What is (-1-3)^2\div (-2)^3+(-4)^0 ?

The solution to (-1-3)^2\div (-2)^3+(-4)^0 is -1

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(-1-3)^2\div (-2)^3+(-4)^0

(- 1 - 3)^{2} \div (- 2)^{3} + (- 4)^{0}

- 1

Solution steps

(- 1 - 3)^{2} \div (- 2)^{3} + (- 4)^{0}

Follow the PEMDAS order of operations

Calculate within parentheses

(- 1 - 3) : - 4

- 1 - 3

- 1 - 3 = - 4

= - 4

= (- 4)^{2} \div (- 2)^{3} + (- 4)^{0}

Calculate exponents

(- 4)^{2} : 16

(- 4)^{2}

Apply exponent rule:

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

n

is even

(- 4)^{2} = 4^{2} = 16

= 16

= 16 \div (- 2)^{3} + (- 4)^{0}

Calculate exponents

(- 2)^{3} : - 8

(- 2)^{3}

Apply exponent rule:

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

n

is odd

(- 2)^{3} = - 2^{3} = - 8

= - 8

= 16 \div (- 8) + (- 4)^{0}

Calculate exponents

(- 4)^{0} : 1

(- 4)^{0}

Apply exponent rule:

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

n

is even

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

= 1

= 16 \div (- 8) + 1

Multiply and divide (left to right)

16 \div (- 8) : - 2

16 \div (- 8)

Apply rule

a \div (- b) = - a \div b

16 \div (- 8) = - 16 \div 8 = - 2

= - 2

= - 2 + 1

Add and subtract (left to right)

- 2 + 1 : - 1

- 2 + 1

- 2 + 1 = - 1

= - 1

= - 1

4 (3 (5) + 1)^{2} - 6 (5)

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

9 \cdot 10 + 6^{2}

20 + {[7 + 5 + (- 9 + 7) + 3]}

\frac{108 + 72 + 12 - 36}{12}

What is (-1-3)^2\div (-2)^3+(-4)^0 ?
The solution to (-1-3)^2\div (-2)^3+(-4)^0 is -1