# How do you divide #( -i-9) / (i-2)# in trigonometric form?

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To divide ((-i-9)) by ((i-2)) in trigonometric form, first, multiply the numerator and denominator by the conjugate of the denominator. The conjugate of (i-2) is (i+2).

[ \frac{-i-9}{i-2} \times \frac{i+2}{i+2} = \frac{(-i-9)(i+2)}{(i-2)(i+2)} ]

Next, expand the numerator and denominator, then simplify.

[ \frac{(-i-9)(i+2)}{(i-2)(i+2)} = \frac{-i^2-2i-9i-18}{i^2-4} = \frac{-(-1)-2i-9i-18}{-4} = \frac{1-11i}{-4} ]

So, (\frac{-i-9}{i-2}) in trigonometric form is (\frac{1-11i}{-4}).

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To divide (-i - 9) by (i - 2) in trigonometric form, we first need to rationalize the denominator by multiplying both the numerator and denominator by the conjugate of the denominator. The conjugate of (i - 2) is (i + 2).

((-i - 9) * (i + 2)) / ((i - 2) * (i + 2)) = ((-i^2 - 2i) - 9i - 18) / (i^2 - 4) = ((1 + 2i) - 9i - 18) / (-3) = (-17 - 7i) / (-3).

So, the division of (-i - 9) by (i - 2) in trigonometric form is (-17 - 7i) / (-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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