Which translation rule can be used to prove that triangle r(10,6), s(8,4), t(6,6) and triangle r'(7,4), s'(5,2) t'(3,4)are congruent? a) (x,y) → (x - 3, y - 2) b) (x,y) → (x + 3, y + 2) c) (x,y) → (x - 3, y + 2) d) (x,y) → (x + 3, y - 2)

a) (x,y) =(x-3, y-2)

a) (x,y) → (x - 3, y - 2)  

Step-by-step explanation:

Comparing the pre-image points, r, s and t, to the image points, r', s' and t', we see that the x-coordinates of the image are all 3 less than the x-coordinates of the image:

10-7 = 3; 8-5 = 3; 6-3 = 3

This means the first part of the translation rule will be x-3.

Now comparing the y-coordinates, we see that the y-coordinates of the image are all 2 less than the y-coordinates of the pre-image:

6-4 = 2; 4-2 = 2; 6-4 = 2

This means the second part of the translation rule will be y-2.

This gives us (x, y)→(x-3, y-2).

Correct option: B -> 36

Step-by-step explanation:

The surface area of a circular cone is given by the formula:

Surface area = pi * radius * (radius + slant height)

If the inicial radius is 4 in and the slant height is 9 in, we have:

Surface area = pi * 4 * (4 + 9) = 52pi in2

The radius and the slant height are multiplied by 6, so we have:

radius = 6 * 4 = 24 in

slant height = 6 * 9 = 54 in

So the new surface area is:

New surface area = pi * 24 * (24 + 54) = 1872pi in2

The factor of the surface areas is:

New surface area / Surface area = 1872pi / 52pi = 36

Correct option: B