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Komok [63]
3 years ago
10

Order the integers from least to greatest. 6,11,16,-8,-5

Mathematics
2 answers:
Fynjy0 [20]3 years ago
8 0
-8, -5, 6, 11, 16

Good luck!!
slega [8]3 years ago
6 0
Least togreatest
-8<-5<6<11<16
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4. This one isn't an equation, but simplifies to 2x
4 0
3 years ago
Solve the equation below
Agata [3.3K]
Set up the equation:

4(8.75x-2)=97
35x-8 =97
35x =97+8
35x =105
x. =3

The value of x is 3.
7 0
3 years ago
(50 Points)
Yanka [14]
C
If you plug in the values in that equation, it works.
5 0
3 years ago
Read 2 more answers
More math help please!
Nata [24]
AS we know that:

V=SA*h

And we have the volume and we can calculate the SA,,,

So that, we can conclude:
h= \frac{V}{SA} = \frac{323.5}{\pi (7)^2} =2.1015~~in

Are you satisfied?




3 0
3 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
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