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SashulF [63]
3 years ago
9

I need help solving |5+2j|=9 please

Mathematics
1 answer:
V125BC [204]3 years ago
3 0

5 + 2j = 9 \\ j = 2
5 + 2j =  - 9 \\ j =  - 7
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Please answer the following 4 questions
Lilit [14]
1. The correct answer should be A
2. The answer should be C
3. I think the answer is D
4. The answer should be D


Hope this helps :)
7 0
3 years ago
Read 2 more answers
Solve for k 5/8(3k+1)-1/4(7k+2)=1
lesya [120]

Answer:

<h2>k = 7</h2>

Step-by-step explanation:

\dfrac{5}{8}(3k+1)-\dfrac{1}{4}(7k+2)=1\qquad\text{multiply both sides by 8}\\\\8\!\!\!\!\diagup^1\cdot\dfrac{5}{8\!\!\!\!\diagup_1}(3k+1)-8\!\!\!\!\diagup^2\cdot\dfrac{1}{4\!\!\!\!\diagup_1}(7k+2)=8\cdot1\\\\5(3k+1)-2(7k+2)=8\qquad\text{use the distributive property}\ a(b+c)=ab+ac\\\\(5)(3k)+(5)(1)+(-2)(7k)+(-2)(2)=8\\\\15k+5-14k-4=8\qquad\text{combine like terms}\\\\(15k-14k)+(5-4)=8\\\\k+1=8\qquad\text{subtract 1 from both sides}\\\\k=7

3 0
3 years ago
I will give u brianliest
lianna [129]

Answer:

10.63

Step-by-step explanation:

8 0
3 years ago
Consider the two data sets below:
alina1380 [7]

Answer:

<u><em>Option c) The data sets will have the same values of their interquartile range.</em></u>

<u><em></em></u>

Explanation:

<u>1. The values are in order: </u>they are in increasing oder, from lowest to highest value.

<u>2. Calculate the interquartile range.</u>

<em />

<em>Interquartile range</em>, IQR, is the third quartile, Q3, less the first quartile Q1:

  • IQR = Q3 - Q1

To find the first and the third quartile, first find the median:

<u>Data Set 1</u>: 19, 25, 35, 38, 41, 49, 50, 52, 59

             [19, 25, 35, 38],  41,  [49, 50, 52, 59]

                                         ↑

                                     median = 41

   

<u>Data Set 2</u>: 19, 25, 35, 38, 41, 49, 50, 52, 99

             [19, 25, 35, 38] , 41,  [49, 50, 52, 99]

                                         ↑

                                      median = 41

Now find the median of each subset: the values below the median and the values above the median.

Data set 1: <u>First quartile</u>

                [19, 25, 35, 38],

                            ↑

                           Q1 = [25 + 35] / 2 = 30

                   <u>Third quartile</u>

                   [49, 50, 52, 59]

                                ↑

                                Q3 = [50 + 52] / 2 = 51

                     IQR = Q3 - Q1 = 51 - 30 = 21

Data set 2: <u> First quartile</u>

                   [19, 25, 35, 38]

                               ↑

                               Q1 = [25 + 35] / 2= 30

                  <u>Third quartile</u>

                   [49, 50, 52, 99]

                                ↑

                                Q3 = [52 + 50]/2 = 51

                   IQR = 51 - 30 = 21

Thus, it is shown that the data sets have will have the same values for the interquartile range: IQR = 21. (option c)

This happens because replacing one extreme value (in this case the maximum value) by other extreme value does not affect the median.

<em>An outlier will change the range</em> because the range is the maximum value less the minimum value.

5 0
3 years ago
^^^^<br> please help :)))
ikadub [295]

2x-1.1/4+x/3=2

2x-1/4+x/3=2

2x-1/4+x/3+1/4=2+1/4

2x+x/3=9/4

3(2x+x/3)=3.94

7x=27/4

7x/7=27/4/7

x=27/8

<h2>PLEASE MARK ME AS BRAINLIEST IF U LIKE MY ANSWER AND SORRY FOR GIVING THE ANSWER LATE BECAUSE I'VE GIVEN U ANSWER FROM MY LAPTOP PLEASE TELL THAT IT'S CORRECT OR NOT</h2>

7 0
3 years ago
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