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MaRussiya [10]
4 years ago
12

A sinusoidal function whose frequency is 1/6pi

Mathematics
2 answers:
Lyrx [107]4 years ago
8 0

Answer:

B. f(x)=9\sin (\frac{x}{3})+3

Step-by-step explanation:

We are given that,

The frequency of the function is \frac{1}{6\pi}

The maximum and minimum value is 12 and -6.

Also, the y-intercept is 3.

From the options, we have,

Options C and D have minimum value 6. So, they does not represent the given function.

We know, 'If a function has a period P, then the function a+f(bx+c) will have the period \frac{P}{|b|}.

Also, 'The frequency is the reciprocal of the period'.

So, function a+f(bx+c) will have the frequency \frac{|b|}{P}.

From the options, we see,

Option B have the frequency, \frac{\frac{1}{3}}{2\pi} i.e. \frac{1}{6\pi}.

Option A have the frequency, \frac{6\pi}{2\pi} i.e. 3

Thus, option A is not correct.

Hence, option B is the required sinusoidal function.

mart [117]4 years ago
4 0

Answer:

The answer is B

Step-by-step explanation:

just took the test

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Answer:

B) 3(7x + 2)

Step-by-step explanation:

6x is 6x and x is 1x but you don't usually put the 1 infront of the x.

3 0
3 years ago
Determine if each statement is True or False. If False, determine what needs to be
kicyunya [14]

Answer:

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c. True.

8 0
3 years ago
0.770000 in scientific notations
Masja [62]
All you have to do is to move the decimal point 1 place to the right and then append E1 to the result:

0.770000 = <span>7.70000E1</span>
6 0
3 years ago
Read 2 more answers
Use the given graph to determine the limit, if it exists. A coordinate graph is shown with a horizontal line crossing the y axis
Lesechka [4]

Answer:

The limit of the function does not exists.

Step-by-step explanation:

From the graph it is noticed that the value of the function is 6 from all values of x which are less than 2. At x=2, the line y=6 has open circle. It means x=2 is not included.

For x<2

f(x)=6

The value of the function is -3 from all values of x which are greater than 2. At x=2, the line y=-3 has open circle. It means x=2 is not included.

For x>2

f(x)=-3

The value of y is 1 at x=2, because of he close circles on (2,1).

For x=2

f(x)=1

Therefore the graph represents a piecewise function, which is defined as

f(x)=\begin{cases}6& \text{ if } x2 \end{cases}

The limit of a function exist at a point a if the left hand limit and right hand limit are equal.

lim_{x\rightarrow a^-}f(x)=lim_{x\rightarrow a^+}f(x)

The function is broken at x=2, therefore we have to find the left and right hand limit at x=2.

lim_{x\rightarrow 2^-}f(x)=6

lim_{x\rightarrow 2^+}f(x)=-3

6\neq-3

Since the left hand limit and right hand limit are not equal therefore the limit of the function does not exists.

6 0
3 years ago
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PLEASE HELP ASAP!!! Evaluate 7p + 6(p ÷ q)^2 - 2q (if p = 6 and q = 3). Show your work and explain each step.
Arturiano [62]

Answer:

60

Step-by-step explanation:

Evaluate 7p + 6(p ÷ q)^2 - 2q (if p = 6 and q = 3)

7p + 6(p \div q)^2 - 2q\\\\7(6) + 6(6 \div 3)^2 - 2(3)\\\\7(6) + 6(2)^2 - 2(3)\\\\7(6) + 6(4) - 2(3)\\\\42 + 24 - 6\\\\66-6\\\\60

Hope this helps!

4 0
3 years ago
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