QuestionIs $f=\{(a, 0),(c, 12),(d, 18),(b, 12)\}$ a function from $A=\{a, b, c, d\}$ ? Explain your answer.

Studdy Solution

STEP 1

Assumptions1. $A=\{a, b, c, d\}$ is the domain of the function. . $f=\{(a,0),(c,12),(d,18),(b,12)\}$ is the set of ordered pairs, where the first element of each pair is from the domain $A$ and the second element is from the range $B$ .

STEP 2

A relation $f$ from a set $A$ to a set $B$ is a function if every element of $A$ is associated with exactly one element of $B$ . In other words, for each element in the domain, there is one and only one corresponding element in the range.

STEP 3

Let's check whether each element of $A$ is associated with exactly one element of $B$ in the relation $f$ .

STEP 4

The element $a$ in $A$ is associated with the element $0$ in $B$ .

STEP 5

The element $b$ in $A$ is associated with the element $12$ in $B$ .

STEP 6

The element $c$ in $A$ is associated with the element $12$ in $B$ .

STEP 7

The element $d$ in $A$ is associated with the element $18$ in $B$ .

STEP 8

From steps4 to7, we can see that each element of $A$ is associated with exactly one element of $B$ . Therefore, $f$ is a function from $A$ to $B$ .
The correct answer is Yes, because each element in the domain is paired with a unique element in the range.

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QuestionIs $f=\{(a, 0),(c, 12),(d, 18),(b, 12)\}$ a function from $A=\{a, b, c, d\}$ ? Explain your answer.

Studdy Solution

STEP 1

Assumptions1. $A=\{a, b, c, d\}$ is the domain of the function. . $f=\{(a,0),(c,12),(d,18),(b,12)\}$ is the set of ordered pairs, where the first element of each pair is from the domain $A$ and the second element is from the range $B$ .

STEP 2

A relation $f$ from a set $A$ to a set $B$ is a function if every element of $A$ is associated with exactly one element of $B$ . In other words, for each element in the domain, there is one and only one corresponding element in the range.

STEP 3

Let's check whether each element of $A$ is associated with exactly one element of $B$ in the relation $f$ .

STEP 4

The element $a$ in $A$ is associated with the element $0$ in $B$ .

STEP 5

The element $b$ in $A$ is associated with the element $12$ in $B$ .

STEP 6

The element $c$ in $A$ is associated with the element $12$ in $B$ .

STEP 7

The element $d$ in $A$ is associated with the element $18$ in $B$ .

STEP 8

From steps4 to7, we can see that each element of $A$ is associated with exactly one element of $B$ . Therefore, $f$ is a function from $A$ to $B$ .
The correct answer is Yes, because each element in the domain is paired with a unique element in the range.

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QuestionIs f={(a,0),(c,12),(d,18),(b,12)}f=\{(a, 0),(c, 12),(d, 18),(b, 12)\}f={(a,0),(c,12),(d,18),(b,12)} a function from A={a,b,c,d}A=\{a, b, c, d\}A={a,b,c,d}? Explain your answer.

Studdy Solution

STEP 1

STEP 2

A relation fff from a set AAA to a set BBB is a function if every element of AAA is associated with exactly one element of BBB. In other words, for each element in the domain, there is one and only one corresponding element in the range.

STEP 3

Let's check whether each element of AAA is associated with exactly one element of BBB in the relation fff.

STEP 4

The element aaa in AAA is associated with the element 000 in BBB.

STEP 5

The element bbb in AAA is associated with the element 121212 in BBB.

STEP 6

The element ccc in AAA is associated with the element 121212 in BBB.

STEP 7

The element ddd in AAA is associated with the element 181818 in BBB.

STEP 8

From steps4 to7, we can see that each element of AAA is associated with exactly one element of BBB. Therefore, fff is a function from AAA to BBB.The correct answer is Yes, because each element in the domain is paired with a unique element in the range.

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QuestionIs $f=\{(a, 0),(c, 12),(d, 18),(b, 12)\}$ a function from $A=\{a, b, c, d\}$ ? Explain your answer.

A relation $f$ from a set $A$ to a set $B$ is a function if every element of $A$ is associated with exactly one element of $B$ . In other words, for each element in the domain, there is one and only one corresponding element in the range.

Let's check whether each element of $A$ is associated with exactly one element of $B$ in the relation $f$ .

The element $a$ in $A$ is associated with the element $0$ in $B$ .

The element $b$ in $A$ is associated with the element $12$ in $B$ .

The element $c$ in $A$ is associated with the element $12$ in $B$ .

The element $d$ in $A$ is associated with the element $18$ in $B$ .

From steps4 to7, we can see that each element of $A$ is associated with exactly one element of $B$ . Therefore, $f$ is a function from $A$ to $B$ .
The correct answer is Yes, because each element in the domain is paired with a unique element in the range.