To determine which polynomial sequence shows that the terms of the 2nd differences are constant, we can examine the sequences provided.
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Calculate the first differences for each sequence:
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For the sequence -6, -1, 14, 39:
- First differences: \(-1 - (-6) = 5\), \(14 - (-1) = 15\), \(39 - 14 = 25\)
- First differences: \(5, 15, 25\)
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For the sequence 2, 6, 11, 14:
- First differences: \(6 - 2 = 4\), \(11 - 6 = 5\), \(14 - 11 = 3\)
- First differences: \(4, 5, 3\)
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For the sequence 0, 4, 16, 34:
- First differences: \(4 - 0 = 4\), \(16 - 4 = 12\), \(34 - 16 = 18\)
- First differences: \(4, 12, 18\)
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For the sequence 1, 7, 12, 19:
- First differences: \(7 - 1 = 6\), \(12 - 7 = 5\), \(19 - 12 = 7\)
- First differences: \(6, 5, 7\)
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Calculate the second differences (the differences of the first differences):
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For the first sequence 5, 15, 25:
- Second differences: \(15 - 5 = 10\), \(25 - 15 = 10\)
- Second differences: \(10, 10\) (constant)
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For the second sequence 4, 5, 3:
- Second differences: \(5 - 4 = 1\), \(3 - 5 = -2\)
- Second differences: \(1, -2\) (not constant)
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For the third sequence 4, 12, 18:
- Second differences: \(12 - 4 = 8\), \(18 - 12 = 6\)
- Second differences: \(8, 6\) (not constant)
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For the fourth sequence 6, 5, 7:
- Second differences: \(5 - 6 = -1\), \(7 - 5 = 2\)
- Second differences: \(-1, 2\) (not constant)
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From this analysis, only the first sequence (-6, -1, 14, 39) has constant second differences. Thus, this sequence is represented by a polynomial of degree 2.
So, the answer is: -6, -1, 14, 39.