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.
80, divide 200 by 2.5 and you get 80. Check by multiplying 2.5 and 80, which equals 200
2x/3-6=4. 2x/3=4+6. 2x/3=10. 2x=10*3. 2x=30. X=30/2. x=15.
-4x - 5y = 7
3x + 5y = -14
You can add these two equations together straightaway since the y-terms have opposite coefficients.
-4x - 5y = 7
3x + 5y = -14
+___________
-x - 0 = -7
-x = -7
x = 7
Substitute 7 for x into either of the original equations and solve algebraically to find y.
3x + 5y = -14
3(7) + 5y = -14
21 + 5y = -14
21 = -14 - 5y
35 = -5y
-7 = y
Finally, check work by substituting both x- and y-values into both original equations.
-4x - 5y = 7
-4(7) - 5(-7) = 7
-28 + 35 = 7
7 = 7
3x + 5y = -14
3(7) + 5(-7) = -14
21 - 35 = -14
-14 = -14
Answer:
x = 7 and y = -7; (7, -7).
Answer:
2x dollars an hour. A.
Step-by-step explanation:
2 * x = 2x
