Help plz help plz fast

Draw and label the final image of △ABC after the given sequence of transformations. Reflect △ABC over the y−axis and then transl

arlik [135]

Answer: The correct option is A.

Explanation:

From the given figure it is noticed that each box in the graph represents the length of 2 units. The coordinates of point A,B and C are (-1,1),(-3,3) and (-2,4) respectively.

If a point reflects over the y-axis then,

$(x,y)\rightarrow(-x,y)$

So after reflecting coordinates of point C become,

$(-2,4)\rightarrow(2,4)$

If a point translated by (a,b), then

$(x,y)\rightarrow(x+a,y+b)$

So, after reflection over y-axis followed by translation by (2,-2), the coordinate of C are,

$(2,4)\rightarrow(2+2,4-2)\\(2,4)\rightarrow(4,2)$

Similarly the image of point A and B are (3,-1) and (5,1) respectively.

The image of C after reflection over y-axis followed by translation by (2,-2) is (4,2), which is shown in only graph A, Therefore A is the correct option.

8 0

3 years ago

Attempt

Pachacha [2.7K]

Answer:

Yes, the triangles are similar by SAS

Step-by-step explanation:

Looking at the picture, we see that there are two triangles with side lengths of different sides. We can deduce that the scale factor is 1 2/3, because 8 divided by 4.8 is 1 2/3, and 10 divided by 6 is the same. Now that we have clarified that the triangles share the same scale factor, we notice that the angle is also the same, as mentioned in the picture. This leads us to say that the triangles are similar by the SAS similarity theorem (Side, Angle, Side). I hope this helped and please don't hesitate to reach out with more questions!

4 0

3 years ago

How to find matrics M

makkiz [27]

$\left[\begin{array}{ccc}9&7\\-4&2\end{array}\right] -4M=\left[\begin{array}{ccc}-3&3\\4&-18\end{array}\right]\qquad(*) \\\\M=\left[\begin{array}{ccc}a&b\\c&d\end{array}\right]\to4M=\left[\begin{array}{ccc}4a&4b\\4c&4d\end{array}\right]$

$(*)\qquad\left[\begin{array}{ccc}9&7\\-4&2\end{array}\right] -\left[\begin{array}{ccc}4a&4b\\4c&4d\end{array}\right]=\left[\begin{array}{ccc}-3&3\\4&-18\end{array}\right]\\\\\left[\begin{array}{ccc}9-4a&7-4b\\-4-4c&2-4d\end{array}\right]=\left[\begin{array}{ccc}-3&3\\4&-18\end{array}\right]\\.\qquad\qquad\qquad\qquad\ \ \ \Downarrow\\(_{11})\qquad9-4a=-3\ \ \ |-9\\-4a=-12\ \ \ \ |:(-4)\\a=3\\\\(_{12})\qquad7-4b=3\ \ \ \ |-7\\-4b=-4\ \ \ \ |:(-4)\\b=1$

$(_{21})\qquad-4-4c=4\ \ \ \ |+4\\-4c=8\ \ \ \ |:(-4)\\c=-2\\\\(_{22})\qquad2-4d=-18\ \ \ \ |-2\\-4d=-20\ \ \ \ \ |:(-4)\\d=5\\\\Answer:\ A.\ M=\left[\begin{array}{ccc}3&1\\-2&5\end{array}\right]$