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.
No he is wrong there will be 150000000 bacteria on it at 2 minutes
WORK:
2 min=120 sec
120 divided by 15=8
15 times 10 (7 times)
u get 150,000,000
The amount of the food budget is $2,509 as it is 13% of $19,300.
Here's one way to do it.
AB ≅ AC . . . . . . . . . . given
∠BAY ≅ ∠CAY . . . . given
AY ≅ AY . . . . . . . . . . reflexive property
ΔBAY ≅ ΔCAY . . . .. SAS congruence
XY ≅ XY . . . . . . . . . . reflexive property
∠AYB ≅ ∠AYC . . . . CPCTC
BY ≅ CY . . . . . . . . . . CPCTC
ΔXYB ≅ ΔXYC . . . .. SAS congruence
Therefore ...
∠XCY ≅ ∠XBY . . . . CPCTC
Answer:
2720 feet squared
Step-by-step explanation:
85 * 32 = 2720
