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.
9 x² + 16 y² = 144 /:144
General formula of ellipse ( the center is at the origin ):
a² = 16, b² = 9
Domain: [-a, a ] = [-4, 4]
Range:[-b, b ]
Answer:
B ) ellipse.Domain: { -4 ≤ x ≤ 4 }Range: { -3 ≤ y ≤ 3 }
Answer:
- The two solutions are:
- The next and every step are below.
Explanation:
1. 
: Given (addition property / add - 3 to both sides)
2. 
: Given (commom factor - 2)
3. 
To obtain the perfect square it was added the square of half of the coefficient of x: (1/2)² = 1/4, inside the parenthesis.
Since, the terms inside the parentthesis are multiplied by - 2, you have to add - 2 (1/4) = - 1/2 to the left side of the equation.
4. Now, you have that the trinomial x² - x + 1/4 is a square perfect trinomial which is factored as (x - 1/2)² and get the expression:
5. Divide both sides by - 2 to get the next expression:
6. The last step is to extract squere root from both sides of the equality:
Answer:
Veronika's weight = 92 pounds
Kate's weight = 46 pounds
Step-by-step explanation:
Let Veronika's weight = x
Let Kate's weight = y
The sum of Veronika's and Kate's weight is 118 pounds.
x + y = 118
Veronika weighs 26 pounds more than Kate.
x = y + 26
How much do they weigh individually?
We substitute y + 26 for x in Equation 1
x + y = 118
y + 26 + y = 118
26 + 2y = 118
2y = 118 - 26
2y = 92
y = 92/2
y = 46 pounds
x = y + 26
x = 46 + 26
x = 92 pounds
Hence,
Veronika's weight = 92 pounds
Kate's weight = 46 pounds
Answer:
134
Step-by-step explanation:
Angle Formed by Two Chords= 1/2(SUM of Intercepted Arcs)
x = 1/2 ( m DH + m BC)
x = 1/2 (144+ 124)
x = 1/2 ( 268)
x = 134
