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Vika [28.1K]
3 years ago
11

What is the slope of a line containing points (2,-1) (3,5)

Mathematics
1 answer:
Kitty [74]3 years ago
8 0

Answer:

slope(m)=6

Step-by-step explanation:

(2 , -1)=(x1 , y1)

(3 , 5)=(x2 , y2)

use formula : y2 - y1/x2 - x1

=5-(-1)/3-2

=5+1/1

=6/1

=6

therefore slope of a line is 6.

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Please help me with this.
Anna71 [15]
It’s B the slope is 1/3 that’s why
3 0
3 years ago
The city's central library has 72,864 books. The library has 72 bookcases. If each bookcase has room for the same number of book
solong [7]

Answer:

There are 1012 books per bookcase

Step-by-step explanation:

To find the number of books per bookcase take the number of bookcases (72) divided by the number of books (72,864)

So;

72,864 / 72 = 1012

Hope this helps!

5 0
3 years ago
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
Does -2(z+3)-z=-z-4(z+2) have one,zero or infinite solutions
Andrew [12]

Hello from MrBillDoesMath!

Answer:

One solution (z = -1)

Discussion:

-2(z+3)-z=-z-4(z+2)        =>

-2z -6 -z = -z -4z - 8       =>

-3z -6 = -5z -8                =>  add 6 to both sides

-2z = -5z -2                     => add 5z to both sides

3z = -5z +5z -3               =>

3z = -3                            =>

z = -1

Thank you,

MrB

6 0
4 years ago
Which number is a solution of the inequality
SashulF [63]

x(7-x)-8>8-8

x(7-x)-8>0


8 0
3 years ago
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