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8090 [49]
3 years ago
13

Determine the equations of the vertical and horizontal asymptotes, if any, for y=x^3/(x-2)^4

Mathematics
2 answers:
klio [65]3 years ago
7 0
ANSWER

The correct answer is A.

EXPLANATION

The given function is,

y = \frac{ {x}^{3} }{ {(x - 2)}^{4} }

To find the vertical asymptote, we equate the denominator to zero.

This implies that,

{(x - 2)}^{4} = 0

x - 2 = 0

x = 2

To find the horizontal asymptote,we take limit to infinity.

lim_{x\rightarrow \infty} \frac{ {x}^{3} }{ {(x - 2)}^{4} } =0

The horizontal asymptotes is

y=0
djverab [1.8K]3 years ago
3 0

Answer:

Option a)

Step-by-step explanation:

To get the vertical asymptotes of the function f(x) you must find the limit when x tends k of f(x). If this limit tends to infinity then x = k is a vertical asymptote of the function.

\lim_{x\to\\2}\frac{x^3}{(x-2)^4} \\\\\\lim_{x\to\\2}\frac{2^3}{(2-2)^4}\\\\\lim_{x\to\\2}\frac{2^3}{(0)^4} = \infty

Then. x = 2 it's a vertical asintota.

To obtain the horizontal asymptote of the function take the following limit:

\lim_{x \to \infty}\frac{x^3}{(x-2)^4}

if \lim_{x \to \infty}\frac{x^3}{(x-2)^4} = b then y = b is horizontal asymptote

Then:

\lim_{x \to \infty}\frac{x^3}{(x-2)^4} \\\\\\lim_{x \to \infty}\frac{1}{(\infty)} = 0

Therefore y = 0 is a horizontal asymptote of f(x).

Then the correct answer is the option a) x = 2, y = 0

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Factor as the product of two binomials x^2-3x-10
cestrela7 [59]

Answer:

(x-5)(x+2)

Step-by-step explanation:

Hello there!

Your expression there has a highest power of x^{2}, so there will be two constants that you need to find.

Your last term is -10, so the constants will have a product of -10. Also your middle term is -3x, so the terms will add up to -3

-5 and 2 fit the mold.

Have a great day!

If I am most helpful, mark me brainliest!

3 0
3 years ago
Use your understanding of the unit circle and trigonometric functions to find the values requested.
vfiekz [6]

Answer:

a) For this case we can use the fact that sin (\pi/3) = \frac{\sqrt{3}}{2}

And for this case since we ar einterested on -\frac{\pi}{3} and we know that the if we are below the y axis the sine would be negative then:

sin (-\pi/3) = -\frac{\sqrt{3}}{2}

b) From definition we can use the fact that tan x= \frac{sin x}{cos x} and we got this:

tan (5\pi/4) = \frac{sin(5\pi/4)}{cos(5\pi/4)}

We can use the notabl angle \pi/4 and we know that :

sin (\pi/4) = cos(\pi/4) = \frac{\sqrt{2}}{2}

Then we know that 5\pi/4 correspond to 225 degrees and that correspond to the III quadrant, and we know that the sine and cosine are negative on this quadrant. So then we have this:

tan (5\pi/4) = \frac{sin(5\pi/4)}{cos(5\pi/4)}= \frac{\frac{sqrt{2}}{2}}{\frac{\sqrt{2}}{2}} = 1

Step-by-step explanation:

For this case we can use the notable angls given on the picture attached.

Part a

For this case we can use the fact that sin (\pi/3) = \frac{\sqrt{3}}{2}

And for this case since we ar einterested on -\frac{\pi}{3} and we know that the if we are below the y axis the sine would be negative then:

sin (-\pi/3) = -\frac{\sqrt{3}}{2}

Part b

From definition we can use the fact that tan x= \frac{sin x}{cos x} and we got this:

tan (5\pi/4) = \frac{sin(5\pi/4)}{cos(5\pi/4)}

We can use the notabl angle \pi/4 and we know that :

sin (\pi/4) = cos(\pi/4) = \frac{\sqrt{2}}{2}

Then we know that 5\pi/4 correspond to 225 degrees and that correspond to the III quadrant, and we know that the sine and cosine are negative on this quadrant. So then we have this:

tan (5\pi/4) = \frac{sin(5\pi/4)}{cos(5\pi/4)}= \frac{\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}} = 1

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3 years ago
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Archy [21]

Answer: 40

Step-by-step explanation:

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makkiz [27]

Answer: lmk

Step-by-step explanation:

8 0
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9966 [12]

what is your clear question? it lacks some details

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