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2 years ago

What is the equation of the line in slope-intercept form?

Mathematics

2 answers:

77julia77 [94]2 years ago

7 0

ANSWER

The equation of the line in slope-intercept form is

$y = \frac{3}{5} x + 3$

EXPLANATION

The slope intercept form of a line is given by
$y = mx + c$
Where $m$ is the slope and $c$
is the y-intercept.

We can read any two points from the graph and use it to determine the slope of the line.

For instance, the graph passes through the points
$(-5,0)$ and $(0,3)$

The slope can be found using the formula,

$m= \frac{y_2-y_1}{x_2-x_1}$

This implies that,

$m = \frac{ 3 - 0}{0 - - 5}$

$m = \frac{ 3 - 0}{0 + 5}$

$m = \frac{ 3 }{ 5}$

Also the y-intercept is

$(0,3)$

Therefore $c = 3$

We substitute all these values in to the above equation to obtain,

$y = \frac{3}{5} x + 3$

olasank [31]2 years ago

5 0

The equation of the line that passes through the points is y=3/5x+3

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shutvik [7]

Answer:

Statements 2, 4, and 5

Step-by-step explanation:

15≥22+x

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x≤-7

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this means your 3 correct statements would be:

x less-than-or-equal-to negative 7,

–6 is part of the solution.

–7 is part of the solution.

the last two are true based on the inequality.

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kondor19780726 [428]

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2 years ago

Read 2 more answers

Convert 6\5 into a mixed number

netineya [11]

6/5 into a mixed number is 5.

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2 years ago

Write the number 3.7 as a mixed number let x

puteri [66]

3.7/1 = 3.7

multiply it through to make it a whole number.
x 10

37
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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