What is 2^4-(2)^3-3(2)^2+7(2)-10 ?

The solution to 2^4-(2)^3-3(2)^2+7(2)-10 is 0

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2^4-(2)^3-3(2)^2+7(2)-10

2^{4} - (2)^{3} - 3 (2)^{2} + 7 (2) - 10

0

Solution steps

2^{4} - (2)^{3} - 3 (2)^{2} + 7 (2) - 10

Follow the PEMDAS order of operations

Calculate exponents

2^{4} : 16

2^{4}

2^{4} = 16

= 16

= 16 - (2)^{3} - 3 (2)^{2} + 7 (2) - 10

Calculate exponents

(2)^{3} : 8

(2)^{3}

(2)^{3} = 8

= 8

= 16 - 8 - 3 (2)^{2} + 7 (2) - 10

Calculate exponents

(2)^{2} : 4

(2)^{2}

(2)^{2} = 4

= 4

= 16 - 8 - 3 \cdot 4 + 7 (2) - 10

Multiply and divide (left to right)

3 \cdot 4 : 12

3 \cdot 4

3 \cdot 4 = 12

= 12

= 16 - 8 - 12 + 7 (2) - 10

Multiply and divide (left to right)

7 (2) : 14

7 (2)

Apply rule:

(a) = a

(2) = 2

= 7 \cdot 2

Multiply the numbers:

7 \cdot 2 = 14

= 14

= 16 - 8 - 12 + 14 - 10

Add and subtract (left to right)

16 - 8 - 12 + 14 - 10 : 0

16 - 8 - 12 + 14 - 10

16 - 8 = 8

= 8 - 12 + 14 - 10

8 - 12 = - 4

= - 4 + 14 - 10

- 4 + 14 = 10

= 10 - 10

10 - 10 = 0

= 0

= 0

10 \times 5 - (- 6)^{2} + (- 8)

3 [2 - 4 (7 - 12)]

\frac{63}{5} \times (1 - \frac{6}{7})

90 + (70% \times 90)

- 2 \cdot (- \frac{3}{2}) + 2

What is 2^4-(2)^3-3(2)^2+7(2)-10 ?
The solution to 2^4-(2)^3-3(2)^2+7(2)-10 is 0