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Sonja [21]
3 years ago
11

Hi everyone , Would you help me please :

Mathematics
1 answer:
Veseljchak [2.6K]3 years ago
8 0
 
\displaystyle  \\ 
 \frac{1}{ \sqrt{2} } = \frac{1 \times \sqrt{2}}{ \sqrt{2} \times \sqrt{2}} = \boxed{\frac{\sqrt{2}}{2}}



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Given that Jon said,

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\begin{gathered} \text{for m = positive integer.} \\ m=5 \\ m-1=5-1=4 \\ 1-m=1-5=-4 \\ So,\text{ } \\ m-1>1-m \\ \text{for m equals positive integer } \end{gathered}

secondly for negative values of m;

\begin{gathered} m=-5 \\ m-1=-5-1=-6 \\ 1-m=1-(-5)=1+5=6 \\ So, \\ m-1

So, the statement "m-1 is always greater than 1-m" is false.

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