All categories

2 years ago

The domain of the function is

Mathematics

1 answer:

lyudmila [28]2 years ago

8 0

Answer:

Unknown

Step-by-step explanation:

We need to know the function first, like a graph or chart or something.

You might be interested in

Use log4(2) = 0.5 to approximate the value of log4(32)

shepuryov [24]

log4(32) = 2.5

Step-by-step explanation:

Given that,

log4(2) = 0.5

To find: log4(32)

32 can be written as 2 power 5

log4(32) = log4(2^5)

= 5 log4(2)

In the logarithms if there is a power it is multiplied and only the log is applied to the base of the powered term.

Hence,

log4(32) = 5log4(2)

We know that log4(2) = 0.5

So,

log4(32) = 5(0.5)

log4(32) = 2.5

4 0

2 years ago

What polynomial identity should be used to prove that 16^2=(10+6)^2? (A. Difference of Cubes; B. Difference of Squares; C. Squar

Anna71 [15]

Not difference of cubes since it is 2nd degree
not difference of squares since it is plus
not sum of cubes because 2nd degreee

answer is square of binomial (why do we even need this property, oh well)

C

3 0

2 years ago

Read 2 more answers

Ysidra paints a room that has 400 square feet of wall space in 2 and a half hours. at this rate, how long will it take her to pa

Pie

1 hour 10 minutes /////

5 0

2 years ago

Can you help me with this

kifflom [539]

Answer:

the anwer is d

I did the math

8 0

1 year 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

2 years ago