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3 years ago

What is the complex conjugate of 13-21i?

2 answers:

4 0

The conjugate of "anything" is pretty much the same binomial, with a different sign.

ducks + geese conjugate -> ducks - geese

cheese - spaghetti conjugate -> cheese + spaghetti

and so on and so forth.

well, for this one, yeap, you guessed it

13 - 21i conjugate -> 13 + 21i.

irga5000 [103]3 years ago

3 0

Answer:

To find a complex conjugate, simply change the sign of the imaginary part (the part with the i). This means that it either goes from positive to negative or from negative to positive.

As a general rule, the complex conjugate of a+bi is a−bi.

Therefore, the complex conjugate of 3−2i is 3+2i.

Hope this helped!!... :D

Please correct me if I'm wrong!!.. :3

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I need your help again guys!

solmaris [256]

ANSWER

$x = 13$

EXPLANATION

Use the Pythagoras Theorem to obtain,

${x}^{2} = {12}^{2} + {5}^{2}$

This implies that,

${x}^{2} = 144 + 25$

${x}^{2} = 169$

Take the positive square root of both sides to get,

$x = \sqrt{169}$

$x = 13$

8 0

3 years ago

Read 2 more answers

Write and equati9n of a l8ne that passes through (1,-2) and is perpendicular to -4x+7y=21

yarga [219]

Given:

A line passes through (1,-2) and is perpendicular to $-4x+7y=21$ .

To find:

The equation of that line.

Solution:

We have, equation of perpendicular line.

$-4x+7y=21$

Slope of this line is

$m_1=-\dfrac{\text{Coefficient of x}}{\text{Coefficient of y}}$

$m_1=-\dfrac{-4}{7}$

$m_1=\dfrac{4}{7}$

Product of slope of two perpendicular lines is -1.

$m_1\times m_2=-1$

$\dfrac{4}{7}\times m_2=-1$

$m_2=-\dfrac{7}{4}$

Now, slope of required line is $-\dfrac{7}{4}$ and it passes through (1,-2). So, the equation of line is

$y-y_1=m(x-x_1)$

where, m is slope.

$y-(-2)=-\dfrac{7}{4}(x-1)$

$4(y+2)=-7(x-1)$

$4y+8=-7x+7$

$4y+7x=7-8$

$7x+4y=-1$

Therefore, the equation of required line is $7x+4y=-1$ .

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By definition,
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Answer:

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