What is (-2)^2-(+1)^3-7^0 ?

The solution to (-2)^2-(+1)^3-7^0 is 2

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(-2)^2-(+1)^3-7^0

(- 2)^{2} - (+ 1)^{3} - 7^{0}

2

Solution steps

(- 2)^{2} - (1)^{3} - 7^{0}

Follow the PEMDAS order of operations

Calculate exponents

(- 2)^{2} : 4

(- 2)^{2}

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

(- 2)^{2} = 2^{2} = 4

= 4

= 4 - (1)^{3} - 7^{0}

Calculate exponents

(1)^{3} : 1

(1)^{3}

(1)^{3} = 1

= 1

= 4 - 1 - 7^{0}

Calculate exponents

7^{0} : 1

7^{0}

7^{0} = 1

= 1

= 4 - 1 - 1

Add and subtract (left to right)

4 - 1 - 1 : 2

4 - 1 - 1

4 - 1 = 3

= 3 - 1

3 - 1 = 2

= 2

= 2

- 2 {1 - 4 [8 + (- 5)]}

- 16 \times 5 \div 20

2 - 5 \cdot 4 - [6 \cdot (3 - 4)]

\frac{7 ( 12 - 9 ) + 3}{4 ^{2} - 4}

4 - 9 \cdot 6 - 7