6 semicircles = 3 circles
The formula of the area of a equilateral triangle:

We have a = |AE| = 3cm. Substitute:

The formula of the area of a circle:

We have 2r = |AE| =3cm → r = 1.5cm. Substitute:

The area of the figure:

Answer:
○![\displaystyle -\frac{2\sqrt{3}}{3}\:[or\:-\frac{2}{\sqrt{3}}]](https://tex.z-dn.net/?f=%5Cdisplaystyle%20-%5Cfrac%7B2%5Csqrt%7B3%7D%7D%7B3%7D%5C%3A%5Bor%5C%3A-%5Cfrac%7B2%7D%7B%5Csqrt%7B3%7D%7D%5D)
Step-by-step explanation:
You have to know the Unit Circle on this one. So, to start off, in the degree-angle range from 180° - 270° [Quadrant III], you must figure out the cosine value that gives you −½, and according to the graph, <em>theta</em><em> </em>will be represented as 240°. So, now that we have our <em>θ</em>, looking at the <em>y-value</em><em> </em>240°, since we want the <em>cosecant</em><em> </em>function, all we have to do is take the multiplicative inverse of
which gives you 
Extended Information

![\displaystyle -\frac{2\sqrt{3}}{3} = [-\frac{\sqrt{3}}{2}]^{-1}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20-%5Cfrac%7B2%5Csqrt%7B3%7D%7D%7B3%7D%20%3D%20%5B-%5Cfrac%7B%5Csqrt%7B3%7D%7D%7B2%7D%5D%5E%7B-1%7D)
I am joyous to assist you anytime.
Since you are working with exponents and roots you would add the exponents together. The ninth root of x is equal to x^(1/9) and since you are using it four times your answer would be x^(4/9).
Answer:
6l², where l is the length of one side.
If the side is increased 10 times,
6(10l)²
=100 x6l²
The surface area increases 100 times. Thus, the new surface area is
272 * 100
= 27,200 sq in
Step-by-step explanation: