The area of an equilateral triangle of side "s" is s^2*sqrt(3)/4. So the volume of the slices in your problem is
(x - x^2)^2 * sqrt(3)/4.
Integrating from x = 0 to x = 1, we have
[(1/3)x^3 - (1/2)x^4 + (1/5)x^5]*sqrt(3)/4
= (1/30)*sqrt(3)/4 = sqrt(3)/120 = about 0.0144.
Since this seems quite small, it makes sense to ask what the base area might be...integral from 0 to 1 of (x - x^2) dx = (1/2) - (1/3) = 1/6. Yes, OK, the max height of the triangles occurs where x - x^2 = 1/4, and most of the triangles are quite a bit shorter...
1. Distribute
5(2n-4)
10n-20
9(n+2)
9n+18
10n-20=9n+18
2. Combine like terms
combine the terms with “n”, and then combine the other numbers.
Hope this helps!
So if we take 1/4 to be the 100%, what is 2/3 in percentage off of it then?
yes, is a large value, because 2/3 is indeed larger than 1/4, more than twice as large.
Parallelograms, rectangles, rhombus, or square
