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kvasek [131]
3 years ago
10

Can someone please help me on this one!!!!

Mathematics
1 answer:
VashaNatasha [74]3 years ago
5 0
This appears to be about rules of exponents as much as anything. The applicable "definitions, identities, and properties" are
  i^0 = 1 . . . . . as is true for any non-zero value to the zero power
  i^1 = i . . . . . . as is true for any value to the first power
  i^2 = -1 . . . . . from the definition of i
  i^3 = -i . . . . . = (i^2)·(i^1) = -1·i = -i
  i^n = i^(n mod 4) . . . . . where "n mod 4" is the remainder after division by 4


1. = -3^4·i^(3·2+0+2·4) = -81·i^14 = 81

2. = i^((3-5)·2+0 = i^-4 = 1

3. = -2^2·i^(4+2+2+(-1+1+5)·3+0) = -4·i^23 = 4i

4. = i^(3+(2+3+4+0+2+5)·2) = i^35 = -i

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The answer to your question is: Yes, it is a solution

Step-by-step explanation:

           

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Process, replace the point in the line

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5 0
3 years ago
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Sladkaya [172]
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2 years ago
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Step-by-step explanation:

7 0
2 years ago
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Step-by-step explanation:

<u>Step 1:  Find the answer </u>

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