Identify the horizontal asymptote of f(x) =x2+5x-3/4x-1

1 answer:

5 0

since the numerator is x² + 5x - 3, and therefore has a degree of 2, whilst the denominator, 4x¹ - 1, has a degree of 1, therefore, there's no horizontal asymptote.

recall, we only get a horizontal asymptote if the denominator's expression degree is equals or greater than that of the numerator's.

You might be interested in

The number of revolutions made by a figure skater for each type of axel jump is given. Determine the measure of the angle genera

trapecia [35]

For radians, multiply the number of jump revolutions by 2 pi

<span>For degrees, multiply the number of revolutions by 360

Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.
</span>

4 0

3 years ago

Choose the correct simplification of (4x − 3)(3x2 − 4x − 3).

tester [92]

$(4x - 3)(3x^2 - 4x - 3)$

$=4x\cdot \:3x^2+4x\left(-4x\right)+4x\left(-3\right)-3\cdot \:3x^2-3\left(-4x\right)-3\left(-3\right)$

$=4x\cdot \:3x^2-4x\cdot \:4x-4x\cdot \:3-3\cdot \:3x^2+3\cdot \:4x+3\cdot \:3$

$\mathrm{Simplify}\:4x\cdot \:3x^2-4x\cdot \:4x-4x\cdot \:3-3\cdot \:3x^2+3\cdot \:4x+3\cdot \:3:\quad 12x^3-25x^2+9$

$=12x^3-25x^2+9$

8 0

3 years ago

Class records at rockwood college indicate that a student selected at random has probability 0.75 of passing french 101. for the

Kipish [7]

To get the probability of two individual events both occurring, you have to multiply the probabilities of their individual events occurring. Therefore in this problem, the probability that a student selected at random will pass both French 101 and French 102 is 0.683 (.75 x .91). The answer is already rounded to three decimal places.

4 0

4 years ago

Read 2 more answers

What is 10 1/2 divded by 2 1/4?

Fittoniya [83]

Answer:

4.66666666667

Step-by-step explanation:

8 0

2 years ago

Read 2 more answers

Which is equivalent to √60n^11/256n^4 after it has been simplified completely? Assume n =0

motikmotik

Answer:

$\large\boxed{\dfrac{n^3}{8}\sqrt{15n}}$

Step-by-step explanation:

$\sqrt{\dfrac{60n^{11}}{256n^4}}\qquad\text{use}\ \dfrac{a^n}{a^m}=a^{n-m}\ \text{and}\ \sqrt{\dfrac{a}{b}}=\dfrac{\sqrt{a}}{\sqrt{b}}\\\\=\sqrt{\dfrac{60}{256}n^{11-4}}=\dfrac{\sqrt{60n^7}}{\sqrt{256}}=\dfrac{\sqrt{4n^{6+1}\cdot15}}{16}\qquad\text{use}\ a^n\cdot a^m=a^{n+m}\\\\=\dfrac{\sqrt{4n^6\cdot15n}}{16}\qquad\text{use}\ \sqrt{ab}=\sqrt{a}\cdot\sqrt{b}\\\\=\dfrac{\sqrt{4n^{3\cdot2}}\cdot\sqrt{15n}}{16}\qquad\text{use}\ (a^n)^m=a^{nm}$

$=\dfrac{\sqrt{4(n^3)^2}\cdot\sqrt{15n}}{16}=\dfrac{\sqrt{4}\cdot\sqrt{(n^3)^2}\cdot\sqrt{15n}}{16}=\dfrac{2\!\!\!\!\diagup^1\sqrt{15n}}{16\!\!\!\!\!\diagup_8}=\dfrac{n^3\sqrt{15n}}{8}$

5 0

4 years ago