Let be the statement ""x has a cat,"" let be the statement ""x has a dog,"" and let be the statement ""x has a ferret."" Express each of these statements in terms ofquantifiers, and logical connectives. Let the domain consist of all students in your class. A) Some student in your class has a cat and a ferret, but not a dog.
B) No student in your class has a cat, a dog, and a ferret.

Complete question is:

Let C(x) be the statement “x has a cat,” let D(x) be the statement “x has a dog”, and let F(x) be the statement “x has a ferret”. Express each of these statements in terms of C(x), D(x), F(x), quantifiers, and logical connectives. Let the domain consist of all students in your class.

a) A student in your class has a cat, a dog, and a ferret.

b) All students in your class have a cat, a dog, or a ferret.

c) Some student in your class has a cat and a ferret, but not a dog.

d) No student in your class has a cat, a dog, and a ferret.

e) For each of the three animals, cats, dogs, and ferrets, there is a student in your class who has this animal as a pet.

a) ∃x(C(x) ∧ D(x) ∧ F(x))

b) ∀x(C(x) ∨ D(x) ∨ F(x))

c) ∃x(C(x) ∧ D(x) ¬ F(x))

d) ¬∃x(C(x) ∧ D(x) ∧ F(x))

e) (∃xC(x)) ∧ (∃xD(x)) ∧ (∃xF(x))

Step-by-step explanation:

We are given;

C(x) = “x has a cat”

D(x) = “x has a dog”

F(x) = “x has a ferret”

We will make use of the following interpretation symbols;

Negation ¬ A : Not A

Disjunction A ∨ B : A or B

Conjunction A ∧ B : A and B

Existential quantification ∃ x A(x) : There exists an element x in the domain such that A(x)

Universal Quantification ∀x A(x) : A(x) for all values of x in the domain

So, using the above interpretations, we can answer the questions as;

a) ∃x(C(x) ∧ D(x) ∧ F(x))

b) ∀x(C(x) ∨ D(x) ∨ F(x))

c) ∃x(C(x) ∧ D(x) ¬ F(x))

d) ¬∃x(C(x) ∧ D(x) ∧ F(x))

e) (∃xC(x)) ∧ (∃xD(x)) ∧ (∃xF(x))

(1,4) is on graph of f(x)

f(1) = 4

so 1/4 f(x)

1/4 f(1)

1/4. ×4

1

so point for function 1/4. f(x) is (1,1)