What is -3^0+(-2)^0+(-2)^0+4^0 ?

The solution to -3^0+(-2)^0+(-2)^0+4^0 is 2

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-3^0+(-2)^0+(-2)^0+4^0

- 3^{0} + (- 2)^{0} + (- 2)^{0} + 4^{0}

2

Solution steps

- 3^{0} + (- 2)^{0} + (- 2)^{0} + 4^{0}

Follow the PEMDAS order of operations

Calculate exponents

3^{0} : 1

3^{0}

3^{0} = 1

= 1

= - 1 + (- 2)^{0} + (- 2)^{0} + 4^{0}

Calculate exponents

(- 2)^{0} : 1

(- 2)^{0}

Apply exponent rule:

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

n

is even

(- 2)^{0} = 2^{0} = 1

= 1

= - 1 + 1 + (- 2)^{0} + 4^{0}

Calculate exponents

(- 2)^{0} : 1

(- 2)^{0}

Apply exponent rule:

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

n

is even

(- 2)^{0} = 2^{0} = 1

= 1

= - 1 + 1 + 1 + 4^{0}

Calculate exponents

4^{0} : 1

4^{0}

4^{0} = 1

= 1

= - 1 + 1 + 1 + 1

Add and subtract (left to right)

- 1 + 1 + 1 + 1 : 2

- 1 + 1 + 1 + 1

- 1 + 1 = 0

= 0 + 1 + 1

0 + 1 = 1

= 1 + 1

1 + 1 = 2

= 2

= 2

8 \times 3^{2} - 5 \times 4 + 9

- 5 (- 2)^{2}

- 2 (1) - 4

(- 5 + (- 3) \times 7) \div (- 2)

2 \times 2 - 1