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Usimov [2.4K]
2 years ago
7

A rectangle is transformed according to the rule R0, 90º. The image of the rectangle has vertices located at R'(–4, 4), S'(–4, 1

), P'(–3, 1), and Q'(–3, 4). What is the location of Q?

Mathematics
2 answers:
zalisa [80]2 years ago
3 0
We are said that a rectangle has been transformed into the one indicated in Figure 1 according to this rule:

R_0, \ 90^{\circ}

We know that the center of rotation is origin and two rules are applied to rotate a point 90 degrees, namely:

1. Clockwise

In this case, the rule to transform a point is:

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

This rule was already applied to form the image, so all we need to do is to reverse the answer using this formula, therefore:

For \ Q(-3,4): \\ \\ (y,-x)=(-3,4) \\ \\ \therefore y=-3 \ and \ -x=4 \therefore x=-4 \\ \\ Thus, \ the \ point \ is: \\ \\ \boxed{Q(-4,-3)}

2. Counterc
lockwise

Applying the same previous concept but with the new rules for this case:

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

By reversing the answer, we have:

For \ Q(-3,4): \\ \\ (-y,x)=(-3,4) \\ \\ \therefore -y=-3 \therefore y=3 \ and \ x=4 \\ \\ Thus, \ the \ point \ is: \\ \\ \boxed{Q(4,3)}

lisov135 [29]2 years ago
3 0

Answer:

Q(4,3)

Step-by-step explanation:

One type of transformation is rotations, which are done counter-clockwise direction.

In this case, we have a rotation of 90° around the origin (0,0), that can be expressed as

(x,y) \implies (-y,x)

Which means a 90° rotation would be done by changing coordinates positions and inverting the sign of y-coordinate.

However, the problem is giving the transformed coordinates where Q'(-3,4).

So, applying the rule described above, the original coordinate is Q(4,3).

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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
Furkat [3]

Answer:

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:

We know that T: IR^{2}  → IR^{2} is a linear transformation that maps e_{1} into y_{1} ⇒

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

And also maps e_{2} into y_{2}  ⇒

T(e_{2})=y_{2}

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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larisa [96]

Answer:

r=8

Step-by-step explanation:

Because 16 divided by 2 is 8

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Read 2 more answers
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