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telo118 [61]
3 years ago
13

Find the value of given expression

ula1" title=" \sqrt{645 \times 645} " alt=" \sqrt{645 \times 645} " align="absmiddle" class="latex-formula">
​
Mathematics
2 answers:
antoniya [11.8K]3 years ago
4 0

Answer:

√{645×645}=√645²=±645 is your answer

Sergio039 [100]3 years ago
3 0

208012.5

Step-by-step explanation:

\sqrt{645 \times 645}

=  \sqrt{416025}

= 208012.5

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When working with mean, what steps do you do to find the mean?
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3 0
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Vlad [161]
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4 0
3 years ago
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.

6 0
3 years ago
No link need right answer 100 points
maks197457 [2]

Answer:

13 1/8

Step-by-step explanation:

3/2* 7/2*5/2

= 105/8

= 13 1/8

8 0
3 years ago
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