Answer:
1. Figure C has reflection symmetry.
2. A and D have reflection symmetry.
Step-by-step explanation:
1. When you cut the arrow horizontally, you get the same shape on each side.
2. When you divide them, you should have symmetry, meaning they are the same when divided/cut.
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.
Answer:
choice a.) degree 3, constant term -1 leading coefficient 1
Step-by-step explanation:
expand the binomial product
(x - 1)(xx + 1) = xxx - 1 + x - xx
Answer:
53.85
Step-by-step explanation:
Idk this is hard sorry I couldn't help you but you just helped me
