Pre-Algebra Examples

$\frac{- 4^{2} - 2 (5^{2} - 3^{3})}{6 - | 2 - 12 |}$

Step 1

Simplify the numerator.

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Step 1.1

Raise $4$ to the power of $2$ .

$\frac{- 1 \cdot 16 - 2 (5^{2} - 3^{3})}{6 - | 2 - 12 |}$

Step 1.2

Multiply $- 1$ by $16$ .

$\frac{- 16 - 2 (5^{2} - 3^{3})}{6 - | 2 - 12 |}$

Step 1.3

Simplify each term.

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Step 1.3.1

Raise $5$ to the power of $2$ .

$\frac{- 16 - 2 (25 - 3^{3})}{6 - | 2 - 12 |}$

Step 1.3.2

Raise $3$ to the power of $3$ .

$\frac{- 16 - 2 (25 - 1 \cdot 27)}{6 - | 2 - 12 |}$

Step 1.3.3

Multiply $- 1$ by $27$ .

$\frac{- 16 - 2 (25 - 27)}{6 - | 2 - 12 |}$

Step 1.4

Subtract $27$ from $25$ .

$\frac{- 16 - 2 \cdot - 2}{6 - | 2 - 12 |}$

Step 1.5

Multiply $- 2$ by $- 2$ .

$\frac{- 16 + 4}{6 - | 2 - 12 |}$

Step 1.6

Add $- 16$ and $4$ .

$\frac{- 12}{6 - | 2 - 12 |}$

Step 2

Simplify the denominator.

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Step 2.1

Subtract $12$ from $2$ .

$\frac{- 12}{6 - | - 10 |}$

Step 2.2

The absolute value is the distance between a number and zero. The distance between $- 10$ and $0$ is $10$ .

$\frac{- 12}{6 - 1 \cdot 10}$

Step 2.3

Multiply $- 1$ by $10$ .

$\frac{- 12}{6 - 10}$

Step 2.4

Subtract $10$ from $6$ .

$\frac{- 12}{- 4}$

Step 3

Divide $- 12$ by $- 4$ .

$3$