Describe the sequence of transformations that maps PQR into P'Q'R' use the drop-down menus to explain your answer

Answer:

Rotation of 180° clockwise around the origin

Step-by-step explanation: if you rotate the first triangle 90 degrees, it will be in the lower quarter, so you need to rotate it 1 more time.

The transformations are clockwise rotation 180° and translation down by 1 unit.

Rotations are transformations that turn a shape around a fixed point. To rotate a shape we need: a centre of rotation. an angle of rotation (given in degrees) a direction of rotation – either clockwise or anti-clockwise.

From the given graph, the coordinates of triangle PQR are P(2, 1), Q(4, 6) and R(4, 1).

Rotating a point that has the coordinates (x,y) 180° about the origin in a clockwise direction, we have (-x,-y).

Here,

P(2, 1) → P'(-2, -1)

Q(4, 6) → Q'(-4, -6)

R(4, 1) → R'(-4, -1)

The translation down by 1 unit, we get (x, y-k)

P'(-2, -1) → P'(-2, -2)

Q'(-4, -6) → Q'(-4, -7)

R'(-4, -1) → R'(-4, -2)

The coordinates of triangle P'Q'R' are P'(-2, -2), Q(-4, -7) and R'(-4, -2)

Therefore, the transformations are clockwise rotation 180° and translation down by 1 unit.

Learn more about the rotation here:

brainly.com/question/1571997.

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one number added to twice another number is 23. Four times the first number added to twice the other number is 38. What are the numbers?

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y = 3/4x + 12
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The system of linear equations will have
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1. Find the equation of the parabola satisfying the given conditions.

Focus: (3,6); Directrix: x=−1

A. (x−1)2=8(y−6)

B. (y−6)2=8(x−1)

C. (x−1)2=−8(y−6)

D. (y−6)2=−8(x−1)


2. Find the equation of the parabola satisfying the given conditions.

Focus: (−6,3); Directrix: y=1

A. (y−2)2=4(x+6)

B. (x+6)2=4(y−2)

C. (x+6)2=−4(y−2)

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3. Find the equation of an ellipse that has foci at (−1,0) and (4,0), where the sum of the distances between each point on the ellipse and the two foci is 9.

A. (x+1)2+y2−−−−−−−−−−−√+(x−4)2+y2−−−−−−−−−−−√=9

B. (x−1)2+y2−−−−−−−−−−−√+(x+4)2+y2−−−−−−−−−−−√=9

C. (x+1)2+y2−−−−−−−−−−−√+(x−4)2+y2−−−−−−−−−−−√=81

D. (x−1)2+y2−−−−−−−−−−−√+(x+4)2+y2−−−−−−−−−−−√=81


4. Find the equation of a hyperbola that has foci at (−1,0) and (4,0), where the difference of the distances between each point on the ellipse and the two foci is 5.

A. (x+1)2+y2−−−−−−−−−−−√−(x−4)2+y2−−−−−−−−−−−√=25

B. (x−1)2+y2−−−−−−−−−−−√−(x+4)2+y2−−−−−−−−−−−√=5

C. (x−1)2+y2−−−−−−−−−−−√−(x+4)2+y2−−−−−−−−−−−√=25

D. (x+1)2+y2−−−−−−−−−−−√−(x−4)2+y2−−−−−−−−−−−√=5

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