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3 years ago

Can anyone understand this? I certinaly don't!

Mathematics

1 answer:

lara [203]3 years ago

6 0

Answer:

y-3=6(x+8)

Step-by-step explanation:

Ugh here goes

the point-slope form of an equation is y=mx+c

we have m with is the slope and we have x and y ( -8,3) we neec C which is the y intercept so put all the information in the equation and solve for c

3=6(-8)+c

3=-48+c

0=-48-3+c

c=51

so the equation is y=6x+51

take 3 from 51 and put it on the other side

y-3=6x+48

y-3=6(x+8)

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Match the scenarios to the equations.

Diano4ka-milaya [45]

Answer:

first one is -4+y=20 second is 4+y=-20

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2 years ago

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Legend has it that a bull sasquatch in rut will howl approximately 9 times per hour when it is 40°f outside and only 5 times per

ahrayia [7]

The answer is 13.

To solve this question, you need to determine the slope of the function. if howl is y and the temperature is x of the graph, the slope would be:

9-5/(40-70)= 4/-30= -2/15

Using one of the point and the slope, you can find the number of howl

-2/15= 9-y/40-(-20)

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3 years ago

Four more than three times a number is thirteen. What is this number?

Ostrovityanka [42]

Answer:

Step-by-step explanation:

13-4=9

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2 years ago

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Find the Fourier series of f on the given interval. f(x) = 1, ?7 < x < 0 1 + x, 0 ? x < 7

Zolol [24]

$f(x)=\begin{cases}1&\text{for }-7$

The Fourier series expansion of $f(x)$ is given by

$\dfrac{a_0}2+\displaystyle\sum_{n\ge1}a_n\cos\frac{n\pi x}7+\sum_{n\ge1}b_n\sin\frac{n\pi x}7$

where we have

$a_0=\displaystyle\frac17\int_{-7}^7f(x)\,\mathrm dx$
$a_0=\displaystyle\frac17\left(\int_{-7}^0\mathrm dx+\int_0^7(1+x)\,\mathrm dx\right)$
$a_0=\dfrac{7+\frac{63}2}7=\dfrac{11}2$

The coefficients of the cosine series are

$a_n=\displaystyle\frac17\int_{-7}^7f(x)\cos\dfrac{n\pi x}7\,\mathrm dx$
$a_n=\displaystyle\frac17\left(\int_{-7}^0\cos\frac{n\pi x}7\,\mathrm dx+\int_0^7(1+x)\cos\frac{n\pi x}7\,\mathrm dx\right)$
$a_n=\dfrac{9\sin n\pi}{n\pi}+\dfrac{7\cos n\pi-7}{n^2\pi^2}$
$a_n=\dfrac{7(-1)^n-7}{n^2\pi^2}$

When $n$ is even, the numerator vanishes, so we consider odd $n$ , i.e. $n=2k-1$ for $k\in\mathbb N$ , leaving us with

$a_n=a_{2k-1}=\dfrac{7(-1)-7}{(2k-1)^2\pi^2}=-\dfrac{14}{(2k-1)^2\pi^2}$

Meanwhile, the coefficients of the sine series are given by

$b_n=\displaystyle\frac17\int_{-7}^7f(x)\sin\dfrac{n\pi x}7\,\mathrm dx$
$b_n=\displaystyle\frac17\left(\int_{-7}^0\sin\dfrac{n\pi x}7\,\mathrm dx+\int_0^7(1+x)\sin\dfrac{n\pi x}7\,\mathrm dx\right)$
$b_n=-\dfrac{7\cos n\pi}{n\pi}+\dfrac{7\sin n\pi}{n^2\pi^2}$
$b_n=\dfrac{7(-1)^{n+1}}{n\pi}$

So the Fourier series expansion for $f(x)$ is

$f(x)\sim\dfrac{11}4-\dfrac{14}{\pi^2}\displaystyle\sum_{n\ge1}\frac1{(2n-1)^2}\cos\frac{(2n-1)\pi x}7+\frac7\pi\sum_{n\ge1}\frac{(-1)^{n+1}}n\sin\frac{n\pi x}7$

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2 years ago