How do you evaluate #c=-2c+1/3rs# when s = 3 and c =5?
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Jace AndersonWhen ( s = 3 ) and ( c = 5 ), the expression ( c = -2c + \frac{1}{3}rs ) evaluates to ( c = -2(5) + \frac{1}{3}(3)(5) ), which simplifies to ( c = -10 + 5 = -5 ).
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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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