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2 years ago

What are the steps on how to copy an angle? (Geometry)

Mathematics

1 answer:

Juli2301 [7.4K]2 years ago

4 0

Step-by-step explanation:

Make a point P that will be the vertex of the new angle.

From P, draw a ray PQ. ...

Place the compasses on point A, set to any convenient width.

Draw an arc across both sides of the angle, creating the points J and K as shown.

Hope this helps :)

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Bhavik bought 3 liters of milk and 5 loaves of bread for a total of $11. A month later, he bought 4 liters of milk

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Answer:

Step-by-step explanation:

A liter of milk cost $.75

A loaf of bread cost $1.75

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3 years ago

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exis [7]

Answer:

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Step-by-step explanation:

Divided both sides by the numeric factor on the left side, then solve.

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3 years ago

How many miles are in 400 kilometers? How many miles are in 400 kilometers? Follow 3 answers 3 Report Abuse Are you sure you wan

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Hello there. :)

<span>How many miles are in 400 kilometers

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2 years ago

What is 2+2736=344×2484848÷484848=

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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:

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

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

$C^{-1}b=\frac{1}{17} \left[\begin{array}{ccc}5&-18&12\\2&3&-2\\-1&7&1\end{array}\right]\left[\begin{array}{c}10&1&3\end{array}\right]=\left[\begin{array}{c}4&1&0\end{array}\right]$

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3 years ago