Answer:
The coordinates of the vertices of △JKL△JKL are J(1,4)J(1,4), K(6,4)K(6,4), and L(1,1)L(1,1). The coordinates of the vertices of
△J′K′L′△J′K′L′ are J′(0,−4)J′(0,−4), K′(−5,−4)K′(−5,−4), and L′(0,−1)L′(0,−1). What is the sequence of transformations that maps △JKL△JKL to △J′K′L′△J′K′L′ ?
2 answers:
Solution: The ΔJKL reflect along the x-axis, then reflects along the y-axis after that translate 1 unit right to transform ΔJ'K'L'.
Explanation:
The vertices of ΔJKL lies in first quadrant because the coordinated of the vertices are positive. It means the vertices are in the form of (x,y).
If the ΔJKL reflects along the x-axis then the vertices are in the form of (x,-y). Now the ΔJKL reflects along the y-axis then the vertices are in the form of (-x,-y).
The given coordinates of the the vertices after transformation is J'(0,-4), K'(-5,-4), L'(0,-1). But after reflection along x and y axis the vertices of ΔJKL are J'(-1,-4), K'(-6,-4), L'(-1,-1). Since the y coordinates are same so we have to tralanate horizontally. Since the difference between x coordinates is positive 1, so we have translate the triangle 1 unit right.
Therefore, the ΔJKL reflect along the x-axis, then reflects along the y-axis after that translate 1 unit right to transform ΔJ'K'L'.
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Answer: Order will be F,D,C and Fourth Option is correct and x = 10
Step-by-step explanation:
Since we have given that
We first transpose the square root to the right , so it becomes square of 8,i.e.
Now, transpose 4 to the right so it will get subtract from 64 i.e.
Since 6 is multiplied to x on tranposing it will get divided by 60 i.e.
Hence, on simplification, we get x=10.
Hence , the order is F,D,C.