Answer:
I gave my answer in chat and imma try and do this to see if it works with a few words now.
x.x
Answer:
![\begin{equation} \sqrt{3}-1,2 \sqrt{10} \div 5, \sqrt{14}, 3 \sqrt{2}, \sqrt{19}+1,6 \end{equation}](https://tex.z-dn.net/?f=%5Cbegin%7Bequation%7D%0A%5Csqrt%7B3%7D-1%2C2%20%5Csqrt%7B10%7D%20%5Cdiv%205%2C%20%5Csqrt%7B14%7D%2C%203%20%5Csqrt%7B2%7D%2C%20%5Csqrt%7B19%7D%2B1%2C6%0A%5Cend%7Bequation%7D)
Explanation:
Given the irrational numbers:
![$3 \sqrt{2}, \sqrt{3}-1, \sqrt{19}+1,6$, $2 \sqrt{10} \div 5,\sqrt{14}$](https://tex.z-dn.net/?f=%243%20%5Csqrt%7B2%7D%2C%20%5Csqrt%7B3%7D-1%2C%20%5Csqrt%7B19%7D%2B1%2C6%24%2C%20%242%20%5Csqrt%7B10%7D%20%5Cdiv%205%2C%5Csqrt%7B14%7D%24)
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}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%203%5Csqrt%7B2%7D%3D3%5Ctimes1.414%5Capprox4.242%20%5C%5C%20%5Csqrt%7B3%7D-1%5Capprox1.732-1%3D0.732%20%5C%5C%20%5Csqrt%7B19%7D%2B1%5Capprox4.3589%2B1%3D5.3589%20%5C%5C%206%3D6%20%5C%5C%202%5Csqrt%7B10%7D%5Cdiv5%3D2%283.1623%29%5Cdiv5%3D1.2649%20%5C%5C%20%5Csqrt%7B14%7D%3D3.7147%20%5Cend%7Bgathered%7D)
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}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20%5Csqrt%7B3%7D-1%5Capprox1.732-1%3D0.732%20%5C%5C%202%5Csqrt%7B10%7D%5Cdiv5%3D2%283.1623%29%5Cdiv5%3D1.2649%20%5C%5C%20%5Csqrt%7B14%7D%3D3.7147%20%5C%5C%203%5Csqrt%7B2%7D%3D3%5Ctimes1.414%5Capprox4.242%20%5C%5C%20%5Csqrt%7B19%7D%2B1%5Capprox4.3589%2B1%3D5.3589%20%5C%5C%206%3D6%20%5Cend%7Bgathered%7D)
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}](https://tex.z-dn.net/?f=%5Cbegin%7Bequation%7D%0A%5Csqrt%7B3%7D-1%2C2%20%5Csqrt%7B10%7D%20%5Cdiv%205%2C%20%5Csqrt%7B14%7D%2C%203%20%5Csqrt%7B2%7D%2C%20%5Csqrt%7B19%7D%2B1%2C6%0A%5Cend%7Bequation%7D)
Note: In your solution, you can make the conversion of each irrational begin on a new line.
The area of a sphere is : 4π.R²
Here R(radius) = 10 ===R²=100
& The Area is 400π inches²
The diagonals of the parallelogram bisect each other,
so this statement is true