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4 years ago

What is 27.539 rounded to the nearest tenth

Mathematics

2 answers:

kondaur [170]4 years ago

6 0

The answer to your question is 28 if you round the give up you get 28

Genrish500 [490]4 years ago

3 0

The answer of this is 27.5

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PolarNik [594]

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Step-by-step explanation:

Hi there.

Well, let's translate it from Latex. Then expanding it:

$\left(\sqrt{x}+\dfrac5x\right)^{9}=$ $(\sqrt{x})^9(5/x)^{0} +(\sqrt{x})^{8}(5/x)^{1}+(\sqrt{x})^{7}(5/x)^{2} \\+(\sqrt{x})^{6}(5/x)^{3}+(\sqrt{x})^{5}(5/x)^{4}+(\sqrt{x})^{4}(5/x)^{5}\\+(\sqrt{x})^{3}(5/x)^{6}+(\sqrt{x})^{2}(5/x)^{7}+(\sqrt{x})^{1}(5/x)^{8}+(\sqrt{x})^{0}(5/x)^{9} =$

$\frac{\left(5+x^{\frac{3}{2}}\right)^9}{x^9}$

Finding a constant term, in a Binomial has its cases.

In this case, there's no constant as a real number. Also, this an univariate binomial. So to find this Binomial expansion's constant term. We must follow this formula, (for a 9th degree) there are 10 terms:

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$(\sqrt{x})^{5}(5/x)^{4}$

Because this result satisfy a ratio.

$y=\frac{c}{x}$

Where c is the constant term, x is the first term of this binomial

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