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MatroZZZ [7]
3 years ago
10

Find the value of k of this problem

Mathematics
1 answer:
Yuliya22 [10]3 years ago
4 0
You already have the answer.
You might be interested in
Drag the point A to the location indicated in each scenario to complete each statement.
Art [367]

The graph from which the position of the point <em>A</em> can determined following

the multiplication with a scalar is attached.

Responses:

  • If <em>A</em> is in quadrant I and is multiplied by a negative scalar, <em>c</em>, then c·A is in <u>quadrant III</u>
  • If A is in quadrant II and is multiplied by a positive scalar, <em>c</em>, then c·A is in <u>quadrant II</u>
  • If <em>A</em> is in quadrant II and is multiplied by a negative scalar, <em>c</em>, then c·A is in <u>quadrant IV</u>
  • If <em>A</em> is in quadrant III and is multiplied by a negative scalar, <em>c</em>, then c·A is in <u>quadrant I</u>

<h3>Methods by which the above responses are obtained</h3>

Background information;

The question relates to the coordinate system with the abscissa represent the real number and the ordinate representing the imaginary number.

Solution:

If A is in quadrant I; A = a + b·i

When multiplied by a negative scalar, <em>c</em>, we get;

c·A = c·a + c·b·i

Therefore;

c·a is negative

c·b is negative

  • c·A = c·a + c·b·i is in the <u>quadrant III</u> (third quadrant)

If A is quadrant II, we have;

A = -a + b·i

When multiplied by a positive scalar <em>c</em>, we have;

c·A = c·(-a) + c·b·i = -c·a + c·b·i

-c·a is negative

c·b·i is positive

Therefore;

  • c·A = -c·a + c·b·i is in <u>quadrant II</u>

Multiplying <em>A</em> by negative scalar if <em>A</em> is in quadrant II, we have;

c·A = -c·a + c·b·i

-c·a is positive

c·b·i is negative

Therefore;

c·A = -c·a + c·b·i is in <u>quadrant IV</u>

If A is in quadrant III, we have;

A = a + b·i

a is negative

b is negative

Multiplying <em>A</em> with a negative scalar <em>c</em> gives;

c·A = c·a + c·b·i

c·a is positive

c·b  is positive

Therefore;

  • c·A = c·a + c·b·i is in<u> quadrant I</u>

Learn more about real and imaginary numbers here;

brainly.com/question/5082885

brainly.com/question/13573157

4 0
3 years ago
Which relation is not a function?
kvv77 [185]

Answer:

B.

Step-by-step explanation:

You can't have the same X values for two different Y values

in this case 19 goes with 11 and 10 in the pairs (19,11) and (19,10)

6 0
2 years ago
QUESTION 69
bazaltina [42]
The line is parallel to x-axis

the correct answer -

y = –4
8 0
3 years ago
Help ? I don’t know how to do it
ryzh [129]
I think 60°
these angles are very confusing sorry!
4 0
3 years ago
PLEASE HELP
cricket20 [7]
You can use the cosine rule for this - you might have to change some of the letters around for it to make sense though. Hope this helps!

4 0
3 years ago
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