The graph shown is not the correct graph. The graph of y =4x + 3 will have a y-intercept of 3, meaning it will pass through the point (0, 3). It will also have a slope of 4, so the graph will be a line pointing upward rather steeply.
It will pass through the following points:
(-2, -5) ,(-1,-1), (1, 7), (2, 11), (3, 15).

6 0

4 years ago

A student club holds a meeting. The predicate M(x) denotes whether person x came to the meeting on time. The predicate O(x) refe

Novay_Z [31]

Answer:

a) $\exists \, x \in C : O(x) = 0$

b) $\{ x \in C : O(x) = 1 \} \subseteq \{ x \in C : M(x) = 1 \}$

c) $\{ x \in C: M(x) = 1 \} = C$

d) $\{ x \in C : D(x) = 1 \} \, \cup \, \{x \in C : M(x) = 1 \} = C$

e) $\exists \, x \in C : M(x) = 1 \, \wedge D(x) = 1$

f) $\exists \, x \in C : O(x) = 1 \, \wedge M(x) = 0$

Step-by-step explanation:

M(x) = 1 if the person x came to the meeting, and 0 otherwise.

O(x) = 1 if the person is an officer of the club and 0 otherwise.

D(x) = 1 if the person has paid hid/her club dues and 0 otherwise.

Lets also call C the set given by the members of the club. C is the domain of the functions M, O and D.

a) If someone is not an officer, the there should be at least one value x such that O(x) = 0. This can be expressed by logic expressions this way

$\exists \, x \in C : O(x) = 0$

b) If all the officers came on time to the meeting, then for a value x such that O(x) = 1, we also have that M(x) = 1. Thus, the set of officers of the Club is contained on the set of persons which came to the meeting on time, this can be written mathematically this way:

$\{ x \in C : O(x) = 1 \} \subseteq \{ x \in C : M(x) = 1 \}$

c) If everyone was in time for the meeting, then C is equal to the set of persons who came to the meeting on time, or, equivalently, the values x such that M(x) = 1. We can write that this way:

$\{ x \in C: M(x) = 1 \} = C$

d) If everyone paid their dues or came on time to the meeting, then if we take the set of persons who came to the meeting on time and the set of the persons who paid their dues, then the union of the two sets should be the entire domain C, because otherwise there should be a person that didnt pay nor was it on time. This can be expressed logically this way:

$\{ x \in C : D(x) = 1 \} \, \cup \, \{x \in C : M(x) = 1 \} = C$

e) If at least one person paid their dues on time and came on time to the meeting, then there should be a value x on C such that M(x) and D(x) are both equal to 1. Therefore

$\exists \, x \in C : M(x) = 1 \, \wedge D(x) = 1$

f) If there is an officer who did not come on time for the meeting, then there should be a value x in C such that O(x) = 1 (x is an officer), and M(x) = 0. As a result, we have

$\exists \, x \in C : O(x) = 1 \, \wedge M(x) = 0$

I hope that works for you!

7 0

4 years ago