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Vsevolod [243]
3 years ago
8

Complex number in standard form

Mathematics
2 answers:
sladkih [1.3K]3 years ago
5 0

Answer:

The answer is (C) \frac{2}{3}+\frac{17}{2}i

Step-by-step explanation:

∵ \frac{2}{3}+6i+\frac{5}{2}i = \frac{2}{3}+\frac{12}{2}i+\frac{5}{2}i=

                                                  = \frac{2}{3}+\frac{17}{2}i

∴ The answer is \frac{2}{3}+\frac{17}{2}i in standard form

spin [16.1K]3 years ago
3 0

Answer:

Choice C is the answer.

Step-by-step explanation:

We have given an expression.

(\frac{2}{3} +6i)+(\frac{5}{2}i)

We have to simplify it and write it as complex number in standard form.

a+bi is a complex number in standard form.

\frac{2}{3}+6i+\frac{5}{2}i

Adding like terms, we have

\frac{2}{3} + (6+\frac{5}{2})i

\frac{2}{3} +(\frac{12+5}{2})i

\frac{2}{3}+(\frac{17}{2})i

\frac{2}{3}+\frac{17}{2}i which is the answer.

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2 years ago
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2 years ago
Consider the following relation.
Jet001 [13]

Answer:

(0,-1),(1,0),(2,1),(3,2)

Step-by-step explanation:

We are given that a relation

y=x+1y=x+1

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The given function is linear function

Therefore,

Domain of function=R

Range of function=R

x=y-1x=y−1

Now, replace x by y and y replace by x

y=x-1y=x−1

Now, substitute y=f^{-1}(x)=f

−1

(x)

f^{-1}(x)=x-1f

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(x)=x−1

It is linear function and defined for all real values.

Substitute x=0

f^{-1}(0)=-1f

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Substitute x=1

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(2)=2−1=1

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(3)=3−1=2

Therefore, four points contained in the inverse (0,-1),(1,0),(2,1) and (3,2)

6 0
2 years ago
Can someone help me. ​
marshall27 [118]

Answer:

B

Step-by-step explanation:

yeah just pick B man lol

7 0
3 years ago
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luda_lava [24]

Hello There!

Consecutive Angles In Parallelograms Are "Supplementary"

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6 0
2 years ago
Read 2 more answers
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