HELP PLEASE 13. Triangle ABC is dilated by a scale factor of 3. The
vertices of ABC are A(1,5), B(2,3) and C(4,2). What are
the vertices of Triangle A'B'C'.
A. A'(3,15), B'(6,9), C'(12,6)
B. A'(1,15), B'(2,9), C'(4,6)
C. A'(3,5), B’(6,3), C'(12,2)
D. A'(1/3, 5/3), B'(2/3,1), C'(4/3, 2/3)

y = - 6

step-by-step explanation:

given the point (- 2, y) lies on the graph then it is a solution of it's equation.

substitute x = - 2 into the equation for value of y

y = 3(- 2) = 3 × - 2 = - 6

the point on the graph is (- 2, 6)

(sin 42)/31. = sinx)/37

sinx = 37 (sin42)/31

= 37× 0.66/31 = 0.787

x= 51.9

other angle opposite to y = 180-51.9-42

= 180- 93.9 =86.1

so

sin42)/31 = sin 86.1)/y

0.021 = 0.997 /y

y= 0.997/0.021 =47.47

i dont know men

step-by-step explanation: