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Solutions for Session 7, Part C
See solutions for Problems: C1 | C2 | C3 | C4 | C5 | C6| C7 | C8 | C9
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Problem C2 | |
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Here is the completed table:
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Number of dots on side of square | | Total number of dots (square number) |
1 | | 1 |
2 | | 4 |
3 | | 9 |
4 | | 16 |
5 | | 25 |
6 | | 36 |
10 | | 100 |
13 | | 169 |
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<< back to Problem C2
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Problem C3 | |
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It is neither a linear nor an exponential graph. Its successive outputs do not have the same ratio; therefore, it cannot be an exponential graph. It is certainly not a straight line, because successive outputs do not have the same difference, so it cannot be a linear graph.
<< back to Problem C3
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Problem C4 | |
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The rule is O = n2, where O is the output, the total number of dots, and n is the number of dots on the side of a square. A recursive rule is Dn = Dlast + (2n - 1).
<< back to Problem C4
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Problem C6 | |
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Here is the completed table.
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Number of dots on side of triangle | | Total number of dots (triangular number) |
1 | | 1 |
2 | | 3 |
3 | | 6 |
4 | | 10 |
5 | | 15 |
6 | | 21 |
9 | | 45 |
19 | | 190 |
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<< back to Problem C6
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Problem C7 | |
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As with Problem C3, the graph does not demonstrate exponential behavior, because successive terms do not have the same ratio. It's not a linear graph, either, because it is certainly not a straight line. Actually, the graphs of the table of square numbers and the table of triangular numbers look pretty similar.
<< back to Problem C7
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Problem C8 | |
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There is more than one answer, but one is O = (n)(n + 1) / 2. The recursive form is easier to find: Dn = Dlast + n, because n new dots are added in the nth triangle.
<< back to Problem C8
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Problem C9 | |
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Both closed-form rules involve multiplying n by itself at some point, and both recursive rules involve adding something linear to n. Compare this to linear functions, which have only a single use of the variable in their closed-form rules, and a constant in the recursive rule.
<< back to Problem C9
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