Consider the following statement. ∀ integer d, if 6 d is an integer then d = 3. Which of the following is a negation for the statement? ∃ an integer d such that 6 d is an integer and d ≠ 3. ∃ an integer d such that 6 d is not an integer and d = 3. ∀ integer d, if 6 d is not an integer, then d ≠ 3. ∀ integer d, if d ≠ 3, then 6 d is not an integer. ∀ integer d, if d = 3, then 6 d is an integer.

The correct option is the first:

∃ an integer d such that 6d is an integer and d ≠ 3

Step-by-step explanation:

If 6d is an integer, then d = 3

Let's divide the statement into two:

Let "If 6d is an integer" be p and "then d = 3" be q.

The statement is a conditional statement. Therefore, we have

p ⇒ q

The negation of p ⇒q, ¬(p ⇒ q) ≡ p ∧ ¬q

I have attached an image proving the above.

Using this: ¬(p ⇒ q) ≡ p ∧ ¬q

Then, we leave p as it is and negate q.

P is "if 6d is an integer"

q is "then d = 3," hence ¬q is "then d ≠ 3."

Therefore, the negation of "If 6d is an integer, then d = 3" is "If 6d is an integer, then d ≠ 3."

The correct option is the first:

∃ an integer d such that 6d is an integer and d ≠ 3

the direct variation says that:

then the equation is of the form is given by:

            where k is the constant variation.

from the given table:

consider any value of x and m(x)=y

x = 2 and m(x)=y = 3

substitute these value in [1]; we have;

divide both sides by 2 we get;

⇒

check:

put x = -1 and m(x) = -1.5

substitute in the equation:

-1.5 = -1.5                   true

therefore, the function m(x) represents a direct variation

graph of this function as shown below:

the answer is b.)

< 1=< 2