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2 years ago

PLS HELP WILL MARK YOU BRAINLIEST! NO FAKE ANSWERS!

Mathematics

2 answers:

Nataly_w [17]2 years ago

8 0

Answer:

see attached diagram

Step-by-step explanation:

<u>Angle Bisector</u>

The angle bisector bisects the angle into two equal angles.

Measure the angle EDF (I've measured this as 115°)

Divide this by 2: 115 ÷ 2 = 57.5°

Measure an angle of 57.5° from point D on the line DF and draw a line.

<u>Parallel lines</u>

Measure two lines of equal length from points X and Y at an angle of 90° to the line XY. Join the points for the parallel line.

The angle bisector and parallel line are shown in red on the attached diagram.

ohaa [14]2 years ago

6 0

Answer:

1)An angle bisector is basically the division of an angle into two equal parts, so you should just draw a line in the middle after you found the angles using the compass

2) parallel lines is a pair of lines that never intersect, and are always the same distance apart. you should use the compass to draw a 90-degree angle at the end of the line XY, and use the straightedge to draw a line that is parallel to the line XY.

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Express the system as AX = B; then solve using matrix inverses

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Answer with explanation:

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$A=\left[\begin{array}{cc}3&-5\\-1&2\end{array}\right] ,\\\\ X=\left[\begin{array}{c}x&y\end{array}\right],\\\\B=\left[\begin{array}{c}2&0\end{array}\right]$

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$Adj.A=\left[\begin{array}{cc}2&1\\5&3\end{array}\right] ,\\\\ |A|=6-5\\\\|A|=1\\\\\left[\begin{array}{c}x&y\end{array}\right]=\left[\begin{array}{cc}2&5\\1&3\end{array}\right] \times \left[\begin{array}{c}2&0\end{array}\right]\\\\x=4, y=2$

The given equations are

x+y-z=2

x+z=7

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⇒The matrix in the form of , AX=B, is

$A=\left[\begin{array}{ccc}1&1&-1\\1&0&1\\2&1&1\end{array}\right]\\\\ X=\left[\begin{array}{ccc}x\\y\\z\end{array}\right]\\\\B= \left[\begin{array}{ccc}2\\7\\13\end{array}\right]\\\\\rightarrow X=A^{-1}B\\\\\rightarrow X=\frac{Adj.A}{|A|}\times B\\\\a_{11}=-1,a_{12}=1,a_{13}=1,a_{21}=-2,a_{22}=3,a_{23}=1,a_{31}=1,a_{32}=-2,a_{33}=-1\\\\|A|=1\times(0-1)-1\times(1-2)-1\times(1-0)\\\\=-1+1-1\\\\|A|=-1\\\\Adj.A=\left[\begin{array}{ccc}-1&-2&1\\1&3&-2\\1&1&-1\end{array}\right]$

$\frac{Adj.A}{|A|}=\left[\begin{array}{ccc}1&2&-1\\-1&-3&2\\-1&-1&1\end{array}\right]\\\\X=A^{-1}B\\\\\left[\begin{array}{ccc}x\\y\\z\end{array}\right]=\left[\begin{array}{ccc}1&2&-1\\-1&-3&2\\-1&-1&1\end{array}\right]\times\left[\begin{array}{ccc}2\\7\\13\end{array}\right]\\\\x=3,y=3,z=4$

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