How do you simplify i²⁹? (If you can't tell, that's the imaginary number i^29).

1 answer:

6 0

The imaginary number ' i ' is the square root of -1.

i = √-1
i² = -1
i³ = -√-1
i to the 4th power = +1
i to the 5th power = √-1
etc.

and all the higher powers keep going in the same 4-step cycle.

What's 29 divided by 4 ?
It's 7 with a remainder of 1.

So 29 with all the 4s thrown away is 1, and ' i ' to the 29th power is
the same thing as ' i ' to the 1st power = √-1 or just plain ' i '.

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If h(x) = 5 x and k(x)=1/x, which expression is equivalent to (koh)(x)?

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May be there is an operator missing in the first function, h(x). I will solve this in two ways, 1) as if the h(x) = 5x and 2) as if h(x) = 5 + x

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Then (k o h) (x) = k ( h(x) ) = k(5x) = 1/(5x)

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