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sasho [114]
3 years ago
14

Graph the image of the figure after a dilation with a scale factor of 1/2 centered at (−1, 3) . Use the polygon tool to graph th

e triangle by connecting all its vertices.

Mathematics
1 answer:
meriva3 years ago
6 0

Answer:

see attached diagram

Step-by-step explanation:

Given triangle has vertices at points B(-5,3), C(-5,7) and D(1,7). The center of dilation is point A(-1,3) and the factor of dilation is \dfrac{1}{2}. The dilation by a factor \dfrac{1}{2} decreases lengths AB, AC and AD twice, then image triangle vertices are midpoints of segments AB, AC and AD.

1. The midpoint E of the segment AB has coordinates

E\left(\dfrac{-5+(-1)}{2},\dfrac{3+3}{2}\right)\Rightarrow E(-3,3).

2. The midpoint F of the segment AC has coordinates

F\left(\dfrac{-5+(-1)}{2},\dfrac{7+3}{2}\right)\Rightarrow F(-3,5).

3. The midpoint G of the segment AD has coordinates

G\left(\dfrac{1+(-1)}{2},\dfrac{3+7}{2}\right)\Rightarrow G(0,5).

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tatiyna

Answer:

The solution to the system of equations is

\begin{gathered} x=\frac{179}{13} \\  \\ y=-\frac{279}{39} \\  \\ z=-\frac{48}{13} \end{gathered}

Explanation:

Giving the system of equations:

\begin{gathered} x+3y-z=-4\ldots\ldots\ldots\ldots\ldots\ldots..........\ldots\ldots\ldots\ldots.\ldots\text{.}\mathrm{}(1) \\ 2x-y+2z=13\ldots\ldots...\ldots\ldots\ldots\ldots..\ldots..\ldots\ldots\ldots\ldots\ldots.(2) \\ 3x-2y-z=-9\ldots\ldots\ldots.\ldots\ldots\ldots\ldots....\ldots\ldots.\ldots\ldots\ldots\text{.}\mathrm{}(3) \end{gathered}

To solve this, we need to first of all eliminate one variable from any two of the equations.

Subtracting (2) from twice of (1), we have:

5y-4z=-21\ldots\ldots\ldots\ldots\ldots.\ldots.\ldots..\ldots..\ldots\ldots.\ldots..\ldots\text{...}\mathrm{}(4)

Subtracting (3) from 3 times (1), we have

3y-5z=-3\ldots\ldots...\ldots\ldots..\ldots\ldots\ldots\ldots\ldots.\ldots\ldots\ldots\ldots\ldots..\ldots\ldots(5)

From (4) and (5), we can solve for y and z.

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\begin{gathered} 13z=-48 \\  \\ z=-\frac{48}{13} \end{gathered}

Using the value of z obtained in (5), we have

\begin{gathered} 3y-5(-\frac{48}{13})=-3 \\  \\ 3y+\frac{240}{13}=-3 \\  \\ 3y=-3-\frac{240}{13} \\  \\ 3y=-\frac{279}{13} \\  \\ y=-\frac{279}{39} \end{gathered}

Using the values obtained for y and z in (1), we have

\begin{gathered} x+3(-\frac{279}{39})-(-\frac{48}{13})=-4 \\  \\ x-\frac{279}{13}+\frac{48}{13}=-4 \\  \\ x-\frac{231}{13}=-4 \\  \\ x=-4+\frac{231}{13} \\  \\ x=\frac{179}{13} \end{gathered}

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Answer:

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Step-by-step explanation:

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