What is (4)+(5)(2)+(40)\div (10)-7 ?

The solution to (4)+(5)(2)+(40)\div (10)-7 is 11

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(4)+(5)(2)+(40)\div (10)-7

(4) + (5) (2) + (40) \div (10) - 7

11

Solution steps

(4) + (5) (2) + (40) \div (10) - 7

Follow the PEMDAS order of operations

Multiply and divide (left to right)

(5) (2) : 10

(5) (2)

(5) (2) = 5 \cdot 2

(5) (2)

Apply rule:

(a) = a

(5) = 5

= 5 (2)

Apply rule:

(a) = a

(2) = 2

= 5 \cdot 2

= 5 \cdot 2

Multiply the numbers:

5 \cdot 2 = 10

= 10

= (4) + 10 + (40) \div (10) - 7

Multiply and divide (left to right)

(40) \div (10) : \frac{( 40 )}{( 10 )}

(40) \div (10)

(40) \div (10) = \frac{( 40 )}{( 10 )}

= \frac{( 40 )}{( 10 )}

= (4) + 10 + \frac{( 40 )}{( 10 )} - 7

Add and subtract (left to right)

(4) + 10 + \frac{( 40 )}{( 10 )} - 7 : 11

(4) + 10 + \frac{( 40 )}{( 10 )} - 7

(4) + 10 = 14

= 14 + \frac{( 40 )}{( 10 )} - 7

14 + \frac{( 40 )}{( 10 )} = 18

14 + \frac{40}{10}

Divide the numbers:

\frac{40}{10} = 4

= 14 + 4

Add the numbers:

14 + 4 = 18

= 18

= 18 - 7

18 - 7 = 11

= 11

= 11

- 3 (8 - 30)

4 + 4^{2} \div 2

\frac{9}{7 ^{2} - 16}

3 (- 1) + 6

(- 2)^{2} + 8

What is (4)+(5)(2)+(40)\div (10)-7 ?
The solution to (4)+(5)(2)+(40)\div (10)-7 is 11