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Goryan [66]
2 years ago
12

HELP: due sep 18!!

Mathematics
1 answer:
damaskus [11]2 years ago
4 0

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 |

The input, x = 4, therefore;

y  = - \left | 3\times 4 \right | = -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

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Step-by-step explanation:

Given that,

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The length of the metal box be = (15-2x) inches.

The width of the metal box be =(10-2x) inches

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∴At x=0.1528 inch, the volume of the metal box will be maximum.

Therefore the dimensions of the square should be 0.1528 inch by 0.1528 inch so, the box  has largest volume.

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