For δabc, ∠a = x + 30, ∠b = 2x - 4, and ∠c = 4x. if δabc undergoes a dilation by a scale factor of 1 2 to create δa'b'c' with ∠a' = 2x + 8, ∠b' = x + 18, and ∠c' = 5x - 22, which confirms that δabc∼δa'b'c by the aa criterion?

ABC
Solve for x
x + 30 + 2x - 4 + 4x = 180 All triangles have 180°. collect like terms.
7x + 26 = 180 subtract 26 to both sides.
7x = 154 Divide by 7
x = 154/7
x = 22
A = x + 30 = 22 + 30 = 52
B = 2x - 4 = 2(22) + 30 = 44 - 4 = 40
C = 4x = 4*22 = 88

Check
<A + <B + <C = 180
<52 + <40 + <88 = 180°
180 = 180

It checks and we are done with this triangle.

A'B'C' 
2x + 8 + x + 18 + 5x - 22 = 180 Collect the like terms.
8x + 4 = 180 Subtract 4 from both sides.
8x = 180 - 4
8x = 176 Divide by 8
x = 176 / 8
x = 22
 
A' = 2x + 8
A' = 2(22) + 8 = 44 + 8 = 52
B' = x + 18 = 22 + 18 = 40
C' = 5x - 22 = 5*22 - 22 = 110 - 22 = 88

Conclusion
A = A'
B = B'
C = C'

All three angles of one triangle are equal to all three angles of the other. So by AA the two triangles are similar.

Comment
Note: the dilation of 12 has nothing to do with this answer. If you were asked about similarity of sides then the 12 would be something to do with the problem. 

Solution: beginning of the unit = - 2end of the unit = beginning of the unit + 4end of the unit = - 2 + 4end of the unit = 2final answer: therefore, aila was able to reach 2 ( insert name of unit of measurement ) at the end of the unit.hope this ! : )have a lovely day! < 3