∆XYZ has vertices X(0, 3), Y(2, 0), and Z(4, 2). If you reflect ∆XYZ across the line x = 4, what will be the coordinates of the vertices of the image ∆X′Y′Z′?

X′(4, 3), Y′(2, 0), and Z′(4, 2)

X′(8, 3), Y′(6, 0), and Z′(4, 2)

X′(3, 0), Y′(0, 2), and Z′(2, 4)

X′(4, 3), Y′(6, 0), and Z′(4, 2)

answer: answer is c!

ok i got to organize the

3/15 and 12/55

3*

aha

3*4 equals 12 but not

a: no

b:

8/24 12/35

no

gautie

it's c

5/18 = 25/90

5*5=25

18*5 = 90

hope that you!

one more happy customer!

< 3

step-by-step explanation:

c. 27 units.

step-by-step explanation:

we are told that in circle a, ∠bae ≅ ∠dae.

we can see from our given diagram that in and ;

, as these are radii of our given circle.

therefore, by sas congruence postulate.

hence, side be will be equal to side de as corresponding parts of congruent triangles are congruent. so we can set an equation to find the value of x as:

now let us substitute x=17 in the expression for the length of be.

therefore, length of be will be 27 units and option c is the correct choice.

here is a big fat