Question

What is [[3^3-7)·5)/10]+((1+3+6+9)-)]

Answer 1

Answer:

$\left(\left(3^3-7\right)\frac{5}{10}\right)+\left(\left(1+3+6+9\right)-1\right)=28$

Step-by-step explanation:

As far as I am able to observe from the statement of your question, the expression is:

$\left[[\left(3^3-7\right)\cdot \frac{5}{10}\right]+\left(\left(1+3+6+9\right)-1\right)$

So, lets solve this expression, which anyways would clear your concept

Considering the expression

$\left[[\left(3^3-7\right)\cdot \frac{5}{10}\right]+\left(\left(1+3+6+9\right)-1\right)$

$\mathrm{Remove\:parentheses}:\quad \left(a\right)=a$

$=\left(3^3-7\right)\frac{5}{10}+1+3+6+9-1$

Lets first solve $\left(3^3-7\right)\frac{5}{10}$

As $3^3=27$

$=\frac{5}{10}\left(27-7\right)$

$=\frac{1}{2}\left(27-7\right)$

$=20\cdot \frac{1}{2}$

$=10$

So,

$\left(3^3-7\right)\frac{5}{10}+1+3+6+9-1 = 10+1+3+6+9-1$  

                                                 =  $28$

Therefore,

$\left(\left(3^3-7\right)\frac{5}{10}\right)+\left(\left(1+3+6+9\right)-1\right)=28$

Keywords: Expression solving

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