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What is the product of complex conjugates? (1 point)

Leno4ka [110]

Answer:

A. the product of complex conjugates is a difference of two squares and is always a real number

Step-by-step explanation:

Given a complex number z1 = x+iy where x is the real part and y is the imaginary part.

The conjugate of a complex number is the negative form of the complex number z1 above i.e z2= x-iy (The conjugate is gotten by mere changing of the plus sign in between the terms to a minus sign.

Taking the product of the complex number and its conjugate will give;

z1z2 = (x+iy)(x-iy)

z1z2 = x(x) - ixy + ixy - i²y²

z1z2 = x² - i²y²

.since i² = -1

z1z2 = x²-(-1)y²

z1z2 = x²+y²

The product gave a real function x²+y² since there is no presence of complex number 'i'

It can also be seen that the product of the complex numbers z1z2 is like taking the difference of two square. An example of difference of two square is that of two values a and b which is (a+b)(a-b).

From the above solution, it can be concluded that the product of complex conjugates is a difference of two squares and is always a real number.

3 0

3 years ago

A student raises her average grade from a 75 to a 90. What is the percent of the increase in the students average grade

tensa zangetsu [6.8K]

Since grades are out of 100, you can just subtract 75 from 90 which is 15. the percent of increase in the student's grade is 15%

5 0

3 years ago

Below are the jersey numbers of 11 players randomly selected from a football team. Find the range, variance, and standard devia

jeyben [28]

Answer:

The range is 77.

The variance is 739.8.

The standard deviation is 27.2.

Step-by-step explanation:

The data provided, in ascending order is as follows:

S = {11 , 14 , 23 , 31 , 40 , 60 , 63 , 68 , 75 , 77 , 88}

The range of a data set is the difference between the maximum and minimum values.

From the above data set we know,

Maximum = 88

Minimum = 11

Compute the range as follows:

Range = Maximum - Minimum

= 88 - 11

= 77

The range is 77.

Compute the mean of the data as follows:

$\bar x=\frac{1}{n}\sum x_{i}=\frac{1}{11}\times [11+14+23+...+88]=50$

Compute the variance as follows:

$\sigma^{2}=\frac{1}{n-1}\sum (x_{i}-\bar x})^{2}$

$=\frac{1}{11-1}\times [(11-50)^{2}+(14-50)^{2}+...+(88-50)^{2}]\\\\=\frac{1}{10}\times 7398\\\\=739.8$

The variance is 739.8.

Compute the standard deviation as follows:

$\sigma=\sqrt{\frac{1}{n-1}\sum (x_{i}-\bar x})^{2}}=\sqrt{739.8}=27.199265\approx 27.2$

The standard deviation is 27.2.

4 0

2 years ago

The graph of a cosine function is shown. Which two points on the midline of the function are separated by a distance of one peri

Ray Of Light [21]

Answer: (pi/8, 1) and (5*pi/8, 1)

Step-by-step explanation:

First, the midline is a horizontal line that cuts the graph in two halves.

In this case, we can see that the maximum of the function is y = 2.5, and the minimum is y = -0.5

The difference is:

Diff = (2.5 - (-0.5)) = 3

If we subtract half of the difference (3/2 = 1.5) to the maximum, we get the midline.

Then:

midline: y = 2.5 - (3/2) = 1

the midline is the line y = 1

Now, a period T is such that:

Sin(x) = Sin(x + T)

Two points that are in the midline and that are separated by a period will be two points on the line y = 1, and that are in corresponding parts of the graph.

Here the two points are, counting from left to right, the third and the 11th.

(both of them are in the line y = -1 and in the negative slope part).

The points are:

(pi/8, 1) and (5*pi/8, 1)

6 0

3 years ago

Read 2 more answers

I-Ready

Slav-nsk [51]

Answer:

32

Step-by-step explanation:

4 0

2 years ago