ABC is a equilateral triangle .
Proof :-
Let's assume both circles as C1 and C2 [ as shown in the figure ]
- AB is the radius of circle C1
- AB is the radius of Circle C2
AC is the radius of circle C1.
BC is the radius of circle C2 .
AB and AC both are radius of circle C1 so both are equal ie AB = AC .
AB and BC both are radius of circle C 2 so both are equal ie AB = BC .
Hence we conclude that .
AB = BC = AC.
So the triangle is equilateral triangle.
I just tried to help someone with the same problems. I hope this helps you
Hello, we need to see it as a difference of two squares as we know that for all a and b real numbers

So, here

Voila !
Thank you
For domain 2x sqrt(2+x)>0
x>0,2+x>0,x>-2 combining
we get x>2
f'(x)=[1/{2x sqrt(2+x)}][{2x/(2 sqrt(2+x))}+2 sqrt(2+x)]
There are 108° in each interior angle of<span> a </span><span>regular pentagon.</span>