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leonid [27]
3 years ago
15

Two roots of a 3-degree polynomial equation are 5 and -5. Which of the following cannot be the third root of the equation?

Mathematics
2 answers:
cricket20 [7]3 years ago
7 0

Answer:

5i and -5i

Step-by-step explanation:

Both of these would have to be in the equation for either of them to work, and as the problem already told us that there is only a 3rd root, neither will be our answer.

Have a great day, and feel free to comment with questions!

:)

Pavel [41]3 years ago
6 0

Answer:

5i and -5i

Step-by-step explanation:

Imaginary roots come in pairs, so the third root must be a real number.  Therefore, neither 5i nor -5i can be the third root.

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Step-by-step explanation:

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7(x+5)=-7(x+5) solving equations with variable on both sides​
sergij07 [2.7K]

Answer:

Step-by-step explanation:

7x+5=7x+5

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3 years ago
Graph the solution set of this inequality 3x-2y <_ 12
nadezda [96]

The graph of the solution set for the inequality can be seen below.

<h3>How to graph the solution set?</h3>

Here we have the inequality:

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Now, we just need to graph the line y = (3/2)x + 6 with a dashed line (because the points on the line are not solutions).

And then we need to shade the region above the line.

The graph of the solution set can be seen below.

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8 0
2 years ago
A past survey of students taking a standardized test revealed that ​% of the students were planning on studying engineering in c
Angelina_Jolie [31]

Complete Question

The  complete question is shown on the first uploaded image  

Answer:

The  95% confidence interval is  -0.00870

Step-by-step explanation:

From the question we are told that

     The first sample  size  is  n_1  =  1068000

     The first proportion  \r p_1 = 0.084

     The second  sample size is  n_2  =  1476000

     The  second  proportion is  \r p_2 =  0.092

Given that the confidence level is  95%  then the level of significance is mathematically represented as

      \alpha =  (100 - 95)\%

     \alpha =  0.05

From the normal distribution table  we obtain the critical value of  \frac{ \alpha }{2}  the value is  

      Z_{\frac{\alpha }{2} } =z_c=  1.96

Now using the formula from the question to construct the 95% confidence interval we have  

  (\r p_1 - \r p_2  )- z_c \sqrt{ \frac{\r p_1 \r q_1 }{n_1} + \frac{\r p_2 \r q_2 }{n_2} }

Here \r q_1 =  1 - \r p_1

  =>   \r q_1 =  1 - 0.084

 =>    \r q =  0.916

and  

   \r q_2 =  1 - \r p_2

 =>   \r q_2 =  1 - 0.092

=>   \r q_2 = 0.908

So  

 (0.084 - 0.092 )- (1.96)*  \sqrt{ \frac{0.092* 0.916 }{1068000} + \frac{0.084* 0.908 }{1476000} }

  -0.00870

 

3 0
3 years ago
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