$i^7 + \dfrac 1{i^4}\\\\=i^2 \cdot i^2 \cdot i^2 \cdot i + \dfrac 1{i^2 \cdot i^2}\\\\=(-1)(-1)(-1)i + \dfrac 1{(-1)(-1)} ~~~;[ i^2 = -1] \\\\\ = (-1)^3 i+ \dfrac 1{(-1)^2} \\\\=-i+1$

7 0

2 years ago

What is square root of 167?

enot [183]

The square root of 167 would be 12.922847. If you need it rounded to the nearest tenth it would be 12.9

If you need t rounded to the nearest hundredth, it would be 12.92.

Hope this helps :)

7 0

2 years ago

(−8k+1)(−8k+1) standard form

kifflom [539]

Answer:

64k^2 - 16k +1

Step-by-step explanation:

We can rewrite this as

(-8k+1) ^2

We know that (a+b)^2 = a^2 +2ab +b^2

Let a = -8k and b = 1

(-8k+1) = (-8k)^2 +2*(-8k)(1) + 1^2

=64k^2 - 16k +1

3 0

3 years ago

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