Answer:
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Step-by-step explanation:
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Answer:
Sin of x does not change
Step-by-step explanation:
Whenever a triangle is dilated, the angle remains the same as well as the ratio for sides of triangle. For smshapes with dimensions, when shapes are dilated the dimensions has increment with common factor.
From trigonometry,
Sin(x)=opposite/hypotenose
Where x=4/5
Sin(4/5)= opposite/hypotenose
But we were given the scale factor of 2 which means the dilation is to two times big.
Then we have
Sin(x)=(2×opposite)/(2×hypotenose)
Then,if we divide by 2 the numerator and denominator we still have
Sin(x)=opposite/hypotenose
Which means the two in numerator and denominator is cancelled out.
Then we still have the same sin of x. as sin(4/5)
Hence,Sin of x does not change
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.
9/21 because it can still be reduced to 3/7
