What is 1\div 9-4\div 9 ?

The solution to 1\div 9-4\div 9 is -1/3

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1\div 9-4\div 9

1 \div 9 - 4 \div 9

- \frac{1}{3}

Solution steps

1 \div 9 - 4 \div 9

Follow the PEMDAS order of operations

Multiply and divide (left to right)

1 \div 9 : \frac{1}{9}

1 \div 9

1 \div 9 = \frac{1}{9}

= \frac{1}{9}

= \frac{1}{9} - 4 \div 9

Multiply and divide (left to right)

4 \div 9 : \frac{4}{9}

4 \div 9

4 \div 9 = \frac{4}{9}

= \frac{4}{9}

= \frac{1}{9} - \frac{4}{9}

Add and subtract (left to right)

\frac{1}{9} - \frac{4}{9} : - \frac{1}{3}

\frac{1}{9} - \frac{4}{9}

\frac{1}{9} - \frac{4}{9} = - \frac{1}{3}

\frac{1}{9} - \frac{4}{9}

Apply rule

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{1 - 4}{9}

Subtract the numbers:

1 - 4 = - 3

= \frac{- 3}{9}

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{3}{9}

Cancel the common factor:

3

= - \frac{1}{3}

= - \frac{1}{3}

= - \frac{1}{3}

1 3^{2} - 22 (13) + 120

7 \times 3 - 5 \times 4 + 18 \div 6

5^{2} \cdot 2^{0}

(- 150) \div (+ 5) \times (19 - 21)

(2 \cdot 1 + 3)^{2}