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frutty [35]
1 year ago
9

What are the equations of the asymptotes for the function? f(x)=−3/x−1 −1

Mathematics
1 answer:
Delvig [45]1 year ago
8 0

The vertical asymptote is x = 1 and the horizontal asymptote is y = -1 if the equation of the hyperbola is f(x)=−3/x−1 −1

<h3>What is hyperbola?</h3>

It's a two-dimensional geometry curve with two components that are both symmetric. In other words, the number of points in two-dimensional geometry that have a constant difference between them and two fixed points in the plane can be defined.

We have the equation of the hyperbola:

\rm  f(x)=\dfrac{-3}{x-1} -1

The asymptotes will be

Vertical asymptote:        x -1 = 0 ⇒ x = 1

Horizontal asymptote:   y + 1= 0 ⇒ y = -1

Thus, the vertical asymptote is x = 1 and the horizontal asymptote is y = -1 if the equation of the hyperbola is f(x)=−3/x−1 −1

Learn more about the hyperbola here:

brainly.com/question/12919612

#SPJ1

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The equation x²– 8x – 5 = 0 can be transformed into the equation (x - p)² = q, where p and q are real numbers. What are the valu
Juliette [100K]

The form (x - p)^2 = q is the form that completing the square will leave us with.

---Add 5 to both sides

x^2 - 8x = 5

---Divide the b term by 2, and square it. Then, add that number to both sides.

-8/2 = -4

(-4)^2 = 16

x^2 - 8x + 16 = 5 + 16

x^2 - 8x + 16 = 21

---Factor!

(x - 4)^2 = 21

p = 4

q = 21

Hope this helps!! :)

5 0
2 years ago
2(8r+5)-3=4(4r-1)+11 is it an only solution or a no solution or infinite solution
pogonyaev

Answer:

Infinitely many solutions.

Step-by-step explanation:

Let's begin by carrying out the indicated multiplications, which must be done before any addition or subtraction:

2(8r+5)-3=4(4r-1)+11  becomes  16r + 10 - 3 = 16r - 4 + 11.

Subtracting 16r from both sides, we get 10 - 3 = - 4 + 11, or 7 = 7

This is always true, so we can conclude that this equation has infinitely many solutions.

3 0
2 years ago
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Answer:

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Step-by-step explanation:

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Step-by-step explanation:

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I am pretty sure your answer is going to be 243
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