Look at this set of ordered pairs: (6, 19) (6, 20) (14, 12) Is this relation a function? Choices: (A)yes (B)no

Define Function: Understand the definition of a function in terms of ordered pairs. A set of ordered pairs is a function if for every x-value (first element of the pair), there is exactly one y-value (second element of the pair) associated with it. This means that no x-value is repeated with different y-values.Examine Ordered Pairs: Examine the given set of ordered pairs to see if any x-value is repeated with different y-values. The given set of ordered pairs is: (6, 19) (6, 20) (14, 12) We can see that the x-value 6 is associated with two different y-values, 19 and 20.Determine Functionality: Determine if the relation is a function based on the examination in Step 2. Since the x-value 6 is associated with more than one y-value, the relation is not a function.

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Look at this set of ordered pairs: $\newline$ $(6, 19)$ $\newline$ $(6, 20)$ $\newline$ $(14, 12)$ $\newline$ Is this relation a function? $\newline$ Choices: $\newline$ (A) yes $\newline$ (B) no

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Q. Look at this set of ordered pairs:

\newline

(6, 19)

\newline

(6, 20)

\newline

(14, 12)

\newline

Is this relation a function?

\newline

Choices:

\newline

(A) yes

\newline

(B) no

Define Function: Understand the definition of a function in terms of ordered pairs. $\newline$ A set of ordered pairs is a function if for every $x$ -value (first element of the pair), there is exactly one $y$ -value (second element of the pair) associated with it. This means that no $x$ -value is repeated with different $y$ -values.
Examine Ordered Pairs: Examine the given set of ordered pairs to see if any $x$ -value is repeated with different $y$ -values. $\newline$ The given set of ordered pairs is: $\newline$ $(6, 19)$ $\newline$ $(6, 20)$ $\newline$ $(14, 12)$ $\newline$ We can see that the $x$ -value $6$ is associated with two different $y$ -values, $19$ and $20$ .
Determine Functionality: Determine if the relation is a function based on the examination in Step $2$ . $\newline$ Since the $x$ -value $6$ is associated with more than one $y$ -value, the relation is not a function.