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).

area of part 1:

a1=base*height/2=(8+2)*(8-2-2-1)/2=10*3/2=15 ft^2

area of part 2:

a2=length*width=(8+2)*1=10 ft2

area of part 3:

a3=length*width=8*2=16 ft2

total shaded area : a1+a2+a3=41 ft2

total area of the reactangle : a=16*8=128 ft2

total area of the nonshaded region : 128-41=87 ft2

on monday rodney ate 1/4th of the sandwich = 3/12

on tuesday he ate 1/3rd of the sandwich = 4/12

therefore in total he has eaten = 3/12 + 4/12 = 7/12

thus we can say that he has eaten 7/12 of the sandwich on monday and tuesday alltogether.

happy to

pls mark as brainliest.