Two transformations that can be used to show that polygon ABCDE is congruent to polygon GFJIH : <u>rotation -90° and translation (0,2)</u>
<h3>Further explanation</h3>
There are several forms of transformation, including translation, dilation, rotation or reflection
There are 2 shapes of geometry:
Similar figures:
the same shape but different in size,
Congruent figures:
the same size and the same shape
Both polygons will appear congruent if, after being transformed, the two polygons will map exactly to each other
A sequence of two transformations e can do is :
1. rotation -90°
Let point H(4,1) from polygon GFJIH , we rotate -90°
![\left[\begin{array}{ccc}x'\\y'\end{array}\right]=\left[\begin{array}{ccc}0&1\\-1&0\end{array}\right]\left[\begin{array}{ccc}4-6\\1+1\end{array}\right]+\left[\begin{array}{ccc}6\\-1\end{array}\right]\\\\=\left[\begin{array}{ccc}0&1\\-1&0\end{array}\right]\left[\begin{array}{ccc}-2\\2\end{array}\right]+\left[\begin{array}{ccc}6\\-1\end{array}\right]\\\\=\left[\begin{array}{ccc}8\\1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx%27%5C%5Cy%27%5Cend%7Barray%7D%5Cright%5D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D0%261%5C%5C-1%260%5Cend%7Barray%7D%5Cright%5D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D4-6%5C%5C1%2B1%5Cend%7Barray%7D%5Cright%5D%2B%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D6%5C%5C-1%5Cend%7Barray%7D%5Cright%5D%5C%5C%5C%5C%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D0%261%5C%5C-1%260%5Cend%7Barray%7D%5Cright%5D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-2%5C%5C2%5Cend%7Barray%7D%5Cright%5D%2B%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D6%5C%5C-1%5Cend%7Barray%7D%5Cright%5D%5C%5C%5C%5C%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D8%5C%5C1%5Cend%7Barray%7D%5Cright%5D)
2. translation (0,2)
![\left[\begin{array}{ccc}8\\1\end{array}\right]+\left[\begin{array}{ccc}0\\2\end{array}\right]=\left[\begin{array}{ccc}8\\3\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D8%5C%5C1%5Cend%7Barray%7D%5Cright%5D%2B%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D0%5C%5C2%5Cend%7Barray%7D%5Cright%5D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D8%5C%5C3%5Cend%7Barray%7D%5Cright%5D)
<h3 /><h3>Learn more</h3>
brainly.com/question/13715226
Keywords: transformation, translation, rotation
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Taking point (0,-2) of graph one and replacing in the equation;
2x+3y=6
2(0)+3(-2)=6
-6=6
So graph isn't the required graph.
Taking point (0,2) of graph 2;
2x+3y=6
2(0)+3(2)=6
6=6
So, graph 2 is the required graph.
The answer is 2.6 because 2.6+2=4.6
Replace x with y and solve for y~
y=2*
x=2^y
log x = y log 2
y=log x / log 2
Hope this helps and leave a brainliest to help me reach expert ;)
Answer:
c. m∠1 + m∠6 = m∠4 + m∠6
Step-by-step explanation:
Given: The lines l and m are parallel lines.
The parallel lines cut two transverse lines. Here we can use the properties of transverse and find the incorrect statements.
a. m∠1 + m∠2 = m∠3 + m∠4
Here m∠1 and m∠2 are supplementary angles add upto 180 degrees.
m∠3 and m∠4 are supplementary angles add upto 180 degrees.
Therefore, the statement is true.
b. m∠1 + m∠5 = m∠3 + m∠4
m∠1 + m∠5 = 180 same side of the adjacent angles.
m∠3 + m∠4 = 180, supplementary angles add upto 180 degrees.
Therefore, the statement is true.
Now let's check c.
m∠1 + m∠6 = m∠4 + m∠6
We can cancel out m∠6, we get
m∠1 = m∠4 which is not true
Now let's check d.
m∠3 + m∠4 = m∠7 + m∠4
We can cancel out m∠4, we get
m∠3 = m∠7, alternative interior angles are equal.
It is true.
Therefore, answer is c. m∠1 + m∠6 = m∠4 + m∠6
