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labwork [276]
3 years ago
7

Is 0 a complex, pure imaginary, and a nonreal complex number?

Mathematics
1 answer:
lisabon 2012 [21]3 years ago
7 0

Answer:

A

Step-by-step explanation:

First, 0 doesn't have any i values, so 0 is not pure imaginary.

Also, 0 <em>is</em> indeed a real number, so it is not a "non-real complex number."

Therefore, the only option left is that 0 is a complex number, and 0 is indeed a complex number.

The answer is A.

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Step-by-step explanation:

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Step-by-step explanation:

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Read 2 more answers
Dilution Problem
LUCKY_DIMON [66]

Answer:

  76.7 liters

Step-by-step explanation:

You have ...

  C1×V1 = C2×V2

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4 0
3 years ago
A population of mice is increasing exponentially. On Monday there were 120 most. One month later there were 132 most. Ready func
djyliett [7]

Answer:

y=120 * 1.1^x --- function

The monthly rate is 10%

Step-by-step explanation:

Given

Let

x \to months

y \to mice

So, we have:

(x_1,y_1) = (0,120) --- Monday

(x_2,y_2) = (1,132) --- One month later

Required

The function

The function is represented as:

y=ab^x

In (x_1,y_1) = (0,120), we have:

120 = a * b^0

120 = a * 1

120 = a

a=120

In (x_2,y_2) = (1,132), we have:

132 = a * b^1

132 = a * b

Substitute: a=120

132 = 120 * b

Solve for b

b = \frac{132}{120}

b = 1.1

So, the function is:

y=ab^x

y=120 * 1.1^x

To calculate the monthly rate (r), we have:

y =a(1 + r)^x

Compare to: y =ab^x

1 + r = b

Make r the subject

r = b-1

Substitute b = 1.1

r = 1.1-1

r = 0.1

Express as percentage

r = 0.1*100\%

r = 10\%

5 0
2 years ago
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