Answer:
[tex]\left(\left(3^3-7\right)\frac{5}{10}\right)+\left(\left(1+3+6+9\right)-1\right)=28[/tex]
Step-by-step explanation:
As far as I am able to observe from the statement of your question, the expression is:
[tex]\left[[\left(3^3-7\right)\cdot \frac{5}{10}\right]+\left(\left(1+3+6+9\right)-1\right)[/tex]
So, lets solve this expression, which anyways would clear your concept
Considering the expression
[tex]\left[[\left(3^3-7\right)\cdot \frac{5}{10}\right]+\left(\left(1+3+6+9\right)-1\right)[/tex]
[tex]\mathrm{Remove\:parentheses}:\quad \left(a\right)=a[/tex]
[tex]=\left(3^3-7\right)\frac{5}{10}+1+3+6+9-1[/tex]
Lets first solve [tex]\left(3^3-7\right)\frac{5}{10}[/tex]
As [tex]3^3=27[/tex]
[tex]=\frac{5}{10}\left(27-7\right)[/tex]
[tex]=\frac{1}{2}\left(27-7\right)[/tex]
[tex]=20\cdot \frac{1}{2}[/tex]
[tex]=10[/tex]
So,
[tex]\left(3^3-7\right)\frac{5}{10}+1+3+6+9-1 = 10+1+3+6+9-1[/tex]
= [tex]28[/tex]
Therefore,
[tex]\left(\left(3^3-7\right)\frac{5}{10}\right)+\left(\left(1+3+6+9\right)-1\right)=28[/tex]
Keywords: Expression solving
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