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dalvyx [7]
3 years ago
12

PLEASE HELP

Mathematics
1 answer:
77julia77 [94]3 years ago
4 0

Answer:

The answer to your question is:

 p₁ = - 2 + \sqrt{-2}  

p₂ = -2 - \sqrt{-2}

Step-by-step explanation:

                                            p² + 4p = -6

                                           p² + 4p + (2)² = - 6 + (2)²

                                          (p + 2)² = - 2

                                          p + 2 = ± \sqrt{-2}

                                          p = -2 ± \sqrt{-2}

           p₁ = - 2 + \sqrt{-2}               p₂ = -2 - \sqrt{-2}

                                       

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[10] In the following given system, determine a matrix A and vector b so that the system can be represented as a matrix equation
irina1246 [14]

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A=\left[\begin{array}{cccc}1&1&2&2\\-7&-3&5&-8\\4&1&1&1\\3&7&-1&1\end{array}\right]

And the vector B is formed with the solution of each equation of the system:b=\left[\begin{array}{c}3\\-3\\6\\1\end{array}\right]

To apply the Cramer's rule, take the matrix A and replace the column assigned to the variable that you need to solve with the vector b, in this case, that would be the second column. This new matrix is going to be called A_{2}.

A_{2}=\left[\begin{array}{cccc}1&3&2&2\\-7&-3&5&-8\\4&6&1&1\\3&1&-1&1\end{array}\right]

The value of y using Cramer's rule is:

y=\frac{det(A_{2}) }{det(A)}

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y==\frac{\left|\begin{array}{cccc}1&3&2&2\\-7&-3&5&-8\\4&6&1&1\\3&1&-1&1\end{array}\right|}{\left|\begin{array}{cccc}1&1&2&2\\-7&-3&5&-8\\4&1&1&1\\3&7&-1&1\end{array}\right|} =\frac{158}{-579}

y=-\frac{158}{579}

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3 years ago
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