Answer:
The value of f(z) is not constant in any neighbourhood of D. The proof is as explained in the explaination.
Step-by-step explanation:
Given
For any given function f(z), it is analytic and not constant throughout a domain D
To Prove
The function f(z) is non-constant constant in the neighbourhood lying in D.
Proof
1-Assume that the value of f(z) is analytic and has a constant throughout some neighbourhood in D which is ω₀
2-Now consider another function F₁(z) where
F₁(z)=f(z)-ω₀
3-As f(z) is analytic throughout D and F₁(z) is a difference of an analytic function and a constant so it is also an analytic function.
4-Assume that the value of F₁(z) is 0 throughout the domain D thus F₁(z)≡0 in domain D.
5-Replacing value of F₁(z) in the above gives:
F₁(z)≡0 in domain D
f(z)-ω₀≡0 in domain D
f(z)≡0+ω₀ in domain D
f(z)≡ω₀ in domain D
So this indicates that the value of f(z) for all values in domain D is a constant ω₀.
This contradicts with the initial given statement, where the value of f(z) is not constant thus the assumption is wrong and the value of f(z) is not constant in any neighbourhood of D.
The answer is five! Don't tell me how I got that because my sister did the problem 2 minutes ago because I told her this question! Hope this help!
Since, it is given that the pentagon ABCDE is congruent to pentagon FGHIJ.
By being congruent, the corresponding angles and sides are equal.
Therefore, 
, 
, 
, 
, 
.
By using
So, 
Therefore, the measure of angle 1 is 100 degrees.
So, Option B is the correct answer.
Answer:
a.27
Step-by-step explanation:
Using Pythagoras theorem:
x=sqrt(6^2+26^2)
x=26.683..
By rounding, x=27
I would need more information from the question to give an answer for each trapezoid.
