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Zinaida [17]
3 years ago
10

Which of the following statements are true about the graph of f(x)= csc x? select all that apply

Mathematics
2 answers:
erik [133]3 years ago
6 0

Answer:

c. and d

Step-by-step explanation:

Maurinko [17]3 years ago
3 0

Answer:

c. There is a vertical asymptote at x = pie

d. f(x) is undefined when sin x=0

Step-by-step explanation:

The given function is f(x)=\csc(x).

This is the reciprocal of the sine function;

f(x)=\frac{1}{\sin(x)}

This function is not defined when

sin(x)=0

This implies that there is a vertical asymptote at x=\pi because \sin(\pi)=0.

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NEED HELP FAST‼️‼️
ch4aika [34]

Answer:

Step-by-step explanation:

A ' = (-2, -3)

B ' = (0, -3)

C ' = (-1, 1)

=======================================================

Explanation:

To apply an x axis reflection, we simply change the sign of the y coordinate from positive to negative, or vice versa. The x coordinate stays as is.

Algebraically, the reflection rule used can be written as

Applying this rule to the three given points will mean....

Point A = (-2, 3) becomes A ' = (-2, -3)

Point B = (0, 3) becomes B ' = (0, -3)

Point C = (-1, -1) becomes C ' = (-1, 1)

The diagram is provided below.

Side note: Any points on the x axis will stay where they are. That isn't the case here, but its for any future problem where it may come up. This only applies to x axis reflections.

4 0
3 years ago
Please help me answer these. Its all about upper and lower bounds.
sergeinik [125]

Answer:

red question= 8.5 yellow question=4.25

Step-by-step explanation:

5 0
3 years ago
HELP ME PLEASE !!!!!!!!
anzhelika [568]

Answer:

the second one is the answer

8 0
2 years ago
Read 2 more answers
A design engineer wants to construct a sample mean chart for controlling the service life of a halogen headlamp his company prod
tekilochka [14]

Answer:

C) 515 hours.

D) 500 hours

c) sample 3

Step-by-step explanation:

1. Sample 2 mean = x2`= ∑x2/n2= 2060/4= 515 hours

Sample Service Life (hours)

1                2              3

495      525            470

500         515           480

505        505            460

<u>500         515             470        </u>

<u>∑2000     2060         1880</u>

x1`= ∑x1/n1= 2000/4= 500 hours

x2`= ∑x2/n2= 2060/4= 515 hours

x3`= ∑x3/n3= 1880/4=  470 hours

2. The mean of the sampling distribution of sample means for whenever service life is in control is 500 hours . It is the given mean in the question and the limits are determined by using  μ ± σ , μ±2 σ  or μ ± 3 σ.

In this question the limits are determined by using  μ ± σ .

3. Upper control limit = UCL = 520 hours

Lower Control Limit= LCL = 480 Hours

Sample 1 mean = x1`= ∑x1/n1= 2000/4= 500 hours

Sample 2 mean = x2`= ∑x2/n2= 2060/4= 515 hours

Sample 3 mean = x3`= ∑x3/n3= 1880/4=  470 hours

This means that the sample mean must lie within the range 480-520 hours but sample 3 has a mean of 470 which is out of the given limit.

We see that the sample 3 mean is lower than the LCL. The other  two means are within the given UCL and LCL.

This can be shown by the diagram.

8 0
2 years ago
I NEED HELP ! <br> 2=log(3x+4)
dybincka [34]

2=log(3x+4)


y = loga b    a^y = b

since this is base 10

10^2 =3x+4

100 = 3x+4

subtract 4 from each side

96 =3x

divide by 3

32=x


5 0
3 years ago
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