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Determine whether the relation defines to be a function of . If a function is defined, give its domain and range. If it does not define a function, find two ordered pairs that show a value of that is assigned more than one value of . See Example 2.$$\begin{array}{|c|c|}\hline x & {y} \\\hline 30 & {2} \\{30} & {4} \\{30} & {6} \\{30} & {8} \\{30} & {10} \\\hline\end{array}$$
(30,2) and (30,4)
Algebra
Chapter 3
Graphing Linear Equations and Inequalities in Two Variables; Functions
Section 8
An Introduction to Functions
Graphs and Statistics
Equations and Inequalities
Linear Functions
Systems of Equations and Inequalities
Missouri State University
Oregon State University
McMaster University
University of Michigan - Ann Arbor
Lectures
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in this problem, we're given the table of values that represents a relation, and the first ring were asked to do is determine whoever this table represents a function remember, Ah, function means that every value of X is paired with exactly one why value. But if you know this in our table are for our value of X is 30 is tthe e x valiant, every single case, and it's getting map to a different. Why values time? Because, for example, you have 30 is come mashed up with 2/3 these mashed up before and for six. Therefore, this is not a function now. Our directions then tell us if it's not a function to pick out two ordered pairs where each X value is not is map to the different y value. So essentially what they're saying is the pick out two different order pairs where we have the same X value getting math two different Why values. So for an example, we could use the order pairs 30 and two and 30 and four because they have the same X value. But they're getting map two different y values
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