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2 years ago

Write an equation of the line in slope-intercept form

Mathematics

2 answers:

goldfiish [28.3K]2 years ago

8 0

Answer:

$y = 2x - 2$

Step-by-step explanation:

Starting from the y-intercept of [0, −2], you do $\frac{rise}{run}$ by either moving two blocks south over one block west or two blocks north over one block east [west and south are negatives].

I am joyous to assist you anytime.

Agata [3.3K]2 years ago

5 0

Answer:

y=2x-2

Step-by-step explanation:

Slope intercept form = y=mx+b

Where m=slope and b= y intercept

Reminder: the y intercept is where x=0

Looking at the graph, the y intercept is given as (0,-2)

There are also two points given (0, -2) and (2,2)

Reminder: to find the slope, divide the changes in y by the changes in x or substitute the numbers into this equation: y2-y1/x2-x1

2-(-2)/2-0=4/2

m=2

Now that we have the slope and y intercept, sbstitute into the equation y=mx+b:

y=2x-2

You might be interested in

Complete the square to transform the expression x2 + 4x + 2 into the form a(x − h)2 + k.

Nina [5.8K]

Answer:

Step-by-step explanation:

y = (x^2 + 4x) + 2

Take 1/2 of the linear term 4/2 = 2 and square that result. 2^2 = 4.

Put it after 4x

y = (x^2 + 4x + 4) +2 Subtract what you put inside the brackets on the outside.

y = (x^2 + 4x + 4) + 2 - 4 Combine the right.

y = (x^2 + 4x + 4) - 2 Express the brackets as a square.

y = (x + 2)^2 - 2

That's your answer

a = 1

h = 2

k = -2

3 0

3 years ago

Tape diagram to subtract 33-19

Tpy6a [65]

Well 33 - 19 = 14 and what do u mean tape diagram?

3 0

3 years ago

Can I get help with finding the Fourier cosine series of F(x) = x - x^2

trapecia [35]

Assuming you want the cosine series expansion over an arbitrary symmetric interval $[-L,L]$ , $L\neq0$ , the cosine series is given by

$f_C(x)=\dfrac{a_0}2+\displaystyle\sum_{n\ge1}a_n\cos nx$

You have

$a_0=\displaystyle\frac1L\int_{-L}^Lf(x)\,\mathrm dx$
$a_0=\dfrac1L\left(\dfrac{x^2}2-\dfrac{x^3}3\right)\bigg|_{x=-L}^{x=L}$
$a_0=\dfrac1L\left(\left(\dfrac{L^2}2-\dfrac{L^3}3\right)-\left(\dfrac{(-L)^2}2-\dfrac{(-L)^3}3\right)\right)$
$a_0=-\dfrac{2L^2}3$

$a_n=\displaystyle\frac1L\int_{-L}^Lf(x)\cos nx\,\mathrm dx$

Two successive rounds of integration by parts (I leave the details to you) gives an antiderivative of

$\displaystyle\int(x-x^2)\cos nx\,\mathrm dx=\frac{(1-2x)\cos nx}{n^2}-\dfrac{(2+n^2x-n^2x^2)\sin nx}{n^3}$

and so

$a_n=-\dfrac{4L\cos nL}{n^2}+\dfrac{(4-2n^2L^2)\sin nL}{n^3}$

So the cosine series for $f(x)$ periodic over an interval $[-L,L]$ is

$f_C(x)=-\dfrac{L^2}3+\displaystyle\sum_{n\ge1}\left(-\dfrac{4L\cos nL}{n^2L}+\dfrac{(4-2n^2L^2)\sin nL}{n^3L}\right)\cos nx$

4 0

3 years ago

Factor 2x^2-4x-30 completely. A. 2(x-5)(x+3) B. (2x-5)(x+3) C. (2x+3)(x-5) D. 2(x-3)(x+5)

OLEGan [10]

Answer:

Step-by-step explanation:

$2x^2-4x-30\\\\=2\cdot x^2-2\cdot2x-2\cdot15\\\\=2(x^2-2x-15)\\\\=2(x^2+3x-5x-15)\\\\=2[x(x+3)-5(x+3)]\\\\=2(x+3)(x-5)$