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How do you evaluate #abs(x+y)# when x=3, y=-5?

Answer 1

Gabriel Agustin

#2#

The #color(blue)"absolute value"# denoted by the ' bars' can only ever have a #color(red)"positive value"#

This is because it gives a measure of how far a number is from zero irrespective of it's sign.

That is #|4|=|-4|=4# since 4 and - 4 are both 4 units from zero.

Here #|x+y|=|3+(-5)|=|3-5|=|-2|=2#

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Answer 2

Emery Allen

To evaluate ( \text{abs}(x+y) ) when ( x = 3 ) and ( y = -5 ), substitute the given values into the expression and take the absolute value:

[ \text{abs}(x+y) = \text{abs}(3 + (-5)) ] [ \text{abs}(x+y) = \text{abs}(3 - 5) ] [ \text{abs}(x+y) = \text{abs}(-2) ] [ \text{abs}(x+y) = 2 ]

So, ( \text{abs}(x+y) = 2 ).

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Answer from HIX Tutor

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.