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STALIN [3.7K]
3 years ago
11

Any number in the form of a+-bi, where a and b are real numbers and b is not equal to 0 is considered a pure imaginary number

Mathematics
2 answers:
lilavasa [31]3 years ago
7 0
Wrong those are considered complex numbers. pure imaginary numbers have ,0 for a and are thus in the form. 3i, -2i, 17i. etc
NISA [10]3 years ago
4 0

Answer:

<h2>False.</h2>

Step-by-step explanation:

<em>A pure imaginary number is a complex number that doesn't have a real part.</em>

So, if a is a real number, and it doesn't specify that a is only equal to zero, then the expression a+bi is not a pure imaginary number, it's only a complex number. Examples of imaginary numbers are 2i;3i;4i;....bi where b\inR

In this case, a\in R, that is, <em>a </em>can be 0, \±1, \±2, \±3, ...

<em>Therefore the statement is </em><em>false</em><em>, because a can take any real value.</em>

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3 years ago
1. If RU = 16, UT = 20, and SR = 16, what is the perimeter of SUT?
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Step-by-step explanation:

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3 0
3 years ago
Solve for x,y, and z
Lostsunrise [7]
Equation 1)  3x + 2y - 5z = 3
Equation 2)  4x - 2y - 3z = -10
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Add equation 1 with equation 2.

Equation 4)  7x - 8z = 7

Then subtract equation 3 from equation 2.

Equation 5)  -x -z = 1

Multiply all of equation 5 with 7.

5)  -7x - 7z = 7
4)  7x - 8z = 7

Add equations together.

z = 14

Plug in 14 for z in equation 4.

7x - 8z = 7

7x - 8(14) = 7

7x - 112 = 7

7x = 119

x = 17

Plug in 17 for x in equation 1, and 14 for z.

1)  3x + 2y - 5z = 3

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So, x = 17, y = 11, and z = 14

~Hope I helped!~



3 0
3 years ago
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