What is (2-3)[[4(-2)+1]+1]+1 ?

The solution to (2-3)[[4(-2)+1]+1]+1 is 7

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(2-3)[[4(-2)+1]+1]+1

(2 - 3) [[4 (- 2) + 1] + 1] + 1

7

Solution steps

(2 - 3) [[4 (- 2) + 1] + 1] + 1

Follow the PEMDAS order of operations

Calculate within parentheses

(2 - 3) : - 1

2 - 3

2 - 3 = - 1

= - 1

= - 1 \cdot [[4 (- 2) + 1] + 1] + 1

Calculate within parentheses

[[4 (- 2) + 1] + 1] : - 6

[4 (- 2) + 1] + 1

Calculate within parentheses

[4 (- 2) + 1] : - 7

4 (- 2) + 1

Multiply and divide (left to right)

4 (- 2) : - 8

4 (- 2)

Apply rule

a \cdot (- b) = - a \cdot b

4 (- 2) = - 4 \cdot 2 = - 8

= - 8

= - 8 + 1

Add and subtract (left to right)

- 8 + 1 : - 7

- 8 + 1

- 8 + 1 = - 7

= - 7

= - 7

= - 7 + 1

Add and subtract (left to right)

- 7 + 1 : - 6

- 7 + 1

- 7 + 1 = - 6

= - 6

= - 6

= - 1 \cdot (- 6) + 1

Multiply and divide (left to right)

- 1 \cdot (- 6) : 6

- 1 \cdot (- 6)

Apply rule

- a \cdot (- b) = a \cdot b

- 1 \cdot (- 6) = 1 \cdot 6 = 6

= 6

= 6 + 1

Add and subtract (left to right)

6 + 1 : 7

6 + 1

6 + 1 = 7

= 7

= 7

3 + 6 \div 2 - 5

\frac{( 1 5 ^{3} 3 0 ^{2} 1 8 ^{2} )}{( 9 ^{3} 1 2 ^{4} 5 ^{6} )}

2 (- 5) - 5 - 3^{2}

(2^{3} \times 3^{4})^{5}

11 - 48 \div (8 - 6) 2