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3 years ago

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Which sequence of transformations will map figure K onto figure K′? Two congruent kites, figure K and figure K prime, are drawn

on a coordinate grid. Figure K has vertices at 4, 3, at 6, 5, at 4, 8, and at 2, 5. Figure K prime has vertices at 4, negative 8, at 6, negative 5, at 4, negative 3, and at 2, negative 5 Reflection across x = 4, 180° rotation about the origin, and a translation of (x + 8, y) Reflection across x = 4, 180° rotation about the origin, and a translation of (x − 8, y) Reflection across y = 4, 180° rotation about the origin, and a translation of (x + 8, y) Reflection across y = 4, 180° rotation about the origin, and a translation of (x − 8, y)

2 answers:

elixir [45]3 years ago

7 0

Answer:

The sequence of transformations will map figure K onto figure K′

is the first sequence <u>option (1)</u>

======================================================

Step-by-step explanation:

See the attached figure

as shown in the figure the K' is the image of K by reflection over x-axis

<u>But </u> We need to know which sequence of transformations will give the same result.

So, we will test the options by any point from K and its image from K'

i.e: we will test the options using the points (6,5) , (6,-5)

(6,5) ⇒ (6,-5)

<u>option (1):</u>

Reflection across x = 4, 180° rotation about the origin, and a translation of (x + 8, y)

(6,5) ⇒ (2,5) ⇒(-2,-5) ⇒ (6,-5)

<u>option (2):</u>

Reflection across x = 4, 180° rotation about the origin, and a translation of (x − 8, y)

(6,5) ⇒ (2,5) ⇒(-2,-5) ⇒ (-10,-5)

<u>option (3):</u>

Reflection across y = 4, 180° rotation about the origin, and a translation of (x + 8, y)

(6,5) ⇒ (6,3) ⇒ (-6,-3) ⇒ (2,-3)

<u>option (4):</u>

Reflection across y = 4, 180° rotation about the origin, and a translation of (x − 8, y)

(6,5) ⇒ (6,3) ⇒ (-6,-3) ⇒ (-14,-3)

As shown: The sequence of transformations will map figure K onto figure K′

is the first sequence <u>option (1)</u>

Anika [276]3 years ago

4 0

Answer:

The sequence of transformations will map figure K onto figure K′

Step-by-step explanation:

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Let e1= 1 0 and e2= 0 1 , y1= 4 5 , and y2= −2 7 , and let T: ℝ2→ℝ2 be a linear transformation that maps e1 into y1 and maps

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The image of $\left[\begin{array}{c}4&-4\end{array}\right]$ through T is $\left[\begin{array}{c}24&-8\end{array}\right]$

Step-by-step explanation:

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$T(e_{1})=y_{1}$

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We need to find the image of the vector $\left[\begin{array}{c}4&-4\end{array}\right]$

We know that exists a matrix A from $IR^{2x2}$ (because of how T was defined) such that :

$T(x)=Ax$ for all x ∈ $IR^{2}$

We can find the matrix A by applying T to a base of the domain ( $IR^{2}$ ).

Notice that we have that data :

$B_{IR^{2}}=$ { $e_{1},e_{2}$ }

Being $B_{IR^{2}}$ the cannonic base of $IR^{2}$

The following step is to put the images from the vectors of the base into the columns of the new matrix A :

$T(\left[\begin{array}{c}1&0\end{array}\right])=\left[\begin{array}{c}4&5\end{array}\right]$ (Data of the problem)

$T(\left[\begin{array}{c}0&1\end{array}\right])=\left[\begin{array}{c}-2&7\end{array}\right]$ (Data of the problem)

Writing the matrix A :

$A=\left[\begin{array}{cc}4&-2\\5&7\\\end{array}\right]$

Now with the matrix A we can find the image of $\left[\begin{array}{c}4&-4\\\end{array}\right]$ such as :

$T(x)=Ax$ ⇒

$T(\left[\begin{array}{c}4&-4\end{array}\right])=\left[\begin{array}{cc}4&-2\\5&7\\\end{array}\right]\left[\begin{array}{c}4&-4\end{array}\right]=\left[\begin{array}{c}24&-8\end{array}\right]$

We found out that the image of $\left[\begin{array}{c}4&-4\end{array}\right]$ through T is the vector $\left[\begin{array}{c}24&-8\end{array}\right]$

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Answer

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