What is 3\div 30-5(-4-(-12))+17 ?

The solution to 3\div 30-5(-4-(-12))+17 is -229/10

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3\div 30-5(-4-(-12))+17

3 \div 30 - 5 (- 4 - (- 12)) + 17

- \frac{229}{10}

Solution steps

3 \div 30 - 5 (- 4 - (- 12)) + 17

Follow the PEMDAS order of operations

Calculate within parentheses

(- 4 - (- 12)) : 8

- 4 - (- 12)

Add and subtract (left to right)

Apply rule

- (- a) = + a

- (- 12) = + 12

= - 4 + 12

- 4 + 12 = 8

= 8

= 3 \div 30 - 5 \cdot 8 + 17

Multiply and divide (left to right)

3 \div 30 : \frac{3}{30}

3 \div 30

3 \div 30 = \frac{3}{30}

= \frac{3}{30}

= \frac{3}{30} - 5 \cdot 8 + 17

Multiply and divide (left to right)

5 \cdot 8 : 40

5 \cdot 8

5 \cdot 8 = 40

= 40

= \frac{3}{30} - 40 + 17

Add and subtract (left to right)

\frac{3}{30} - 40 + 17 : - \frac{229}{10}

\frac{3}{30} - 40 + 17

\frac{3}{30} - 40 = - \frac{399}{10}

\frac{3}{30} - 40

Cancel

\frac{3}{30} : \frac{1}{10}

\frac{3}{30}

Cancel the common factor:

3

= \frac{1}{10}

= \frac{1}{10} - 40

Convert element to fraction:

40 = \frac{40 \cdot 10}{10}

= - \frac{40 \cdot 10}{10} + \frac{1}{10}

Since the denominators are equal, combine the fractions:

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

= \frac{- 40 \cdot 10 + 1}{10}

- 40 \cdot 10 + 1 = - 399

- 40 \cdot 10 + 1

Multiply the numbers:

40 \cdot 10 = 400

= - 400 + 1

Add/Subtract the numbers:

- 400 + 1 = - 399

= - 399

= \frac{- 399}{10}

Apply the fraction rule:

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

= - \frac{399}{10}

= - \frac{399}{10} + 17

- \frac{399}{10} + 17 = - \frac{229}{10}

- \frac{399}{10} + 17

Convert element to fraction:

17 = \frac{17 \cdot 10}{10}

= \frac{17 \cdot 10}{10} - \frac{399}{10}

Since the denominators are equal, combine the fractions:

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

= \frac{17 \cdot 10 - 399}{10}

17 \cdot 10 - 399 = - 229

17 \cdot 10 - 399

Multiply the numbers:

17 \cdot 10 = 170

= 170 - 399

Subtract the numbers:

170 - 399 = - 229

= - 229

= \frac{- 229}{10}

Apply the fraction rule:

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

= - \frac{229}{10}

= - \frac{229}{10}

= - \frac{229}{10}

2 4^{2} + 24 + 4

- 4^{3} /2 + 7^{2}

15 - (2 - 10 - 2) 15 + 3 + 2 - 4 - 318 - 2 - 3

2^{2} - 50

70 + \frac{20}{63} \times 60

What is 3\div 30-5(-4-(-12))+17 ?
The solution to 3\div 30-5(-4-(-12))+17 is -229/10