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xxMikexx [17]
3 years ago
7

Are the graphs of the lines in the pair parallel ? Explain.

Mathematics
1 answer:
Julli [10]3 years ago
3 0

<em>So</em><em> </em><em>the</em><em> </em><em>right</em><em> </em><em>answer</em><em> </em><em>is</em><em> </em><em>of</em><em> </em><em>option</em><em> </em><em>D</em>

<em>Look</em><em> </em><em>at</em><em> </em><em>the</em><em> </em><em>attached</em><em> </em><em>picture</em>

<em>H</em><em>ope</em><em> </em><em>it</em><em> </em><em>will</em><em> </em><em>help</em><em> </em><em>you</em><em>.</em><em>.</em><em>.</em>

<em>Good</em><em> </em><em>luck</em><em> </em><em>on</em><em> </em><em>your</em><em> </em><em>assignment</em>

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3 years ago
What is 9.773 divided by 0.12
Verizon [17]

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Gus and Ike carried a total of 84 books to the library. If Gus carried 5 as many books as Ike, how many books did they each carr
Alexus [3.1K]

Answer:

They each carried 25 books each.

Step-by-step explanation:

You get this by doing

84/5=16.8

then you do

84/2/5= 8.4

8.4+16.8=25.2

but round to whatever 5 or higher bump up below 5 keep the same

so your answer results in 25.

3 0
3 years ago
Find the Fourier series of f on the given interval. f(x) = 1, ?7 &lt; x &lt; 0 1 + x, 0 ? x &lt; 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
3 0
2 years ago
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