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3 years ago

Choose correct graph with following function

Mathematics

1 answer:

anyanavicka [17]3 years ago

6 0

Answer:

Step-by-step explanation:

From the function, y = cos (3x)

Inputting values of x (horizontal axis), x = pi

y = cos (3 × pi)

= -1 ( remember you are dealing with values in radians not degree)

So now graph 1 and 3 have the same values. So inputting, x = pi/2 into the function, y = cos (3x)

= cos ((3 × pi)/2)

= 0

View the correct graph below.

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Given z1 and z2, what is the midpoint?<br><br> z1=5-7i and z2=-9+3i.

sveta [45]

Answer:

$7-2i$

Step-by-step explanation:

As you're used to do in the real plane, midpoint is the average of Real and Imaginary part. So $\frac{5+9}2 +i \frac{-7+3}2 = \frac{14}2+i\frac{-4}2 = 7-2i$

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2 years ago

What is the difference of 13.2 - 6.56

vichka [17]

Answer:

6.64

Step-by-step explanation:

Subtract : 13.2 - 6.56

13.2(0)

-6.56

----------

6.64

8 0

3 years ago

Read 2 more answers

The number by which the dividend is being divided? Means?

Ann [662]

That would be called the quotient.

7 0

4 years ago

Read 2 more answers

Solve using Fourier series.

Olin [163]

With $2L=\pi$ , the Fourier series expansion of $f(x)$ is

$\displaystyle f(x)\sim\frac{a_0}2+\sum_{n\ge1}a_n\cos\dfrac{n\pi x}L+\sum_{n\ge1}b_n\sin\dfrac{n\pi x}L$
$\displaystyle f(x)\sim\frac{a_0}2+\sum_{n\ge1}a_n\cos2nx+\sum_{n\ge1}b_n\sin2nx$

where the coefficients are obtained by computing

$\displaystyle a_0=\frac1L\int_0^{2L}f(x)\,\mathrm dx$
$\displaystyle a_0=\frac2\pi\int_0^\pi f(x)\,\mathrm dx$

$\displaystyle a_n=\frac1L\int_0^{2L}f(x)\cos\dfrac{n\pi x}L\,\mathrm dx$
$\displaystyle a_n=\frac2\pi\int_0^\pi f(x)\cos2nx\,\mathrm dx$

$\displaystyle b_n=\frac1L\int_0^{2L}f(x)\sin\dfrac{n\pi x}L\,\mathrm dx$
$\displaystyle b_n=\frac2\pi\int_0^\pi f(x)\sin2nx\,\mathrm dx$

You should end up with

$a_0=0$
$a_n=0$
(both due to the fact that $f(x)$ is odd)
$b_n=\dfrac1{3n}\left(2-\cos\dfrac{2n\pi}3-\cos\dfrac{4n\pi}3\right)$

Now the problem is that this expansion does not match the given one. As a matter of fact, since $f(x)$ is odd, there is no cosine series. So I'm starting to think this question is missing some initial details.

One possibility is that you're actually supposed to use the even extension of $f(x)$ , which is to say we're actually considering the function

$\varphi(x)=\begin{cases}\frac\pi3&\text{for }|x|\le\frac\pi3\\0&\text{for }\frac\pi3$

and enforcing a period of $2L=2\pi$ . Now, you should find that

$\varphi(x)\sim\dfrac2{\sqrt3}\left(\cos x-\dfrac{\cos5x}5+\dfrac{\cos7x}7-\dfrac{\cos11x}{11}+\cdots\right)$

The value of the sum can then be verified by choosing $x=0$ , which gives

$\varphi(0)=\dfrac\pi3=\dfrac2{\sqrt3}\left(1-\dfrac15+\dfrac17-\dfrac1{11}+\cdots\right)$
$\implies\dfrac\pi{2\sqrt3}=1-\dfrac15+\dfrac17-\dfrac1{11}+\cdots$

as required.

5 0

3 years ago

I really need help with this my deadline is very soon

Harman [31]

Answer:

$(3)\ y = -x$