Answer:
The correct order that the machines should be stacked so that when 15 dropped into the first machine, and all four machines have had their effect, the last machine's output is -6 can be written as follows;
4th machine → 2nd machine → 1st machine → 3rd machine
Step-by-step explanation:
Given that the initial input is 15 and the final output is -6, we have;
For the fourth machine;
y = -2·x + 34 = -2×15 + 34 = 4
The output is 4
Next to the second machine
y = (x - 2)²
The input is x = 4, therefore;
y = (4 - 2)² = 2² = 4
The output is 4
Next to the first machine
![y = - \left | 3\cdot x \right |](https://tex.z-dn.net/?f=y%20%20%3D%20-%20%5Cleft%20%7C%203%5Ccdot%20x%20%5Cright%20%7C)
The input, x = 4, therefore;
![y = - \left | 3\times 4 \right | = -12](https://tex.z-dn.net/?f=y%20%20%3D%20-%20%5Cleft%20%7C%203%5Ctimes%204%20%5Cright%20%7C%20%3D%20-12)
The output is -12
Next to the third machine
y = -x/3 -10
The input is x = -12, therefore;
y = -(-12)/3 -10 = 12/3 -10 = 4 - 10 = -6
The output is -6.
Therefore, the correct order that the machines should be stacked so that when 15 dropped into the first machine, and all four machines have had their effect, the last machine's output is -6 can be written as follows;
4th → 2nd → 1st → 3rd