The following is an incomplete paragraph proving that ∠WRS ≅ ∠VQT given the information in the figure where segment UV is parall
el to segment WZ.:
Segments UV and WZ are parallel with line ST intersecting both at points Q and R respectively
According to the given information, segment UV is parallel to segment WZ while angles SQU and VQT are vertical angles. Angle VQT is congruent to angle SQU by the Vertical Angles Theorem. Because angles SQU and WRS are corresponding angles, they are congruent according to the Corresponding Angles Postulate. Finally, angle VQT is congruent to angle WRS by the _____________________.
Which Property of Equality accurately completes the proof?
Reflexive
Substitution
Subtraction
Transitive
Segments UV and WZ are parallel with line ST intersecting both at points Q and R respectively
According to the given information, segment UV is parallel to segment WZ while angles SQU and VQT are vertical angles. Angle VQT is congruent to angle SQU by the Vertical Angles Theorem. Because angles SQU and WRS are corresponding angles, they are congruent according to the Corresponding Angles Postulate. Finally, angle VQT is congruent to angle WRS by the _____________________.
Which Property of Equality accurately completes the proof?
Reflexive
Substitution
Subtraction
Transitive
2 answers:
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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.
Given
but
Since, f(xy) ≠ f(x)f(y)
Therefore, the function is not a homomorphism.
Part B:
Given
Note that in
and
Therefore, the function is a homomorphism.
Part C:
Given
Since, f(x+y) ≠ f(x) + f(y), therefore, the function is not a homomorphism.
Part D:
Given
but
Since, h(ab) ≠ h(a)h(b), therefore, the funtion is not a homomorphism.
Part E:
Given
Then, for any
and
Therefore, the function is a homomorphism.