What are the vertical asymptotes of the function f(x) = the quantity of 2 x plus 8, all over x squared plus 5 x plus 6?

Answers

Answer 1

Answer: You'd find the vertical asymptotes by seeing where the denominator equals zero; you can do so by factoring the denominator.
In this case, you can factor the denominator into (x+3)(x+2), so if you set each of those equal to zero you can find the equations of the vertical asymptotes (x=-3 and x=-2).

Answer 2

Answer:

Answer:

Step-by-step explanation:

Alright, lets get started.

The given rational function is :

$f(x)=(2x+8)/(x^2+5x+6)$

For finding the vertical asymptotes of a rational function, we must set the denominator equal to zero.

So, equaling denominator to zero :

$x^2 +5x+6 = 0$

factoring

$(x+3)(x+2)=0$

This will give two values of x

$x=-3, x=-2$

So, these two are vertical asymptotes : Answer

Hope it will help :)

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You have 60 marbles in a jar, two of which are red. (so 2 red, 58 not red)You pick 10 at random. What is the chance that you draw both red marbles?
Explanation please, not just the answer. Brainliest answer if you calculate it for 2 out of 251 total marbles as well.

Answers

Well, I agonized over this for a while, and I have something that we may want to consider.

The only probability formula I know how to use is

Probability = (number of possible successes) / (total number of possibilities).

Can we work with that ? Let's see . . .

The denominator of that fraction is the total number of ways to draw
10 marbles from a jar of 60 . . .

The first draw can be any one of 60 marbles. For each of those . . .
The second draw can be any one of the remaining 59. For each of those . . .
The third draw can be any one of the remaining 58.
.
.
etc.

So the total number of ways to draw 10 from 60 is

   (60 x 59 x 58 x 57 x 56 x 55 x 54 x 53 x 52 x 51)

That's a very big number. My calculator says something that rounds to
2.736 x 10¹⁷. But my calculator only shows 10 digits, so it can't show
all 18 digits in the number. Fortunately, we don't need to see the whole
number written out. We'll just write it in factorial notation, and go on
to do the numerator of the fraction, which is going to be much harder.

That number that we just found is equal to (60!) / (50!) . It's going to be
the denominator of the big fraction.

Now for the numerator. That's going to be the number of ways that the
two red marbles can be included among the ten marbles drawn.

One red marble can be any one of the 10 marbles pulled out.
   For each of those . . .
The other red marble can be any one of the other 9 that are picked.

So there are (10 x 9) = 90 ways for the selected 10 to include both red ones.

SO ! Now, the probability that the ten that are drawn will include the two
red ones should be (it might be, it could be) . . .

   90 divided by (60! / 50!)   .

Remember, that fraction on the right is just total number of ways
to pick 10 out of 60 . The probability of including both red ones
in the draw is 90 divided by that number. It's very small.

Again from my calculator, it's 3.29 x 10⁻¹⁴ percent.

I have no confidence in my answer, but I invite you to look it over, along with
all the real gurus out there.

If I'm wrong, then I've stolen only 5 points that I'm not entitled to, and at least
I did put some effort into it.

Here you have two proportions. 2/60 and 10/60. I would cross multiply.
Giving you 20/120.
Simplify that to get 1/6. So you have a 1 in 6 chance of getting both the red marbles.
(I'm not 100% sure since it's been a while so you might want to get a second opinion.)
Not sure what you mean by the second part, but hope I provided some help :)

You are challenged to a lucky draw game. If you draw a face card (K, Q, J) from a standard deck of cards, you earn 10 points. If you draw any other card, you lose 2 points. What is the expected value that you can get if you draw another card?

Answers

To problem wants to compute the probability of the expected value that could be earn if you would draw again another card. So base on the fact that if draw a face card of K,Q,J from a standard deck correctly you will earn 10 point in just one draw, but loose 2 points if you another card, the probability that you would draw correct card in just one draw is 30.77%, so the possible value earn you have 2 points only

Answer:

A: 0.77

Step-by-step explanation:

Edg

What is 23 squared A. 1,058 B.129 C. 46 D. 529

Answers

The value of 23 squared is 529

What is square of a number?

A square number or perfect square is an integer that is the square of an integer; in other words, it is the product of some integer with itself.

For example, the 3² is 3 × 3 i.e the product of the 3 by itself , which is 9

Similarly, the square of 23 is 23² which is the same as the 23 × 23

= 529

Therefore the square of 23 is 529. this means that 23² = 529 and the square root of 529 is 23.

learn more about square of numbers from

brainly.com/question/27307830

#SPJ6

23 squared is 529 because 23 x 23 is 529, so the answer is D.

Solve for x
log(x–4)+log(x+3)=2logx

Answers

$D:x-4>0 \wedge x+3>0 \wedge x>0\nD:x>4 \wedge x>-3 \wedge x>0\nD:x>4\n\n\log(x-4)+\log(x+3)=2\log x\n\log(x-4)(x+3)=\log x^2\n(x-4)(x+3)=x^2\nx^2+3x-4x-12=x^2\n-x=12\nx=-12\n-12\not\in D\n\Downarrow\n\boxed{x\in\emptyset}$

This one to its alot

Answers

Chejxhebdhdhe r dhdhebdbdjsbd d euddhsbifif

Evaluate the expression below for x = 4, y = –5, and z = 2. 3x^2 -4y+xza. 12
b. 36
c. 52
d. 76

Answers

3(4)^2 -4(-5) + (4)(2)
= 3(16) + 20 + 8
= 48 + 20 + 8
= 76 (d)

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