What is 6\div 5-4\times 10 ?

The solution to 6\div 5-4\times 10 is -194/5

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6\div 5-4\times 10

6 \div 5 - 4 \times 10

- \frac{194}{5}

Solution steps

6 \div 5 - 4 \times 10

Follow the PEMDAS order of operations

Multiply and divide (left to right)

6 \div 5 : \frac{6}{5}

6 \div 5

6 \div 5 = \frac{6}{5}

= \frac{6}{5}

= \frac{6}{5} - 4 \times 10

Multiply and divide (left to right)

4 \times 10 : 40

4 \times 10

4 \times 10 = 40

= 40

= \frac{6}{5} - 40

Add and subtract (left to right)

\frac{6}{5} - 40 : - \frac{194}{5}

\frac{6}{5} - 40

\frac{6}{5} - 40 = - \frac{194}{5}

\frac{6}{5} - 40

Convert element to fraction:

40 = \frac{40 \cdot 5}{5}

= - \frac{40 \cdot 5}{5} + \frac{6}{5}

Since the denominators are equal, combine the fractions:

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

= \frac{- 40 \cdot 5 + 6}{5}

- 40 \cdot 5 + 6 = - 194

- 40 \cdot 5 + 6

Multiply the numbers:

40 \cdot 5 = 200

= - 200 + 6

Add/Subtract the numbers:

- 200 + 6 = - 194

= - 194

= \frac{- 194}{5}

Apply the fraction rule:

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

= - \frac{194}{5}

= - \frac{194}{5}

= - \frac{194}{5}

- 6 (1)^{2} + 8 (1) + 5

- 2 \times (- 5 + 2) + 10 \div (- 2)

1 1^{2} - 1811 + 81

3 - 2 + 1 \times 2

\frac{1}{2} + 2 - \frac{13}{6} - \frac{1}{12}