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3 years ago

What are the real and imaginary parts of the complex number?

Mathematics

1 answer:

Licemer1 [7]3 years ago

8 0

Answer:

The real part is 2

The imaginary part is -5

Step-by-step explanation:

A complex number consists of a real part and an imaginary part. For example given the complex number z = x+it

x is the real part of the complex number z i.e Re(z) = x

Imaginary part of the complex number z is y i.e Im(z) = y.

Note that the real part are on the x axis of a graph while the y axis is the imaginary axis attached to the complex notation i

Given the complex number 2-5i

Comparing 2-5i to x+iy

x= 2 and y = -5

The real part is 2 (value that is not attached to the complex notation)

The imaginary part is 5(value attached to the complex notation)

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Answer:

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3 years ago

What is -11-3(5-6)^2?

kolbaska11 [484]

Answer:

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2 years ago

Please help having trouble!

zmey [24]

Answer:

Option C is the correct choice that is $y$

Step-by-step explanation:

As this is a multiple choice question we will reduce the options and work on it with the given points $(0,-2),(-2,-3)$

Note:We know that $\leq ,\geq$ where there is $=$ sign associated with it have a straight line graph there is no breaking in the line.

And when there is simply we have a dashed line when we plot it on a graph.

So option B and D are discarded.

Now one-by one we will put the values $(x,y)\ (-2,3)$ to know which equation it satisfies.

If we put $y=3$ then $x=-2$ .

So working with option A.

$y$

Plugging the values.

$3$

And we know that $x$ must be equal to $-2$ so this is not the right answer.

We are left with only one choice that is C .

So option C is the correct option of the above inequality.

7 0

3 years ago

in base ten, the two-digit prime N is 45 more than the number formed by reversing the digits of N, find all the possible values

Drupady [299]

Let $N=10x+y$ , so that $x$ is the digits in the tens place and $y$ is the digit in the ones place. (Clearly $x\neq0$ .)

$10x+y=45+10y+x\implies 9x-9y=45\implies x-y=5$

There are five possible two digits integers that satisfy this relation:

$x=5,y=0\implies N=50$
$x=6,y=1\implies N=61$
$x=7,y=2\implies N=72$
$x=8,y=3\implies N=83$
$x=9,y=4\implies N=94$

But the first, third, and fifth candidates are even, so they are not prime. The remaining are prime, however, so $N=61$ or $N=83$ .

6 0

2 years ago

Any1 can help me wif q12

xenn [34]

I) HCF - use the smallest powers of each common factors
HCF (A,B) = 2^2 × 3^4 × 5^2

LCM - use the highest powers of each factors
LCM (A,B) = 2^4 × 3^6 × 5^2 × 7^2 × 11^16

ii) Add powers together.
A×B = 2^6 × 3^10 × 5^4 × 7^2 × 11^16
sqrt(A × B)
Divide powers by 2.
sqrt(A × B) = 2^3 × 3^5 × 5^2 × 7 × 11^8

iii) C = 3^7 × 5^2 × 7
Ck = (3^7 × 5^2 × 7) × k
B/c Ck should be a product that is a perfect cube, the powers of the products should be divisible by 3.
(3^7 × 5^2 × 7) × k = 3^9 × 5^3 × 7^3

k = (3^9 × 5^3 × 7^3) / (3^7 × 5^2 × 7)
k = 3^(9-7) × 5^(3-2) × 7^(3-1)
k = 3^2 × 5 × 7^2

8 0

3 years ago