<span>14-3y=4y
Add 3y to both sides so that the variable y is only on one side.
14=7y
Divide 7 on both sides so that the only thing remaining on the right side is the variable y.
Final Answer: 2=y</span>
Step-by-step explanation:
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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: y > -1/2x + 2
Step-by-step explanation: first, in order to find the inequalities you should find the gradient by choosing two points from the line and you should you the formula m=y2-y1/x2-x1 to find the gradient.
Next, you should find the y-intercept in order to complete the inequality it can be easily found as the y-intercept is the place where the line crosses the y axis
Then you create your equation { y = -1/2x + 2 } and then if above the line is shaded then it is {> greater than} and if below the line is shaded then it should be {< less than}
(so you should replace the equation with the lesser or greater sign according to the way the graph is shaded)
