Triangle ABC has vertices A(5, 6), B(6,-3) and C(7,2).
TS-2,3 > (triangle ABC = triangle A'B'C'). What are the coordinates of the vertices of triangle A'B'C'?
A. A'(7,3), B'(8,-6), C'(9, -1)
B. A'(3, 3), B'(4, -6), C'(5, -1)
C. A'(2, 4), B'(3. –5), C'(4,0)
D. A'(3, 9), B'(4,0), C'(5, 5)

for b its (t) and c its (d)

step-by-step explanation:

square root of 36/6 = 2.449489743

square root of 16/6 = 1.632993162

square root of 6 = 2.449489743

the answer is square root of 16/6 because the square root of 36/6 is the same answer as square root of 6 = 2.449489743

set each factor equal to 0

4x-3 = 0

2x-1 = 0

solve for x.

4x-3=0

4x=3

x= 3/4

solve for x. 

2x-1= 0

2x= 1

x= 1/2

eventually after a few more steps, your answer should be 1/2 is less than or equal to x and 3/4 is greater than or equal to x