3 years ago

IM STILL NOT OUT ITS ANNOYING P.S. NO QUESTION

Mathematics

2 answers:

Nadusha1986 [10]3 years ago

7 0

they deleted all my points im so triggered oml

s2008m [1.1K]3 years ago

6 0

Answer:

wym

Step-by-step explanation:

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4 years ago

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Mashutka [201]

Answer:

$\begin{equation}
\sqrt{3}-1,2 \sqrt{10} \div 5, \sqrt{14}, 3 \sqrt{2}, \sqrt{19}+1,6
\end{equation}$

Explanation:

Given the irrational numbers:

$3 \sqrt{2}, \sqrt{3}-1, \sqrt{19}+1,6$, $2 \sqrt{10} \div 5,\sqrt{14}$

In order to arrange the numbers from the least to the greatest, we convert each number into its decimal equivalent.

$\begin{gathered} 3\sqrt{2}=3\times1.414\approx4.242 \\ \sqrt{3}-1\approx1.732-1=0.732 \\ \sqrt{19}+1\approx4.3589+1=5.3589 \\ 6=6 \\ 2\sqrt{10}\div5=2(3.1623)\div5=1.2649 \\ \sqrt{14}=3.7147 \end{gathered}$

Finally, sort these numbers in ascending order..

$\begin{gathered} \sqrt{3}-1\approx1.732-1=0.732 \\ 2\sqrt{10}\div5=2(3.1623)\div5=1.2649 \\ \sqrt{14}=3.7147 \\ 3\sqrt{2}=3\times1.414\approx4.242 \\ \sqrt{19}+1\approx4.3589+1=5.3589 \\ 6=6 \end{gathered}$

The given numbers in ascending order is:

$\begin{equation}
\sqrt{3}-1,2 \sqrt{10} \div 5, \sqrt{14}, 3 \sqrt{2}, \sqrt{19}+1,6
\end{equation}$

Note: In your solution, you can make the conversion of each irrational begin on a new line.

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1 year ago