How do you find three consecutive even integers such that the sum of the first, third, and twice the second is -16?
The three consecutive numbers are -6, -4, -2
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Let the three consecutive even integers be ( n ), ( n + 2 ), and ( n + 4 ). According to the given conditions, the sum of the first, third, and twice the second is:
[ n + (n + 4) + 2(n + 2) = -16 ]
Solve for ( n ):
[ n + n + 4 + 2n + 4 = -16 ] [ 4n + 8 = -16 ] [ 4n = -24 ] [ n = -6 ]
So the three consecutive even integers are ( -6 ), ( -4 ), and ( -2 ).
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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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