How do you simplify #7i*3i(-8-6i)#?

Answer 1

#168+126i#

The first step is to multiply 7i and 3i together.

#color(orange)"Reminder " color(red)(bar(ul(|color(white)(a/a)color(black)(i^2=(sqrt(-1))^2=-1)color(white)(a/a)|)))#
#rArr7ixx3i=21i^2=-21#
#rArr-21(-8-6i)=168+126i#
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Answer 2

To simplify (7i \times 3i \times (-8 - 6i)), we first multiply the complex numbers:

[ 7i \times 3i = 21i^2 ]

Recall that (i^2 = -1), so:

[ 21i^2 = 21(-1) = -21 ]

Now, we have:

[ -21 \times (-8 - 6i) = 168 + 126i ]

Therefore, (7i \times 3i \times (-8 - 6i)) simplifies to (168 + 126i).

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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.

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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