Question

which set of ordered pairs represents a function? ,7),(4,8),(5,9)} {(1,2),(2,3),(3,4)} {(2,5),(2,6)} {(1,2),(1,3)}

Explanation:

Step1: Recall Function Definition

A function is a relation where each input (first element of ordered pair) has exactly one output (second element). So, check each set for repeated first elements with different seconds.

Step2: Analyze First Set (Partially Shown)

The first set has pairs like \((4,8)\), \((5,9)\), but the first element (maybe 3? The start is cut) – but let's check others.

Step3: Analyze Second Set \(\{(1,2),(2,3),(3,4)\}\)

First elements: \(1\), \(2\), \(3\) – all unique. Each input has one output.

Step4: Analyze Third Set \(\{(2,5),(2,6)\}\)

First element \(2\) maps to \(5\) and \(6\) – not a function.

Step5: Analyze Fourth Set \(\{(1,2),(1,3)\}\)

First element \(1\) maps to \(2\) and \(3\) – not a function. Also, the second set (and the first partial one, but assuming the first set's first elements are unique, but the second set is clear: all first elements unique. Wait, the first set's first element (the cut one) – but the second set: \((1,2)\), \((2,3)\), \((3,4)\) – each \(x\) (first) has one \(y\) (second). So this is a function. Also, the first set (if the first element is, say, \(3\), \(4\), \(5\) – unique, but the second set is definitely a function. Wait, the options: let's list all:

1. First (partial): \((?,7),(4,8),(5,9)\) – assume first elements are unique (since \(4\), \(5\) are unique, and the first is maybe \(3\) – so unique. But the second set: \(\{(1,2),(2,3),(3,4)\}\) – all first elements \(1\), \(2\), \(3\) unique. Third: \((2,5),(2,6)\) – same \(x=2\) two \(y\). Fourth: \((1,2),(1,3)\) – same \(x=1\) two \(y\). So both first (partial) and second are functions? Wait, maybe the first set's first element is repeated? Wait, the first set's first element is cut, but the second set is clear. Wait, maybe the first set is, e.g., \(\{(3,7),(4,8),(5,9)\}\) – all \(x\) unique. The second set: \(\{(1,2),(2,3),(3,4)\}\) – all \(x\) unique. So both are functions? But maybe the first set has a repeated \(x\)? Wait, the problem's options: let's check the second set: \(\{(1,2),(2,3),(3,4)\}\) – each \(x\) is unique, so it's a function. The third and fourth have repeated \(x\), so not functions. The first set (if \(x\) values are unique) is a function. But maybe the first set's first element is, say, \(1\), and then \((1,7)\), but no, the first set's first element is cut. Wait, maybe the first set is \(\{(1,7),(4,8),(5,9)\}\) – \(x=1\), \(4\), \(5\) – unique. So both first and second? But the options: the second set is \(\{(1,2),(2,3),(3,4)\}\) – which is a function. So the answer is the second set (and the first, but maybe the first set has a repeated \(x\) – but the second is clear.

Answer:

The set \(\{(1,2),(2,3),(3,4)\}\) (and the first partial set if \(x\) are unique, but among the given clear options, the second set) – so the ordered pair set \(\{(1,2),(2,3),(3,4)\}\) represents a function.