How do you simplify $(- 10 i) (9 i) + 8$ $(-10i)(9i)+8$ ?

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Answer 1 · 2018-08-08T01:39:06

$98$ $98$

$(- 10 i) (9 i) + 8$ $(-10i)(9i) + 8$

First, do multiplication:
$= - 90 i^{2} + 8$ $=-90i^2 + 8$

We know that $i^{2}$ $i^2$ equals to $- 1$ $-1$ :
$= - 90 (- 1) + 8$ $=-90(-1) + 8$

Simplify:
$= 90 + 8$ $=90 + 8$

$= 98$ $=98$

Hope this helps!

Answer 2 · 2018-08-08T01:45:08

$(- 10 i) \cdot (9 i) + 8 = 98$ $color(magenta)((-10i) * (9i) + 8 = 98$

$i = \sqrt{- 1}, i^{2} = {(\sqrt{-} 1)}^{2} = - 1$ $color(red)(i = sqrt(-1)), " " color(maroon)(i^2 = (sqrt-1)^2 = -1$

$(- 10 i) \cdot (9 i) + 8 = (- 90 i^{2}) + 8 = (- 90 \cdot - 1) + 8$ $(-10i) * (9i) + 8 = (-90 i^2) + 8 =( -90 * -1) + 8$

$\Rightarrow 90 + 8 = 98$ $=> 90 + 8 = 98$

Answer 3 · 2018-08-08T01:57:30

$98$ $98$

We can multiply first to now get

$- 90 i^{2} + 8$ $-90i^2+8$

Recall that $i^{2} = - 1$ $i^2=-1$ . With this simplification, we now have

$90 + 8$ $90+8$ , which simplifies to $98$ $98$ .

Hope this helps!

How do you simplify (-10i)(9i)+8(−10i)(9i)+8(-10i)(9i)+8?