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

if the eqn of the parabola is f(x) = (x-4)(x+2), we need to mult. this out to find the coordinates of the vertex.

then f(x) = x^2 + 2x - 4x - 8, and f(x) = x^2 - 2x - 8.

one way in which you can find the vertex is to calculate x = -b / (2a).   here, that comes to

        )

x = = 1.   f(1) = 1^2 - 2(1) - 8 = - 9.   vertex is at (1, -9).

        2(1) 

find another point on the curve:   let x=0, for which y=-8.   then (0, -8) is on the curve.   plot the vertex and this point and draw a smooth parabola thru both.

hope this

geometric series converge the common ratio r has magnitude less than one.

which is equivalent to

for the example we need

|2-x| < 1

-1 < 2-x   and 2-x < 1

x < 3 and x > 1

we can write that

answer: 1 < x < 3

1a. common ratio r=250/25=10 > 1     diverges

1b.   r=50/100=1/2 converges

when these converge they converge to

where a is the first term.

answer: 200

1c. r=-1   so it's not the case that |r| < 1.     diverges

2a. x is the common ratio so we converge when |x|< 1

2b.   r=-10x / -2 = 5x.

we converge when

|5x| < 1

|x| < 1/5

2c.   r=3-2x

-1 < 3-2x and 3-2x < 1

-4 < -2x and 2 < 2x

x < 2 and 1 < x

1 < x < 2

2d. [/tex] r=-48x^3/12 = 4x^3[/tex]

|4x^3| < 1

|x^3| < \frac 1 4

|x| < 4^{-1/3}