What is 2^2+3^3-5^0+1^{42} ?

The solution to 2^2+3^3-5^0+1^{42} is 31

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2^2+3^3-5^0+1^{42}

2^{2} + 3^{3} - 5^{0} + 1^{42}

31

Solution steps

2^{2} + 3^{3} - 5^{0} + 1^{42}

Follow the PEMDAS order of operations

Calculate exponents

2^{2} : 4

= 4 + 3^{3} - 5^{0} + 1^{42}

Calculate exponents

3^{3} : 27

= 4 + 27 - 5^{0} + 1^{42}

Calculate exponents

5^{0} : 1

= 4 + 27 - 1 + 1^{42}

Calculate exponents

1^{42} : 1

= 4 + 27 - 1 + 1

Add and subtract (left to right)

4 + 27 - 1 + 1 : 31

= 31

(- 2)^{4} - (- 3)^{3} + [(- 1) (- 3) + 4^{2}]^{2}

1 0^{2} - 3^{2} \times 6 - 3 \times 2

7^{2} + ((46 - 7 \times 2) \div 2^{3})^{3}

26 - 7 (3 + 5) \div 4 + 2

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