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

The solution to 2\times 3^2-4^2\div 2+3^2-(-1)^4 is 18

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

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

18

Solution steps

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

Follow the PEMDAS order of operations

Calculate exponents

3^{2} : 9

3^{2}

3^{2} = 9

= 9

= 2 \times 9 - 4^{2} \div 2 + 3^{2} - (- 1)^{4}

Calculate exponents

4^{2} : 16

4^{2}

4^{2} = 16

= 16

= 2 \times 9 - 16 \div 2 + 3^{2} - (- 1)^{4}

Calculate exponents

3^{2} : 9

3^{2}

3^{2} = 9

= 9

= 2 \times 9 - 16 \div 2 + 9 - (- 1)^{4}

Calculate exponents

(- 1)^{4} : 1

(- 1)^{4}

Apply exponent rule:

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

n

is even

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

= 1

= 2 \times 9 - 16 \div 2 + 9 - 1

Multiply and divide (left to right)

2 \times 9 : 18

2 \times 9

2 \times 9 = 18

= 18

= 18 - 16 \div 2 + 9 - 1

Multiply and divide (left to right)

16 \div 2 : 8

16 \div 2

16 \div 2 = 8

= 8

= 18 - 8 + 9 - 1

Add and subtract (left to right)

18 - 8 + 9 - 1 : 18

18 - 8 + 9 - 1

18 - 8 = 10

= 10 + 9 - 1

10 + 9 = 19

= 19 - 1

19 - 1 = 18

= 18

= 18

2 + (8 \div 4)

38 + 45 - 2

3^{2} - 3^{2}

2^{12} + 8^{4}

(3 + 9) - 2 \times 3

What is 2\times 3^2-4^2\div 2+3^2-(-1)^4 ?
The solution to 2\times 3^2-4^2\div 2+3^2-(-1)^4 is 18