What is 4-(2-9)^0+3^2\div 1+3 ?

The solution to 4-(2-9)^0+3^2\div 1+3 is 15

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4-(2-9)^0+3^2\div 1+3

4 - (2 - 9)^{0} + 3^{2} \div 1 + 3

15

Solution steps

4 - (2 - 9)^{0} + 3^{2} \div 1 + 3

Follow the PEMDAS order of operations

Calculate within parentheses

(2 - 9) : - 7

2 - 9

2 - 9 = - 7

= - 7

= 4 - (- 7)^{0} + 3^{2} \div 1 + 3

Calculate exponents

(- 7)^{0} : 1

(- 7)^{0}

Apply exponent rule:

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

n

is even

(- 7)^{0} = 7^{0} = 1

= 1

= 4 - 1 + 3^{2} \div 1 + 3

Calculate exponents

3^{2} : 9

3^{2}

3^{2} = 9

= 9

= 4 - 1 + 9 \div 1 + 3

Multiply and divide (left to right)

9 \div 1 : 9

9 \div 1

9 \div 1 = 9

= 9

= 4 - 1 + 9 + 3

Add and subtract (left to right)

4 - 1 + 9 + 3 : 15

4 - 1 + 9 + 3

4 - 1 = 3

= 3 + 9 + 3

3 + 9 = 12

= 12 + 3

12 + 3 = 15

= 15

= 15

20 \times 1 0^{2}

- (5)^{2} + 12 (5) - 32

- 4 - 4 - 4 - 4

4^{2} - 11

- 4 (0) + 3