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Which rule describes a composition of transformations that maps pre-image PQRS to image P"Q"R"S"?

FinnZ [79.3K]

The correct option is Option D $\boxed{{r_{y-axis}}o{R_{0,270^\circ }}\left({x,y}\right)}$ .

Further explanation:

A translation is a transformation that transforms the figure with a fixed distance in the same direction.

A rotation is the transformation that rotates the figure with given angles.

Given:

It is given that the two transformations that maps pre-image PQRS to image ${\text{P''Q''R''S''}}$ .

Step by step explanation:

Step 1:

It can be seen from the given figure that that the pre image is in the first quadrant and the image is the third quadrant.

The coordinates in the second quadrant represents as $\left({-x,y}\right)$ and in the third quadrant represents as $\left({-x,-y}\right)$ if $x,y$ are positive.

Therefore, the rotation is in the counter clockwise direction of $270^\circ$ .

Step 2:

The rotation of $270^\circ$ in the counter clockwise direction represents the coordinates as,

$\left({x,y}\right)\to\left({y,-x}\right)$

It can be seen that the coordinate of ${\text{PQRS}}$ are as follows,

$\begin{aligned}P=\left({1,1}\right)\hfill\\Q=\left(1,5}\right)\hfill\\R=\left({3,5}\right)\hfill\\S=\left({3,1}\right)\hfill\\\end{aligned}$

Then after rotation of $270^\circ$ counterclockwise on ${\text{PQRS}}$ $\left( {x,y}\right)\to\left({y,-x}\right)$ as,

$\begin{gathered}P\left({1,1}\right)\to\left({1,-1}\right)\hfill\\Q\left({1,5}\right)\to\left({5,1}\right)\hfill\\R\left({3,5}\right)\to\left({5,-3}\right)\hfill\\S\left({3,1}\right)\to\left({1,-3}\right)\hfill\\\end{gathered}$

Step 3:

Now apply the rule of $y$ axis of reflection ${R_{y-axis}}\left({x,y}\right)\to\left({-x,y}\right)$ on the above transformation as,

$\begin{gathered}\left({1,-1}\right)\to\left({-1,1}\right)=P''\hfill\\\left({5,1}\right)\to\left({-5,-1}\right)=Q''\hfill\\\left({5,-3}\right)\to\left({-5,-3}\right)=R''\hfill\\\left({1,-3}\right)\to\left({-1,-3}\right)=S''\hfill\\\end{gathered}$

Therefore, the given transformation is the rotation of $270^\circ$ counterclockwise followed by $y$ axis of reflection.

Therefore, this is the composition of transformation.

The composition of the given transformation can be written as,

${r_{y-axis}}o{R_{0,270^\circ}}\left({x,y}\right)$

Therefore, option D ${r_{y-axis}}o{R_{0,270^\circ}}\left({x,y}\right)$ is correct.

Learn more:

Learn more about what is the final transformation in the composition of transformations that maps pre-image abcd to image a"b'c"d"? a translation down and to the right a translation up and to the right a 270° rotation about point b' a 180° rotation about point b' brainly.com/question/2480946

Learn more about the transformation of function brainly.com/question/7297858

Learn more about midpoint of the segment brainly.com/question/3269852

Answer details:

Grade: High school

Subject: Mathematics

Chapter: Transformations

Keywords: transformations, dilation, translation, rotation, counterclockwise, angle, clockwise, coordinates, mapping, rigid transformation, right side, left side, quadrant, composition.

3 0

3 years ago

Read 2 more answers

A dry cleaner charges $18 to clean 4 pairs of pants. How many pairs of pants can be cleaned for $27 at the same rate?

blondinia [14]

4 pairs of pants for $18.

Divide price by quantity to get the cost of 1 pair of pants:

18 / 4 = 4.50 per pair.

Divide 27 by 4.50 to find number of pairs:

27 / 4.5 = 6

6 pairs of pants can be cleaned for $27.

5 0

3 years ago

Given h(t) = -2(t+ 5)^2 +4, find h(-8).

bekas [8.4K]

Answer:

Substitute -8 as x into the equation.

h(-8)=-2(-8+5)^2+4

h(-8)=-2(-3)^2+4

h(-8)=-2(9)+4

h(-8)=-18+4

h(-8)=-14

:)

7 0

3 years ago

Suppose that a large mixing tank initially holds 100 gallons of water in which 50 pounds of salt have been dissolved. Another br

Serggg [28]

Answer:

dA/dt = 12 - 2A/(100 + t)

Step-by-step explanation:

The differential equation of this problem is;

dA/dt = R_in - R_out

Where;

R_in is the rate at which salt enters

R_out is the rate at which salt exits

R_in = (concentration of salt in inflow) × (input rate of brine)

We are given;

Concentration of salt in inflow = 4 lb/gal

Input rate of brine = 3 gal/min

Thus;

R_in = 4 × 3 = 12 lb/min

Due to the fact that solution is pumped out at a slower rate, thus it is accumulating at the rate of (3 - 2)gal/min = 1 gal/min

So, after t minutes, there will be (100 + t) gallons in the tank

Therefore;

R_out = (concentration of salt in outflow) × (output rate of brine)

R_out = [A(t)/(100 + t)]lb/gal × 2 gal/min

R_out = 2A(t)/(100 + t) lb/min

So, we substitute the values of R_in and R_out into the Differential equation to get;

dA/dt = 12 - 2A(t)/(100 + t)

Since we are to use A foe A(t), thus the Differential equation is now;

dA/dt = 12 - 2A/(100 + t)

5 0

3 years ago

The circle given by

mixer [17]

Typically, the equation of a circle is given like this: (x-h)^2 + (y-k)^2 = r^2, where r is the radius of the circle and the point (h,k) is the center. To find the value k, you need to put the first equation you have into this form. Standard form requires that you only have the1 y variable, as seen in the parentheses, so you first need to complete the square.

x^2 + y^2 - 2y = 11
x^2 + y^2 - 2y +1 = 11 + 1
x^2 + (y - 1)^2 = 12

And that puts the equation in standard form. k = 1.

5 0

3 years ago

Read 2 more answers