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hodyreva [135]
3 years ago
10

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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ehidna [41]

Answer:

y=2x+5 is the answer ALSO CAN WE TALK ABOUT THE FACT THAT MOO975 IS ON THREE QUESTIONS AT THE SAME TIME

Step-by-step explanation:

6 0
3 years ago
What is -11-3(5-6)^2?
kolbaska11 [484]

Answer:

The answer is -14 :))

5 0
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
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