The verticies of a triangle on the coordinate plane are a(0, 0), b(2, 0) and c(0, 2). what would be the coordinates of triangle a'b'c' if triangle abc was dilated by a factor of 1 3 ? a) a'(0, 0) b'(3, 0) c'(0, 3) b) the vertices would not change. c) a'(0, 0) b'( 2 3 , 0) c'(0, 2 3 ) d) a'( 1 3 , 1 3 ) b'( 7 3 , 1 3 ) c'( 1 3 , 7 3 )

The correct option is C.

Step-by-step explanation:

It is given that the vertices of a triangle on the coordinate plane are A(0, 0), B(2, 0) and C(0, 2).

If a figure dilated by a factor of k about the origin, then

It is given that triangle ABC was dilated by a factor of 1/3 to create the triangle A'B'C'.

If a figure dilated by a factor of 1/3 about the origin, then

The vertices of image are

The coordinates of triangle A'B'C' are A'(0,0), B'(2/3,0) and C'(0,2/3).

Therefore, the correct option is C.

answer: i guess 105 degrees?

step-by-step explanation:

answer: 0

step-by-step explanation: the positive, like 5, + its opposite, like -5 would be 0

5+(-5)=0, or 5-5=0