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.
It is defined as the difference between the largest and smallest values in the middle 50% of a set of data<span>. To compute an </span>interquartile range<span> using this definition, first remove observations from the lower quartile. Then, remove observations from the upper quartile.</span>
Answer:
What is given to you is interval notation. We have the interval start at 0 and end at 8. The value 0 is not included in the interval as indicated by the parenthesis. So we go with a "less than" sign (instead of a "less than or equal to" sign)
The value 8 is included since a square bracket is used here. The use of "or equal to" is needed to make sure the endpoint 8 is included.
So x can be any number between 0 and 8. It can't be 0 but it can be 8.
Answer:
$13.50
Step-by-step explanation:
if the total is $31.50, it is both the blouse and the skirt together, so uf you take away the blouse from the total ($31.50 - $18) then yiu are keft with the price of the skirt
The median is 1.5, hope this was helpful :)
