Simplify: 8i^7+6i^5-5i^3-3i^2-7i-9

1 answer:

5 0

<span>Consider the remainder of the division of the power exponent of i by 4.
For a remainder r equal to 0, 1, 2 or 3, we have a power equal to 1, i, -1 or -i respectively, therefore:

</span> $8i^7+6i^5-5i^3-3i^2-7i-9 =$
$8*(-i) + 6*i - 5*(-i) - 3*(-1) - 7i - 9 =$
$- 8i + 6i + 5i + 3 - 7i - 9 =$
$3 - 9 - 8i + 6i + 5i- 7i =$
$=\boxed{- 6 - 4i}$

Answer:
$\boxed{- 6 - 4i}$

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To find:

The Law of Sines for the given triangle.

Solution:

Law of sines: In triangle ABC,

$\dfrac{\sin A^\circ}{a}=\dfrac{\sin B^\circ}{b}=\dfrac{\sin C^\circ}{c}$