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3 years ago

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PLEASE HELP : A rectangle has a length that is 15 more than its width. A second rectangle with a perimeter of 72 has a width tha

t is 5 wider and a length that is 2 shorter than the first rectangle. Find the dimensions of the first rectangle.

1 answer:

SashulF [63]3 years ago

7 0

Answer:

length = 24, width = 9

Step-by-step explanation:

Let's call the length and width of the first rectangle w + 15 and w respectively. This means that the length and width of the second rectangle is (w + 15) - 2 and w + 5 respectively. Since the perimeter of the second rectangle is 72, we can write:

2 * ((w + 15) - 2 + w + 5) = 72 because perimeter = 2 * (length + width)

(w + 15) - 2 + w + 5 = 36 (Divide by 2)

2w + 18 = 36 (Combine like terms)

2w = 18 (Subtract 18)

w = 9 (Divide by 2)

This means that the length and width are 24 and 9.

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Solve this step by step

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Step-by-step explanation:

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$x \times \pi$

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Equivalent factored form for 10x^2+15x-9=2x^2+x-14

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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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tester [92]

The exact answer would be <span>392.78

An estimated value could be either 390 or 393</span>

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Answer:

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