see that the sides of the square are equal to the radius of the small circle (see attached diagram)
Let r = the radius of the small circle
Using Pythagoras' Theorem 
(where a and b are the legs, and c is the hypotenuse, of a right triangle)
to find the diagonal of the square:



So the diagonal of the square = 
We are told that the radius of the large circle is 1:
⇒ Diagonal of square + r = 1





Using the quadratic formula to calculate r:




As distance is positive,
only
9=30*25+25*53. which = 2975.
You have to split the shape into two smaller ones to make two rectangles and find the area of both and add them together.
With 12, you need to find the area of both shaped and subtract the triangle form the rectangle to get 93.5.
The answer is A because there is a outlier which is $100,000