All of these sequences of transformations would return a shape to its original position except? a. Translate 3 units up, then 3 units down b. Reflect over line p, then reflect over line p again c. Translate 1 unit to the right, then 4 units to the left, then 3 units to the right. d.Rotate 120 degrees around center c then rotate 220 degrees around c again

Except d. Rotate 120 degrees around center c then rotate 220 degrees around again.

Step-by-step explanation:

Let's choose an arbitrary shape to work with the exercise.

For example a square (All the conclusions that we will make can be use with any shape).

For a. Translate 3 units up, then 3 units down is easy to see that this will return the square to its original position. Wherever we translate up and then down the same units a particularly shape this will return to its original position.

For b. Reflect over line p, then reflect over line p again

Wherever we have any particularly shape and we reflect over an arbitrary line twice, the shape will return to its original position. Particularly, the composition of two reflections over the same line is the identity function.

The identity function is the function that doesn't change the shape (It is the analogy of the multiplication by 1 with the common product between real numbers).

c. Translate 1 unit to the right, then 4 units to the left, then 3 units to the right.

The composition of this three translation will return the square to its original position (Same reasoning as a.)

d. Rotate 120 degrees around center c then rotate 220 degrees around c again.

Given that we choose an arbitrary center c and then chosen an arbitrary rotation sense (counterclockwise or clockwise), the composition of the two rotations is a 340 degrees rotation (given that we sum the degrees).

This transformation will not return a shape to its original position

(Of course, it exists some exceptions such as a rotation of a circle around its center. For any value of degrees, the rotation of a circle around its center will return the circle to its original position).

Generally, option d. is the correct option.

option c. square

step-by-step explanation:

points to remember

distance formula

length of a line segment with end points (x1, y1) and (x2, y2) is given by,

distance = √[(x2 - x1)² + (y2 - y1)²]

to find the sides of quadrilateral lmno

l (1, 1), m (4, 4), n (7, 1), and o (4, −2)

lm = √[(4 - 1)² + (4 - 1)²]

  = √[3² + 3²]

  = √(9+ 9) = √18

mn = √[(7 - 4)² + (1 - 4)²]

  = √[3² + (-3²)]

  = √(9+ 9) = √18

on = √[(4 - 7)² + (-2 - 1)²]

  = √[(-3)² + (-3²)]

  = √(9+ 9) = √18

lo = √[(4 - 1)² + (-2 - 1)²]

  = √[3² + (-3²)]

  = √(9+ 9) = √18

to find the type of quadrilateral

here lm = mn = no = lo = √18

all sides are equal.

therefore the correct answer is option c. square