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Tasya [4]
3 years ago
9

Rectangle LMNP is transformed to rectangle L'M'N'P' as shown on the graph below.

Mathematics
1 answer:
olganol [36]3 years ago
3 0

9514 1404 393

Answer:

  2

Step-by-step explanation:

Each image point is twice as far from the origin as its preimage point. Each image segment is twice as long as its preimage segment. (LM=2, L'M'=4, for example)

The scale factor is 2.

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I need help on how to solve the problem and the correct answer :)
lakkis [162]

This is a system of linear equations. First, you can add the two equations together to eliminate the y so that you can solve for x:

-5x + -7x = 0 + -96

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So, x = 8 and y = 5.

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4 years ago
Question 10
EleoNora [17]

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a.(-b) ( third choice)

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-a+b - there is no plus sign for -a.b.

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3 years ago
In a right triangle, the tangent of an angle is equal to the length of the leg opposite the angle divided by the length of the l
timurjin [86]

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3 years ago
The diameter of a soda can is 2.6 inches. What is the total length of the diameters of 5
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3 years ago
Please help if you can pic attatched
MA_775_DIABLO [31]

Answer:

Step-by-step explanation:

CI=\left[\begin{array}{ccc}1&6&0\\0&1&2\\1&-1&3\end{array}\right] \left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right]

Subtract row 3 from row 1:

\left[\begin{array}{ccc}1&6&0\\0&1&2\\0&7&-3\end{array}\right] \left[\begin{array}{ccc}1&0&0\\0&1&0\\1&0&-1\end{array}\right]

Subtract row 3 from 7 times row 2:

\left[\begin{array}{ccc}1&6&0\\0&1&2\\0&0&17\end{array}\right] \left[\begin{array}{ccc}1&0&0\\0&1&0\\-1&7&1\end{array}\right]

Divide row 3 by 17:

\left[\begin{array}{ccc}1&6&0\\0&1&2\\0&0&1\end{array}\right] \left[\begin{array}{ccc}1&0&0\\0&1&0\\\frac{-1}{17} &\frac{7}{17} &\frac{1}{17} \end{array}\right]

Subtract 2 of row 3 from row 2:

\left[\begin{array}{ccc}1&6&0\\0&1&0\\0&0&1\end{array}\right] \left[\begin{array}{ccc}1&0&0\\\frac{2}{17} &\frac{3}{17} &\frac{-2}{17} \\\frac{-1}{17} &\frac{7}{17} &\frac{1}{17} \end{array}\right]

Subtract 6 of row 2 from row 1:

\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right] \left[\begin{array}{ccc}\frac{5}{17}&\frac{-18}{17}&\frac{12}{17}\\\frac{2}{17} &\frac{3}{17} &\frac{-2}{17} \\\frac{-1}{17} &\frac{7}{17} &\frac{1}{17} \end{array}\right]

C^{-1}=\frac{1}{17} \left[\begin{array}{ccc}5&-18&12\\2&3&-2\\-1&7&1\end{array}\right]

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3 0
3 years ago
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