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VARVARA [1.3K]
3 years ago
11

Simplify the expressions. i^23

Mathematics
1 answer:
masha68 [24]3 years ago
8 0

Answer:

-i

Step-by-step explanation:

i^0=1

i^1=i

i^2=-1

i^3=-i

i^4=1

This repeats so we want to see how many 4 factors of i there is in i^(23) which is 5 with a remainder of 3.

So i^(23)=i^3=-i.

i^(23)=i^(5*4+3)=(i^4)^5 * (i^3)=(1)^5 * (-i)=1(-i)=-i.

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<h3>What is the cost per unit with defective units?</h3>

The cost per unit with defective units can be computed by subtracting the sales value of the sold defective units from the total production costs, thereby reducing the total production costs, and then dividing the resulting figure by the number of good units.

<h3>Data and Calculations:</h3>

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Total costs                                                                        420,000

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Thus, the cost per unit of the complete units assuming that defective units are sold at Ksh 20 per unit is Rm 93.73.

Learn more about accounting for defective units at brainly.com/question/10035226

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