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Oksanka [162]
2 years ago
9

Simplify the following​

Mathematics
2 answers:
hoa [83]2 years ago
7 0

Step-by-step explanation:

please I solved the question in the diagram above

den301095 [7]2 years ago
5 0

Answer:

1) 11\sqrt{3}

2) 2\sqrt{2}

3) 20\sqrt{3}  + 15\sqrt{2}

4) 53 + 12\sqrt{10}

5) -2

6) 7\sqrt{2}  - 5\sqrt{3}

Step-by-step explanation:

1) 2\sqrt{12} + 3\sqrt{48} - \sqrt{75}

=(2 × 2\sqrt{3} )+ (3 × 4\sqrt{3}) - 5\sqrt{3}

= 4\sqrt{3} + 12\sqrt{3} - 5\sqrt{3}

= 11\sqrt{3}

2) 4\sqrt{8} -2\sqrt{98} + \sqrt{128}

= (4 × 2\sqrt{2}) - (2 × 7\sqrt{2}) + 8\sqrt{2}

= 8\sqrt{2} - 14\sqrt{2} +8\sqrt{2}

= 2\sqrt{2}

3) 5\sqrt{12\\} - 3\sqrt{18} + 4 \sqrt{72}  +2\sqrt{75}

= 5× 2\sqrt{3} - 3×3\sqrt{2} + 4×6\sqrt{2} + 2×5\sqrt{3}

= 10\sqrt{3} - 9\sqrt{2} +24\sqrt{2} +10\sqrt{3}

= 20\sqrt{3}  + 15\sqrt{2}

4) (2\sqrt{2}  + 3\sqrt{5} )^{2}

= 8 + 12\sqrt{10} + 45

= 53 + 12\sqrt{10}

5) (1+\sqrt{3} ) (1-\sqrt{3} )

= 1 - 3

= -2

6) (2\sqrt{6} -1) (\sqrt{3} -\sqrt{2}  )

= 2\sqrt{18}-2\sqrt{12}  -\sqrt{3}  +\sqrt{2}

= 2×3\sqrt{2} - 2×2\sqrt{3} - \sqrt{3} + \sqrt{2}

= 6\sqrt{2}  - 4\sqrt{3} -\sqrt{3} +\sqrt{2}

= 7\sqrt{2}  - 5\sqrt{3}

Hope the working out is clear and will help you. :)

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Step-by-step explanation:

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Read 2 more answers
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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