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earnstyle [38]
3 years ago
11

-2(6+s) Greater than or equal to -15 - 2s

Mathematics
1 answer:
Eddi Din [679]3 years ago
7 0

Answer:

I'm leaning mostly towards C Because there is a solution, None of them are real numbers and if they were, S would = 5 Especially positive S

~ Zachary

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Which of the following matrices is the solution matrix for the given system of equations? x + 5y = 11 x - y = 5
givi [52]
<h2>The required solution is x = 6 and y = 11 </h2>

Step-by-step explanation:

Given system of equations are

x+5y = 11 and x-y =5

A= \left[\begin{array}{cc}1&5\\1&-1\end{array}\right]                            X=\left[\begin{array}{c}x\\y\end{array}\right]

and          B= \left[\begin{array}{c}11\\5\end{array}\right]

∴AX=B

adj A = \left[\begin{array}{cc}{-1}&{-5}\\{-1}&1\end{array}\right]

A= \left|\begin{array}{cc}1&5\\1&-1\end{array}\right|=-6

∴A^{-1} =\frac{adj A}{|A|}

So,A^{-1} =\frac{ \left[\begin{array}{cc}{-1}&{-5}\\{-1}&1\end{array}\right]}{-6}

A^{-1} ={ \left[\begin{array}{c \c}  {{\frac{1}{6} }}&{\frac{5}{6}}\ \\  {{\frac{1}{6} }}&{\frac{-1}{6}} \end{array}\right]}

X =A^{-1}\times B

⇒\left[\begin{array}{c}x\\y\end{array}\right] ={ \left[\begin{array}{c \c}  {{\frac{1}{6} }}&{\frac{5}{6}}\ \\  {{\frac{1}{6} }}&{\frac{-1}{6}} \end{array}\right]} \times \left[\begin{array}{c}11\\5\end{array}\right]

⇒\left[\begin{array}{c}x\\y\end{array}\right] ={ \left[\begin{array}{c}  {6}\\  {11} \end{array}\right]}

∴ x= 6 and y = 11

The required solution is x = 6 and y = 11

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Need some help, please. :D
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Answer:   (P.s its very hard to write special characters here, i hope this helped though)

((−6d0)(g−3))(h2)

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=

−6d5g2h2

g3h5m3

=

−6d5

gh3m3

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Answer:

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