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

The solution to (1-2)^2+(4+5) is 10

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

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

10

Solution steps

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

Follow the PEMDAS order of operations

Calculate within parentheses

(1 - 2) : - 1

1 - 2

1 - 2 = - 1

= - 1

= (- 1)^{2} + (4 + 5)

Calculate within parentheses

(4 + 5) : 9

4 + 5

4 + 5 = 9

= 9

= (- 1)^{2} + 9

Calculate exponents

(- 1)^{2} : 1

(- 1)^{2}

Apply exponent rule:

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

n

is even

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

= 1

= 1 + 9

Add and subtract (left to right)

1 + 9 : 10

1 + 9

1 + 9 = 10

= 10

= 10

20 - 5 \times 2 + (8 \div 2)^{2}

25 \div 5 + 16 \div 2

- (7 - 5) \cdot 4

2^{2} - 6 (2) + 8

- 3 (9 - 2)