Question

I dont want you to answer question for me, i have already answered it as shown in the picture. I want you to let me know if i have provided an answer worth full marks and if not tell me how i could improve it

Answer 1

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.