What are the coordinates of the endpoints of the segment T'V'?
Triangle TVW is dilated according to the rule
D02(x, y)=(x, y) to create the image triangle T'V"W",
which is not shown.
O T'(-3, 6) and V'(0, 3)
O T'(-3, 6) and V'(0, 1)
O T'(-1, 2) and V'(0,3)
O T'(-1, 2) and V'(0, 1)
0 N
w
2

The coordinates of the endpoints of the segment T'V are; T'(-3, 6) and V'(0, 3)

According to the question;

The given triangle TVW is dilated according to the rule D (3/4)(x,y)=((3/4)x, (3/4)y) to create the image triangle T'V"W", which is not shown.

From the attached image; the coordinates of endpoints T and V are;

Therefore, the result segment of the dilation according to the rule; D (3/4)(x,y)=((3/4)x, (3/4)y) is;

Read more on triangle dilation;

link

A) T'(-3, 6) and V'(0, 3)

Step-by-step explanation:

In the picture attached, triangle TVW is shown.

Transforming points T and V according to the rule (x,y) -> (3/4x, 3/4y), we get:

T(-4, 8) -> (-4*3/4, 8*3/4) -> (-3, 6) which corresponds to T'

V(0, 4) -> (0*3/4, 4*3/4) -> (0, 3) which corresponds to V'

4 candies

step-by-step explanation:

7 x 28 = 196

200 - 196 = 4 candies remaining.