How do you solve #w^2-5=23#?
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To solve the equation ( w^2 - 5 = 23 ), follow these steps:
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Add 5 to both sides: [ w^2 = 23 + 5 ] [ w^2 = 28 ]
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Take the square root of both sides: [ w = \pm \sqrt{28} ]
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Simplify the square root of 28: [ w = \pm \sqrt{4 \cdot 7} ] [ w = \pm 2\sqrt{7} ]
So the solutions to the equation ( w^2 - 5 = 23 ) are ( w = 2\sqrt{7} ) and ( w = -2\sqrt{7} ).
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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.
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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