If you have 1 - 4" circle, and you have N number of .5" circles, let's work the problem out..5" = .5" (1) .5" = 1.0" (2).5" = 1.5" (3).5" = 2.0" (4).5" = 2.5" (5).5" = 3.0" (6).5" = 3.5" (7).5" = 4.0" (8)Therefore it would take 8 - .5" circles to fit into 1 - 4" circle.
It can be only 1 * * * * * Infinitely many concentric circles can be drawn.
yes it has as many as you can fit across it
If you create third circle with radius 2, then all the points on that circle would be equidistant form both circles. So the answer is a circle with radius 2.
If it has height it's not a circle, its a cylinder. PS whats a fit?
a circle has 4 sides
If you have 1 - 4" circle, and you have N number of .5" circles, let's work the problem out..5" = .5" (1) .5" = 1.0" (2).5" = 1.5" (3).5" = 2.0" (4).5" = 2.5" (5).5" = 3.0" (6).5" = 3.5" (7).5" = 4.0" (8)Therefore it would take 8 - .5" circles to fit into 1 - 4" circle.
It can be only 1 * * * * * Infinitely many concentric circles can be drawn.
There are many suffixes. Suffixes are letters that you add to the end of the root word. 1. Circles 2. Circled There are probably many more, but those are the top ones I can think of right now! Hope this helped!
from 1 to infinite.. depends how many you want to put in it.
Letting x be radius of the 4 circles, then (squareroot(2x^2))+x=10, or x(1+sqrt2)=10. Then radius of circle in middle is ((2*10)-4x)/2. So I get radius of circle in middle = 1.715729 approximately.
yes it has as many as you can fit across it
If you create third circle with radius 2, then all the points on that circle would be equidistant form both circles. So the answer is a circle with radius 2.
If it has height it's not a circle, its a cylinder. PS whats a fit?
That is not possible. Circles are round and have no edges, so that is impossible.
You interlock the 3 large circles as in how part of the Olympic rings look, or as in a Venn diagram. So, each large circle shares are a section with each of the other two, and there is a central section that is within all 3 circles. Now, simply place a small circle in each of the 7 sections of the diagram: - the 3 sections that are only in 1 large circle - the 3 sections that are within 2 large circles - the 1 central section that is within all 3 large circles Voila! You have 4 small circles within each large circle.
1^2 or 2^2 or 3^2 or 4^2.... and it goes on...where diameter of circle is equal to the side of square divided by square root of total numbers of circles.. eg : side of square = 10 no of circles = 2^2 square root = 2 therefore diameter = 10/2=5.