I believe the proper way to find the surface area of a sphere is 4 times 3.14 (which is Pi) times the radius squared. Or 4(3.14)*r^2.
Where I found 804.25
Multiply 13 by the 2 above the divide line
13*2 = 26
so the number in the box should be 260
Ok I will let you know tomorrow afternoon when I get to the school
2+sq root of 5
and the sum of zeros = 4
other zero =4-(2+sq root of 5)=2-sq root of 5
Answer:
Yes, double cosets partition G.
Step-by-step explanation:
We are going to define a <em>relation</em> over the elements of G.
Let 
. We say that 
if, and only if, 
, or, equivalently, if 
, for some 
.
This defines an <em>equivalence relation over </em><em>G</em>, that is, this relation is <em>reflexive, symmetric and transitive:</em>
- Reflexivity: (
for all 
.) Note that we can write 
, where 
is the <em>identity element</em>, so 
and then 
. Therefore, . - Symmetry: (If then
.) If then 
for some 
and 
. Multiplying by the inverses of h and k we get that 
and is known that 
and 
. This means that 
or, equivalently, .
- Transitivity: (If and
, then 
.) If and , then there exists 
and 
such that 
and 
. Then, 
where 
and 
. Consequently, 
.
Now that we prove that the relation "
" is an equivalence over G, we use the fact that the <em>different equivalence classes partition </em><em>G.</em><em> </em>Since the equivalence classes are defined by ![[x]=\{y\in G\colon x\sim y\} =\{y\in G \colon y=hgk\ \text{for some } h\in H, k\in K \}=HxK](https://tex.z-dn.net/?f=%5Bx%5D%3D%5C%7By%5Cin%20G%5Ccolon%20x%5Csim%20y%5C%7D%20%3D%5C%7By%5Cin%20G%20%5Ccolon%20y%3Dhgk%5C%20%5Ctext%7Bfor%20some%20%7D%20h%5Cin%20H%2C%20k%5Cin%20K%20%5C%7D%3DHxK)
, then we're done.
