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valkas [14]
4 years ago
8

The graph below shows the function f(x)=x-3/x^2-2x-3 which statement is true

Mathematics
2 answers:
marishachu [46]4 years ago
7 0

Answer:

Its A. I just passed the final exam with 5 minutes left of the SESSION.

Step-by-step explanation:

Ugo [173]4 years ago
4 0

Answer:

The correct option is A.

Step-by-step explanation:

Domain:

The expression in the denominator is x^2-2x-3

x² - 2x-3 ≠0

-3 = +1 -4

(x²-2x+1)-4 ≠0

(x²-2x+1)=(x-1)²

(x-1)² - (2)² ≠0

∴a²-b² =(a-b)(a+b)

(x-1-2)(x-1+2) ≠0

(x-3)(x+1) ≠0

x≠3 for all x≠ -1

So there is a hole at x=3 and an asymptote at x= -1, so Option B is wrong

Asymptote:

x-3/x^2-2x-3

We know that denominator is equal to (x-3)(x+1)

x-3/(x-3)(x+1)

x-3 will be cancelled out by x-3

1/x+1

We have asymptote at x=-1 and hole at x=3, therefore the correct option is A....

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Alekssandra [29.7K]

Answer:

31 men are enrolled

Step-by-step explanation:

1 to 5

Divide 186 by 6. Then you would get 31 since that is the unit rate.

To check, multiply 31 by 5 and you would get 155, add 155 with 31, you would get 186. I don't really know how to explain it but thats my answer.

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3 years ago
Solving algebraic equationsSolve:1/5m = 4 - 3/10m
sertanlavr [38]

EXPLANATION

Given the equation:

1/5m = 4-3/10m (Adding +3/10 to both sides)

1/5m +3/10m = 4 (Adding 1/5 + 3/10)

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3 0
1 year ago
Elimination method 3x+4y=18 9x-5y=3
tatiyna

Answer:

The solution would be (10, -3)

Step-by-step explanation:

In order to solve using elimination, multiply the top equation by -3 and then add the equations together.

-9x - 12y = -54

9x - 5y = 3

---------------

17y = -51

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Now that we have a value for y, we can put it into either equation and solve for x.

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4 0
3 years ago
Algebra 2 help please!
mezya [45]

Answer:

h(-8)=1 and -12

Step-by-step explanation:

In this problem, we will need to plug in "-8" for any x we see.

h(-8)=|-8+4|-3

h(-8)= |-4|-3

h(-8)=4-3

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Yay! We got an answer. However, since there is an absolute value, there is another case -- if x is negative.

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When \theta terminates in quadrant III, both \cos\theta and \sin\theta are negative, and

\sin^2\theta+\cos^2\theta=1\implies\cos\theta=-\sqrt{1-\sin^2\theta}=-\dfrac{12}{13}

When \varphi terminates in quadrant II, \cos\varphi is negative and \sin\varphi is positive, so

1+\tan^2\varphi=\sec^2\varphi\implies\sec\varphi=-\dfrac{17}{15}

which gives

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\tan\varphi=\dfrac{\sin\varphi}{\cos\varphi}=-\dfrac8{15}\implies\sin\varphi=\dfrac8{17}

Now,

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7 0
3 years ago
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