Question

Can someone please evaluate,,
-i^8 ( negative of iota power 8)
And
-i^5(negative of iota power5)

Answer 1

Answer:

-1 & -i

we know that i^2= -1

1) -i^8 = - (i^2)^4 = -(-1)^4 = -(1) = -1.........

2) -i^5 = - (i^2)^2 × i = -(-1)^2 × i = -(1)×i = -i.........

Answer 2

Answer:

In △ABC AB=AC

⇒∠B=∠C (Angles opposite to equal sides are equal)

Now using angle sum property

∠A+∠B+∠C=180

∘

⇒80

∘

+∠C+∠C=180

∘

⇒2∠C=180

∘

−80

∘

⇒∠C=

2

100

∘

=50

∘

now ∠C+∠x=180

∘

(Angles made on straight line (AC) are supplementary)

⇒50

∘

+∠x=180

∘

⇒∠x=180

∘

−50

∘

=130

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