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Nadya [2.5K]
3 years ago
12

Is 2i a real or complex root?

Mathematics
1 answer:
Maurinko [17]3 years ago
3 0

By definition, i = sqrt(-1) because it is used to solve the equation x^2+1 = 0 or x^2 = -1, which has no real number solutions.

Since i is imaginary or complex, then so is 2i

You can express it as 2i = 0+2i to be in the form a+bi with a = 0 as the real part and b = 2 as the imaginary component.

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Is it cool?

8 0
3 years ago
What is the answer to this?​
Pie

Answer:

the third one

Step-by-step explanation:

p n p is p

p n q is p

q n q is q

q n p is p

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got it?

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8 0
3 years ago
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monitta
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Check your work

8 /(-0.66666666667) = 12- 3(8)
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Hope this helps :)
8 0
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