Enter the ordered pair for the vertices for (Ry-axis o T(2,0))(QRST).

Q' = (-3, 3)

R' = (-5, -3)

S' = (-2, -2)

T' = (0, 1)

Step-by-step explanation:

The coordinates of the vertices of the quadrilateral are;

T(-2, 1), Q(1, 3), S(0, -2), R(3, -3)

Therefore, taking the operation on the right first, we have;

T(2, 0)QRST, gives;

T(2, 0)QRST→ T(-2 + 2, 1), Q(1 + 2, 3), S(0 + 2, -2), R(3 + 2, -3)

T(2, 0)QRST→ T(0, 1), Q(3, 3), S(2, -2), R(5, -3)

For the next process R y-axis which is the reflection about the y-axis, the preimage (x, y) becomes the image (-x, y)

Therefore, we have;

R y-axis(T(0, 1), Q(3, 3), S(2, -2), R(5, -3)) = T'(0, 1), Q'(-3, 3), S'(-2, -2), R'(-5, -3)

From which we get (Ry-axis T(2, 0))(QRST =  T'(0, 1), Q'(-3, 3), S'(-2, -2), R'(-5, -3).

(b) 240

step-by-step explanation:

first, we use the pythagorean formula using the given side lengths to find x. then we use x to find the side lengths. then we add the side lengths to find the perimeter.

pythagorean theorem formula:

a^2 + b^2 = c^2

(x - 20)^2 + (x - 40)^2 = x^2

x^2 - 40x + 400 + x^2 - 80x + 1600 = x^2

x^2 -120x + 2000 = 0

(x - 100)(x - 20) = 0

x - 100 = 0 or x - 20 = 0

x = 100 or x = 20

we see that the solution x = 20 must be discarded because it will give a negative side length and a side length of 0. the only valid solution is x = 100.

perimeter = sum of the lengths of the sides

perimeter = x + x - 20 + x - 40

perimeter = 3x - 60

replace x with 100.

perimeter = 3(100) - 60

perimeter = 300 - 60

perimeter = 240

14

step-by-step explanation:

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