What is 7^2-40+18\div 3^2-10^0 ?

The solution to 7^2-40+18\div 3^2-10^0 is 10

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7^2-40+18\div 3^2-10^0

7^{2} - 40 + 18 \div 3^{2} - 1 0^{0}

10

Solution steps

7^{2} - 40 + 18 \div 3^{2} - 1 0^{0}

Follow the PEMDAS order of operations

Calculate exponents

7^{2} : 49

7^{2}

7^{2} = 49

= 49

= 49 - 40 + 18 \div 3^{2} - 1 0^{0}

Calculate exponents

3^{2} : 9

3^{2}

3^{2} = 9

= 9

= 49 - 40 + 18 \div 9 - 1 0^{0}

Calculate exponents

1 0^{0} : 1

1 0^{0}

1 0^{0} = 1

= 1

= 49 - 40 + 18 \div 9 - 1

Multiply and divide (left to right)

18 \div 9 : 2

18 \div 9

18 \div 9 = 2

= 2

= 49 - 40 + 2 - 1

Add and subtract (left to right)

49 - 40 + 2 - 1 : 10

49 - 40 + 2 - 1

49 - 40 = 9

= 9 + 2 - 1

9 + 2 = 11

= 11 - 1

11 - 1 = 10

= 10

= 10

\frac{1}{3} \times 5^{2} \times 10

- 19 + (- 9 \times 12 + 8) \div (- 8 + 33)

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

(3 \times + 7) (3 \times (- 7))

5^{4} \times 5