Let f (x, y) be the statement "x can fool y," where the domain consists of all people in the world. use quantifiers to express each of these statements. a) everybodycanfoolfred. b) evelyn can fool everybody. c) everybody can fool somebody. d) there is no one who can fool everybody. e) everyone can be fooled by somebody. f) . g) .

a) ∀x ∃f F(x, f)

b) ∃e ∀y F(e, y)

c) ∀x ∃y F(x, y)

d) ¬∃x ∀y F(x, y) ≡ ∃x ∀y ¬F(x, y)

e) ∃x ∃y F(x, y)

f) ¬∃x ∃f ∃j [F(x, f) ∧ F(x, j)] ≡ ∃x ∃f ∃j ¬[F(x, f) ∧ F(x, j)]

g) ∃n ∃a ∃b [F(n, a) ∧ F(n, b)]

Step-by-step explanation: F (x, y) "x can fool y,"

domain consists of all people in the world

a) Everybody can fool Fred.

Let's say Fred = f ∈ y

∀x ∃f F(x, f)

b) Evelyn can fool everybody.

Let's say Evelyn = e ∈ x

∃e ∀y F(e, y)

c) Everybody can fool somebody.

∀x ∃y F(x, y)

d) There is no one who can fool everybody.

¬∃x ∀y F(x, y) ≡ ∃x ∀y ¬F(x, y)

e) Everyone can be fooled by somebody.

∃x ∃y F(x, y)

f) No one can fool both Fred and Jerry.

Let's say Fred = f ∈ y

Let's say Jerry = j ∈ y

¬∃x ∃f ∃j [F(x, f) ∧ F(x, j)] ≡ ∃x ∃f ∃j ¬[F(x, f) ∧ F(x, j)]

g) Nancy can fool exactly two people.

Let's say Nancy = n ∈ x

Let's say person 1 = a ∈ y

Let's say person 2 = b ∈ y

∃n ∃a ∃b [F(n, a) ∧ F(n, b)]

x ≥ 18 is your answer

step-by-step explanation:

let "a number" = x

(2/3)x + 17 ≥ 29

isolate the variable (x). do the opposite of pemdas (parenthesis, exponent (& root), multiplication, division, addition, subtraction).

first, subtract 17 from both sides.

(2/3)x + 17 (-17) ≥ 29 (-17)

(2/3)x ≥ 29 - 17

(2/3)x ≥ 12

next, multiply 3/2 to both sides to isolate the x

(3/2)(2/3)x ≥ 12(3/2)

x ≥ 12(3/2)

simplify. multiply, then divide

x ≥ (12 * 3)/2

x ≥ (36)/2

x ≥ 18

x ≥ 18 is your answer

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step-by-step explanation: