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kvasek [131]
3 years ago
10

The perimeter of a parallelogram is 72 meters. The width of the parallelogram is 4 meters less than its length. Find the length

and the width of the parallelogram.
Mathematics
1 answer:
mafiozo [28]3 years ago
7 0
Perimeter of a parallelogram = 2(l + w). But w = l - 4
Therefore, 72 = 2(l + l - 4) = 2(2l - 4) = 4l - 8
4l = 72 + 8 = 80
l = 80/4 = 20
w = 20 - 4 = 16

Therefore, length = 20 meters and width = 16 meters
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Simplify 2/4 please and thank you
klio [65]

Answer:

1/2

Step-by-step explanation:

2/4 is equal to 1/2, because if you divide 4 by 2 you get 2, and if you divide 2 by 2 you get one.

Basically, the simplification process would look something like this...

2/4 ÷ 2/2 = 1/2

6 0
3 years ago
Calcule:<br> a)-400-(-700)=
Juliette [100K]

-400-(-700)

-140+700

= 300


I hope that's help ! If you have question(s) please let me know



8 0
3 years ago
If -3 1/4 is subtracted from 1 1/3. What will be the result?
Alenkinab [10]

1_1/3 = 4/3

-3_1/4 = -13/4

(4/3) - (-13/4)

(4/3) + (13/4)

We can see that (4/3) + (13/4) leads to a positive answer.

Answer: choice D

4 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
3 years ago
NO LINKS!!!!
mixer [17]

Hope this help!!!

Have a nice day!!!

7 0
3 years ago
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