Solve (3^3-5):10-[40/20-40/50]+0!=

$$( 3 ^ { 3 } - 5 ) : 10 - [ \frac { 40 } { 20 } - \frac { 40 } { 50 } ] + 0 ! =$$

Evaluate

$$2$$

Solution Steps

Calculate 3 to the power of 3 and get 27.

$$\frac{27-5}{10}-\left(\frac{40}{20}-\frac{40}{50}\right)+0!$$

Subtract 5 from 27 to get 22.

$$\frac{22}{10}-\left(\frac{40}{20}-\frac{40}{50}\right)+0!$$

Reduce the fraction \frac{22}{10} to lowest terms by extracting and canceling out 2.

$$\frac{11}{5}-\left(\frac{40}{20}-\frac{40}{50}\right)+0!$$

Divide 40 by 20 to get 2.

$$\frac{11}{5}-\left(2-\frac{40}{50}\right)+0!$$

Reduce the fraction \frac{40}{50} to lowest terms by extracting and canceling out 10.

$$\frac{11}{5}-\left(2-\frac{4}{5}\right)+0!$$

Convert 2 to fraction \frac{10}{5}.

$$\frac{11}{5}-\left(\frac{10}{5}-\frac{4}{5}\right)+0!$$

Since \frac{10}{5} and \frac{4}{5} have the same denominator, subtract them by subtracting their numerators.

$$\frac{11}{5}-\frac{10-4}{5}+0!$$

Subtract 4 from 10 to get 6.

$$\frac{11}{5}-\frac{6}{5}+0!$$

Since \frac{11}{5} and \frac{6}{5} have the same denominator, subtract them by subtracting their numerators.

$$\frac{11-6}{5}+0!$$

Subtract 6 from 11 to get 5.

$$\frac{5}{5}+0!$$

Divide 5 by 5 to get 1.

$$1+0!$$

The factorial of 0 is 1.

$$1+1$$

Add 1 and 1 to get 2.

$$2$$

Factor

$$2$$