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vichka [17]
4 years ago
8

Which function's graph has asymptotes located at the values x= ±nπ?

Mathematics
1 answer:
Marta_Voda [28]4 years ago
4 0

Answer:

y=csc(\theta)

y=cot(\theta)

Step-by-step explanation:

I will just change

x=\pm n\pi, n \in \mathbb{Z}_{\ge 0}

For

\theta = \pi n, n\in\mathbb{Z}

Also, note that sine and cosine function don't have asymptotes.

The vertical asymptotes of cosecant occur every \pi n, n\in\mathbb{Z}

It happens because the cosecant function is undefined for those values.

The cotangent function has asymptotes located at every integer multiple of \pi.

On the other hand, the vertical asymptotes of tangent function occur at:  

$\theta=\frac{\pi}{2} +n \pi, n \in \mathbb{Z}$

It happens because the tangent function is undefined for those values.

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How are the coordinates of the extreme value and the equation from part B in the form y=(x-h)^2-c related.
True [87]

The extreme value of y = (x - h)² - c is the vertex of the equation.

<h3>What are extreme values of a function?</h3>

The extreme values of a function are either the maximum or minimum values of the function.

<h3>The equation of a parabola in vertex form</h3>

The equation of a parabola in vertex form with vertex (h', k) is given by

y = a(x - h')² + k and its extreme values are

  • if a > 0  (h', k) is a minimum point and
  • if a < 0 (h', k) is a maximum point.

Since y = (x - h)² - c is the equation of a parabola in vertex form, comparing with y = a(x - h')² + k,

  • h = h'
  • -c = k and
  • a = 1.

So, the cooordinates of the vertex of y = (x - h)² - c is (h, -c).

Now, since a = 1 > 0, (h, -c) is a minimum point.

So, the extreme value of y = (x - h)² - c is the vertex of the equation.

Learn more about extreme value of a function here:

brainly.com/question/13464558

#SPJ1

4 0
2 years ago
A researcher studied the radioactivity of asbestos. She sampled 81 boards of asbestos, and found a sample mean of 193.2 bips, an
Ksivusya [100]

Answer:

a) Confidence interval is (182.86,203.54).

b) (i) Yes, 200 bips is a true mean as it lie in the interval.

(ii) No, 210 bips is not a true mean as it doesn't lie in the interval.

Step-by-step explanation:

Given : A researcher studied the radioactivity of asbestos. She sampled 81 boards of asbestos, and found a sample mean of 193.2 bips, and a sample standard deviation of 49.5 bips.

To find : (a) Obtain the 94% confidence interval for the mean radioactivity. (b) (i) According the interval that you got, is 200 bips a plausible value for the true mean? (ii) What about 210 bips?

Solution :

a) The confidence interval formula is given by,

\bar{x}-z\times \frac{\sigma}{\sqrt{n}}

We have given,            

The sample mean \bar{x}=193.2 bips

s is the standard deviation \sigma=49.5 bips

n is the number of sample n=81

z is the score value, at 94% z=1.88

Substitute all the values in the formula,

193.2-1.88\times \frac{49.5}{\sqrt{81}}

193.2-1.88\times 5.5

193.2-10.34

182.86

Confidence interval is (182.86,203.54).    

b) (i) According the interval 182.86

200 bips a plausible value for the true mean as it lies in the interval.

(ii) 210 bips not lie in the confidence interval so it is not a true mean.

3 0
4 years ago
A catering company needs 8.75 lbs of shrimp for a small party. They buy 3 2/3 lbs of jumbo shrimp, 2 5/8 lbs of medium-sized shr
Brums [2.3K]
8.75 - (3 2/3 + 2 5/8) 
8.75 - (11/3 + 21/8)
8.75 - (88/24 + 63/24)
8.75 - 151/24
210/24 - 151/24
59/24 = 2 11/24 lbs of mini shrimp <==
3 0
3 years ago
Read 2 more answers
Find the value of x if x/3-2=6
PolarNik [594]

Step-by-step explanation:

hope it is helpful to you

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6 0
3 years ago
Read 2 more answers
Parallel and Perpendicular Lines
larisa86 [58]
The answer is B I’m pretty sure
6 0
3 years ago
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