What is 5^0+4^1+(-1)^2+(-2)^3 ?

The solution to 5^0+4^1+(-1)^2+(-2)^3 is -2

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5^0+4^1+(-1)^2+(-2)^3

5^{0} + 4^{1} + (- 1)^{2} + (- 2)^{3}

- 2

Solution steps

5^{0} + 4^{1} + (- 1)^{2} + (- 2)^{3}

Follow the PEMDAS order of operations

Calculate exponents

5^{0} : 1

5^{0}

5^{0} = 1

= 1

= 1 + 4^{1} + (- 1)^{2} + (- 2)^{3}

Calculate exponents

4^{1} : 4

4^{1}

4^{1} = 4

= 4

= 1 + 4 + (- 1)^{2} + (- 2)^{3}

Calculate exponents

(- 1)^{2} : 1

(- 1)^{2}

Apply exponent rule:

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

n

is even

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

= 1

= 1 + 4 + 1 + (- 2)^{3}

Calculate exponents

(- 2)^{3} : - 8

(- 2)^{3}

Apply exponent rule:

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

n

is odd

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

= - 8

= 1 + 4 + 1 - 8

Add and subtract (left to right)

1 + 4 + 1 - 8 : - 2

1 + 4 + 1 - 8

1 + 4 = 5

= 5 + 1 - 8

5 + 1 = 6

= 6 - 8

6 - 8 = - 2

= - 2

= - 2

- 6^{2} + 7 \times (- 6) - 5

(\frac{3}{5} - \frac{1}{2}) \div \frac{3}{10}

(- 4)^{3} - 9 (- 4)^{2} - 10 (- 5)

2^{2} \times (- 9) - 9

10 \cdot 2^{5}