# What is #(- 6+ | - 3| ) \cdot ( 4- 3\cdot 2) \div ( | - 2| )#?

3

Let's rewrite the equation

Let's solve the LHS bracket.

Now, the middle bracket.

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((-6 + |-3|) \cdot (4 - 3 \cdot 2) \div (|-2|) = 0).

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[ (-6 + |-3|) \cdot (4 - 3 \cdot 2) \div (|-2|) = (-6 + 3) \cdot (4 - 6) \div 2 = (-3) \cdot (-2) \div 2 = 6 \div 2 = 3 ]

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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