Please awnser the question below:

Order of the given expression into smallest to largest are as follow:
[tex]\frac{ \sqrt{\frac{8}{2} } +4 }{6\times (4-12)} < \frac{ \frac{\sqrt{8} }{2} +4}{6\times (4-12)} < \frac{ \frac{\sqrt{8} }{2} +4}{6\times 4-12} < \frac{ \sqrt{\frac{8}{2} } +4 }{6\times 4-12}[/tex]
As given in the question,
Given expressions are as follow:
[tex]\frac{ \sqrt{\frac{8}{2} } +4 }{6\times 4-12},\\\\\frac{ \frac{\sqrt{8} }{2} +4}{6\times 4-12}\\\\\frac{ \sqrt{\frac{8}{2} } +4 }{6\times (4-12)} \\ \\\frac{ \frac{\sqrt{8} }{2} +4}{6\times (4-12)}[/tex]
Arrange them in the order of smallest to largest by simplifying them,
[tex]\frac{ \sqrt{\frac{8}{2} } +4 }{6\times 4-12}= \frac{2+4}{24-12\\}[/tex]
= 0.5
[tex]\frac{ \frac{\sqrt{8} }{2} +4}{6\times 4-12}= \frac{4+\sqrt{2} }{12}[/tex]
=0.45
[tex]\frac{ \sqrt{\frac{8}{2} } +4 }{6\times (4-12)} = \frac{2+4}{-48}[/tex]
= -0.125
[tex]\frac{ \frac{\sqrt{8} }{2} +4}{6\times (4-12)}= \frac{4+\sqrt{2} }{-48}[/tex]
= -0.1128
-0.125 < -0.1128 < 0.45 < 0.5
Therefore, order of the given expression into smallest to largest are as follow:
[tex]\frac{ \sqrt{\frac{8}{2} } +4 }{6\times (4-12)} < \frac{ \frac{\sqrt{8} }{2} +4}{6\times (4-12)} < \frac{ \frac{\sqrt{8} }{2} +4}{6\times 4-12} < \frac{ \sqrt{\frac{8}{2} } +4 }{6\times 4-12}[/tex]
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