Triangle rst has vertices r(2, 0), s(4, 0), and t(1, –3). the image of triangle rst after a rotation has vertices r'(0, –2), s'(0, –4), and t'(–3, –1). which rule describes the transformation?

By the rotation rule of 270 of counter clock wise ( x ,y) →→ ( y , -x )

Step-by-step explanation:

Given  : Triangle RST has vertices R(2, 0), S(4, 0), and T(1, –3). The image of triangle RST after a rotation has vertices  R'(0, –2), S'(0, –4), and T'(–3, –1).

To find :  Which rule describes the transformation.

Solution : We have given

Parent  vertices R(2, 0), S(4, 0), and T(1, –3).

Transformed  vertices  R'(0, –2), S'(0, –4), and T'(–3, –1).

By the rotation rule of 270 of counter clock wise ( x ,y) →→ ( y , -x )

R(2, 0)→→   R'(0, –2)

S(4, 0)→→  S'(0, –4)

T(1, –3)→→  T'(–3, –1).

We can see y coordinate change in to x and x coordinate become - y.

Therefore ,By the rotation rule of 270 of counter clock wise ( x ,y) →→ ( y , -x )

a is 8.b is 10.c is 10.5. d is -.5

step-by-step explanation:

just minus the smallest number from the biggest.