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2 years ago

What is the slope and y-intercept of the equation 3x -y=3

Mathematics

2 answers:

kozerog [31]2 years ago

4 0

Answer:

Slope = 3

Y Intercept = 3

Step-by-step explanation:

3x - y = 3

(subtract 3x from both sides - we have to get y by itself)

-y = -3x - 3

(divide by -1 from both side - we need to make y positive)

y = 3x + 3

amm18122 years ago

4 0

The slope is 3..................

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X-2y=18 2x+y=6 parallel perpendicular or neither

sergey [27]

<h3>Therefore they are perpendicular.</h3>

Step-by-step explanation:

A equation of line is

y =mx +c

Here the slope of the line is m.

Given equations are

x - 2y = 18

⇔-2y = -x +18

$\Leftrightarrow y =\frac{1}{2} x -9$ ............(1)

and 2x + y = 6

⇔y = -2x +6 ............(2)

Therefore the slope of equation (1) is $(m_1)$ = $\frac{1}{2}$

Therefore the slope of equation (2) is $(m_2)$ = -2

If two lines are perpendicular, when we multiply their slope we get -1.

therefore,

$m_1. m_2$ = $\frac{1}{2}$ . (-2) = -1

Therefore they are perpendicular.

4 0

2 years ago

kfc has 11 more then twice as many customers as when they started selling food . kfc now has 73 customers . how may did they hav

matrenka [14]

Answer:

<h2>73 - 11 equals 62. <em><u>THE ANSWER IS 31.</u></em></h2><h2><em><u>PLEASE ADD AS BRAINEST.</u></em></h2>

Step-by-step explanation:

3 0

2 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

2 years ago

Answer quick for brainliest!

Gnom [1K]

Step-by-step explanation:

8 0

3 years ago

Which statement describes the relationship between x and y?

stich3 [128]

Answer:

as x increases, y increases

Step-by-step explanation:

3 0

3 years ago