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3 years ago

What is equivalent to sq root -27

Mathematics

2 answers:

Katena32 [7]3 years ago

7 0

Negative numbers don’t have real square roots since a square is either positive or 0

lesya692 [45]3 years ago

6 0

$\sqrt{-27}=\sqrt{27}\sqrt{-1}=\boxed{3\sqrt{3}i}$ .

Result is a complex irrational number.

Hope this helps.

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Answer:

(See explanation for further details)

Step-by-step explanation:

a) Let consider the polynomial $p(x) = 5\cdot x^{3} +2\cdot x^{2} - 6 \cdot x +17$ . The polynomial is in standards when has the form $p(x) = \Sigma \limit_{i=0}^{n} \,a_{i}\cdot x^{i}$ , where $n$ is the order of the polynomial. The example has the following information:

$n = 3$ , $a_{0} = 17$ , $a_{1} = -6$ , $a_{2} = 2$ , $a_{3} = 5$ .

b) The closure property means that polynomials must be closed with respect to addition and multiplication, which is demonstrated hereafter:

Closure with respect to addition:

Let consider polynomials $p_{1}$ and $p_{2}$ such that:

$p_{1} = \Sigma \limits_{i=0}^{m} \,a_{i}\cdot x^{i}$ and $p_{2} = \Sigma \limits_{i=0}^{n}\,b_{i}\cdot x^{i}$ , where $m \geq n$

$p_{1}+p_{2} = \Sigma \limits_{i=0}^{n}\,(a_{i}+b_{i})\cdot x^{i} + \Sigma_{i=n+1}^{m}\,a_{i} \cdot x^{i}$

Hence, polynomials are closed with respect to addition.

Closure with respect to multiplication:

Let be $p_{1}$ a polynomial such that:

$p_{1} = \Sigma \limits_{i=0}^{m} \,a_{i}\cdot x^{i}$

And $\alpha$ an scalar. If the polynomial is multiplied by the scalar number, then:

$\alpha \cdot p_{1} = \alpha \cdot \Sigma \limits_{i = 0}^{m}\,a_{i}\cdot x^{i}$

Lastly, the following expression is constructed by distributive property:

$\alpha \cdot p_{1} = \Sigma \limits_{i=0}^{m}\,(\alpha\cdot a_{i})\cdot x^{i}$

Hence, polynomials are closed with respect to multiplication.

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3 years ago

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Answer:

obtuse is the anwer

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Answer:

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Step-by-step explanation:

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Answer:

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