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AlladinOne [14]
4 years ago
5

What is the solution i^401?

Mathematics
1 answer:
alexandr1967 [171]4 years ago
6 0
There is a set of rules to solve i to the power.
First off, you need to know this:
{i}^{1} = i \\ i^{2} = - 1 \\ {i}^{3} = - i \\ {i}^{4} = 1
After the power of 4, it just starts to loop from 1. Knowing this, we can say the following:
i ^{4k} = 1 \\ {i}^{4k + 1} = i \\ {i}^{4k + 2} = - 1 \\ {i}^{4k + 3} = - i
Applying this, all we need to do is divide the power by 4 and determine the remainder.
i ^{401} = \\ 401 \times 4 = 100 \: (1 \: left) \\ i^{401} = i^{4 \times 100 + 1} \\ k = 100 \\ {i}^{401} = i^{4 \times 100 + 1} = {i}^{4k + 1} = i
So the answer is:
{i}^{401} = i
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hope it helps mark me brainlieast!

Step-by-step explanation:

<em>For triangle ABC with sides  a,b,c  labeled in the usual way, </em>

<em> </em>

<em>c2=a2+b2−2abcosC  </em>

<em> </em>

<em>We can easily solve for angle  C . </em>

<em> </em>

<em>2abcosC=a2+b2−c2  </em>

<em> </em>

<em>cosC=a2+b2−c22ab  </em>

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<em>C=arccosa2+b2−c22ab  </em>

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<em> </em>

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<em> </em>

<em>0∘≤C≤180∘  </em>

<em> </em>

<em>We needed to include the degenerate triangle angles,  0∘  and  180∘,  among the triangle angles to capture the full range of the cosine. Degenerate triangles aren’t triangles, but they do correspond to a valid configuration of three points, namely three collinear points. </em>

<em> </em>

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<em> </em>

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<em> </em>

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<em> </em>

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