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3 years ago

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

2 answers:

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:

A pure imaginary number is a complex number that doesn't have a real part.

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, a can be $0, \±1, \±2, \±3, ...$

Therefore the statement is false, because a can take any real value.

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Solve the system of equations using elimination. 4x - 7y = 5 9x – 7y = –15

Anni [7]

Step-by-step explanation:

We have to get one positive 7y so,

4x-7y=5

9x-7y=-15

multiply anyone of the equation by -1. I choose the first one so,

4x+7y=-5

9x-7y=-15

we can now cancel the y's so the equation will be left with

-4x=-5

9x=-15

add the equations

5x=-20

divide by 5 n u get

x=-4

now plug in the x value in any one of the

equations

I choose the first one so,

4(-4)-7y=5

-16-7y=5

add 16 to both sides

-7y=21

divide by -7

y=-3.

finally check.

8 0

3 years ago

In the United States, we measure milk by the gallon. If we were using the metric system, what metric unit would we use

Lynna [10]

Answer:

The metric unit used for milk is the liter, L.

3 0

2 years ago

Solve: I kept trying but I can’t do it Thank you a lot for you help

scoray [572]

Answer:

5.25

Step-by-step explanation:

15/10=1.5

9*1.5=13.5

13.5-3=10.5

10.5/2=5.25

3 0

3 years ago

Read 2 more answers

Given that f(x) = x + 3 a) Find f(2)

Nataly_w [17]

${ \qquad\qquad\huge\underline{{\sf Answer}}}$

Here we go ~

In the above question, it is given that :

$\qquad \sf \boxed{ \sf f(x) = \frac{x + 3}{2} }$

A.) Find f(2) :

$\qquad \sf \dashrightarrow \: f(2) = \dfrac{2 + 3}{2}$

$\qquad \sf \dashrightarrow \: f(2) = \dfrac{5}{2}$

$\qquad \sf \dashrightarrow \: f(2) = 0.5$

B.) Find ${ \sf {f}^{-1}(x) }$ :

$\qquad \sf \dashrightarrow \: let \: y = f (x)$

so, we can write it as :

$\qquad \sf \dashrightarrow \: y = \dfrac{x + 3}{2}$

$\qquad \sf \dashrightarrow \: 2y = x + 3$

$\qquad \sf \dashrightarrow \: x = 2y - 3$

Now, put x = ${ \sf {f}^{-1}(x) }$ , and y = x and we will get our required inverse function ~

$\qquad \sf \dashrightarrow \: f {}^{ - 1}(x) = 2x- 3$

C.) Find ${ \sf {f}^{-1}(12) }$ :

$\qquad \sf \dashrightarrow \: f {}^{ - 1}(12) = 2(12)- 3$

$\qquad \sf \dashrightarrow \: f {}^{ - 1}(12) = 36- 3$

$\qquad \sf \dashrightarrow \: f {}^{ - 1}(12) = 33$

7 0

10 months ago

Find two positive real numbers whose product is a maximum. (enter your answers as a comma-separated list.) the sum of the first

Vladimir [108]

the sum of the first and three times the second is 54.

x + 3y = 54

3y = 54 - x

y= 54 /3 - x / 3 = 18 -x /3

xy = x (18 -x /3) = 18x - x^2 / 3 = - 1/3 x ^ 2 + 18 x


since a<0, it is concave downward, and the vertex is the maximum value. That vertex occurs at x= - b /2a = (-18) / (-2/3) = 12


The first number is x=12.
The second is y = 18 -x /3 = 18- 12/ 3 = 18 - 4 = 14

5 0

3 years ago