Answer:
So easy just try your best if you cant contact me
Sin (x) = 12/13 this is because sine is opposite divided by hypotenuse.
cos (x) = 5/13 this is because cosine is adjacent divided by hypotenuse.
tan (x) = 12/5 this is because tangent is opposite divided by adjacent.
Answer:
Where is the question or problem that you want answered?
Step-by-step explanation:
I won't be of much help if I do not know what is it that you are asking help for.
The way to find the factores is by seeking the polonomial's roots, or zeros!
To find the roots, we set the polinomial equation to zero:
x^2+16y^2=0
Then we solve this equation:
x^2=-16y^2
This is only true, if x=y=0.
Part A:
Given

defined by


but

Since, f(xy) ≠ f(x)f(y)
Therefore, the function is not a homomorphism.
Part B:
Given

defined by

Note that in

, -1 = 1 and f(0) = 0 and f(1) = -1 = 1, so we can also use the formular


and

Therefore, the function is a homomorphism.
Part C:
Given

, defined by


Since, f(x+y) ≠ f(x) + f(y), therefore, the function is not a homomorphism.
Part D:
Given

, defined by


but

Since, h(ab) ≠ h(a)h(b), therefore, the funtion is not a homomorphism.
Part E:
Given

, defined by
![\left([x_{12}]\right)=[x_4]](https://tex.z-dn.net/?f=%5Cleft%28%5Bx_%7B12%7D%5D%5Cright%29%3D%5Bx_4%5D)
, where
![[u_n]](https://tex.z-dn.net/?f=%5Bu_n%5D)
denotes the lass of the integer

in

.
Then, for any
![[a_{12}],[b_{12}]\in Z_{12}](https://tex.z-dn.net/?f=%5Ba_%7B12%7D%5D%2C%5Bb_%7B12%7D%5D%5Cin%20Z_%7B12%7D)
, we have
![f\left([a_{12}]+[b_{12}]\right)=f\left([a+b]_{12}\right) \\ \\ =[a+b]_4=[a]_4+[b]_4=f\left([a]_{12}\right)+f\left([b]_{12}\right)](https://tex.z-dn.net/?f=f%5Cleft%28%5Ba_%7B12%7D%5D%2B%5Bb_%7B12%7D%5D%5Cright%29%3Df%5Cleft%28%5Ba%2Bb%5D_%7B12%7D%5Cright%29%20%5C%5C%20%20%5C%5C%20%3D%5Ba%2Bb%5D_4%3D%5Ba%5D_4%2B%5Bb%5D_4%3Df%5Cleft%28%5Ba%5D_%7B12%7D%5Cright%29%2Bf%5Cleft%28%5Bb%5D_%7B12%7D%5Cright%29)
and
![f\left([a_{12}][b_{12}]\right)=f\left([ab]_{12}\right) \\ \\ =[ab]_4=[a]_4[b]_4=f\left([a]_{12}\right)f\left([b]_{12}\right)](https://tex.z-dn.net/?f=f%5Cleft%28%5Ba_%7B12%7D%5D%5Bb_%7B12%7D%5D%5Cright%29%3Df%5Cleft%28%5Bab%5D_%7B12%7D%5Cright%29%20%5C%5C%20%5C%5C%20%3D%5Bab%5D_4%3D%5Ba%5D_4%5Bb%5D_4%3Df%5Cleft%28%5Ba%5D_%7B12%7D%5Cright%29f%5Cleft%28%5Bb%5D_%7B12%7D%5Cright%29)
Therefore, the function is a homomorphism.