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Nonamiya [84]
3 years ago
11

Which sequence of transformations proves that shape I is similar to shape II?

Mathematics
2 answers:
Serhud [2]3 years ago
4 0

Answer:

B. a reflection across X axis and then a dilation by a scale factor of 1.5

Step-by-step explanation:

As clear from the graph all the coordinates of image 2 are 1.5 times of image 1 so shape 2 is dilated by the scale factor 1.5.

It is evident from the graph that lines AO and A"O"are parallel to each other so shape 2 is the reflection of shape 1.

Now we calculate the magnitude of line AB

m1 = (y2-y1)/(x2-x1)

    =(-3+6)/(-4.5=9)

    =3/4.5

    =1/1.5

    =2/3

Next we calculate magnitude m2 of A"B"  

m2= (2-4) /(-3+6)

     =(-2/3)

Then we know Tan(180-∅) = -tan∅

similarly if m1=(-m2)

then the one line having magnitude m2 is the rotated image through X axis by 180° of the line having magnitude m1.

So the answer is B.

Grace [21]3 years ago
3 0

Answer:

b

Step-by-step explanation:

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Check the first picture below.


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if two lines are parallel, they have the same slope, namely the <u>same rise and run</u>.

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now, for part atop

Check the 2nd picture, in a parallelogram, opposite sides are parallel and opposite angles are equal, so if WLP is 144°, then PNW is also 144°.

in a parallelogram, diagonals bisect each other, so each cut the other in halves, so if PM is 9 then MW is also 9 and thus PW is just 9+9=18.


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notice what we said about a parallelogram above, thus


\bf 8x+13=11x-23\implies 8x+36=11x\implies 36=3x \\\\\\ \cfrac{36}{3}=x\implies \boxed{12=x} \\\\[-0.35em] ~\dotfill\\\\ 2y+12=4y-4\implies 2y+16=4y\implies 16=2y \\\\\\ \cfrac{16}{2}=y\implies \boxed{8=y}


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∡T ==> 180 - 109 ==> 71°.

3 0
3 years ago
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Answer:

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