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Mila [183]
2 years ago
14

You only have to get this second answer right (30 points)

Mathematics
2 answers:
ololo11 [35]2 years ago
6 0

Answer:

6 with 4 leftover

Step-by-step explanation:

Just some division

DanielleElmas [232]2 years ago
5 0

Answer:

6 nights with 4 logs left over

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icang [17]
38000 you should use a calculator.
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3 years ago
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Use the theorem in Sec. 28 to show that if f (z) is analytic and not constant throughout a domain D, then it cannot be constant
alexandr1967 [171]

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.

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Find the volume of the cone. Round your answer to the nearest tenth if necessary. Use 3.14 for π.
Doss [256]

Answer:

I think the answer is 287.98

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84211

Step-by-step explanation:

Got it right

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How could I Answer With Work
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16 inches for the answer
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