Answer:
6xy+4x
Step-by-step explanation: You start off by gathering the like terms (which are 9xy & -3xy) and then you just keep 4x there. You then subtract 9xy-3xy=6xy and bring down the 4 since it has no like terms. So it’s 6xy+4x.
A pint container because there are 2 cups in a pint, whereas there are 6 cups in 1/2 gallon and 4 cups in a quart.
Take ln to both sides
ln 14^x = ln 5
then you'll get
x ln 14 = ln 5
divide both sides by ln 14
x = 0.6098533345
after rounding
x = 0.6099
Answer:

Step-by-step explanation:
Given

First, we need to list the multiples of 5

Then, multiples of 3
Next, is to list out the common elements in both


The required probability is then calculated as thus:



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.