What is $(4^{2} - 8) \cdot 2^{3} - 8 \cdot 5 \div 10$ $(4^2-8)2^3-85-:10$ ?

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What is $(4^{2} - 8) \cdot 2^{3} - 8 \cdot 5 \div 10$ $(4^2-8)2^3-85-:10$ ?

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Answer 1 · 2018-06-07T04:51:23

$[(16 - 8) \times 8] - \frac{40}{10} = 64 - 4 = 60$ $[(16-8)xx8]-40/10= 64-4=60$

Explanation:

It's actually extremely simple if you know the order of operations.
PEMDAS!

Parentheses
Exponent
Multiplication
Division
Addition
Subtraction

It also goes left to right! So beginning with the parentheses, we focus on what is in the parentheses

$(4^{2} - 8)$ $( 4^2 -8)$

meaning $16$ $16$ , which is the square of $4$ $4$ , subtracted by $8$ $8$ . which turns in to $8$ $8$ .

$(4^{2} - 8) = 16 - 8 = 8$ $( 4^2 -8) = 16 - 8 = 8$

After the paratheses, you multiply by $8$ $8$ the cube of $2$ $2$ , which is then subtracted by $4$ $4$ .

$8 \times 2^{3} = 8 \times 8 = 64$ $8 xx 2^3 = 8 xx 8 = 64$

Since, if you follow PEMDAS and multiply then divide from left to right, it's $\frac{40}{10}$ $40/10$ , which is $4$ $4$ . So the answer, in the end, is

$64 - \frac{40}{10} = 64 - 4 = 60$ $64 - 40/10 = 64 - 4 = 60$

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Explanation:

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