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PLEASE HELP ME . how can the converse of the Pythagorean theorem help you determine whether the roped off area is in the shape o

trapecia [35]

Step-by-step explanation:

From the Pythagorean theorem we have:

$a^2+b^2=c^2$ <em>subtract b² from both sides</em>

$a^2=c^2-b^2\to a=\sqrt{c^2-b^2}$

Same:

$b=\sqrt{c^2-a^2}$

4 0

3 years ago

The perimeter of a rectangle is 46 millimeters. The length of the rectangle can be represented by (3x+3), and its width can be r

Tamiku [17]

Answer:

The dimension of this rectangle = 2mm by 21mm

Where 2 mm = Width of the rectangle

21 mm = Length of the rectangle.

Step-by-step explanation:

The formula for the perimeter of a rectangle = 2(L + W)

= 2L + 2W

The perimeter = 46 mm

L = 3x + 3

W = x - 4

P = 2L + 2W

46mm = 2(3x + 3) + 2(x - 4)

46 = 6x + 6 + 2x - 8

Collect like terms

46 +8 -6 = 6x + 2x

48 = 8x

x = 48/8

x = 6

Since the Length = 3x + 3

= 3(6) + 3

= 18 + 3

= 21mm

Since the Width = x - 4

= 6 - 4

= 2mm

Therefore, the dimension of this rectangle = 2mm by 21mm

Where 2 mm = Width of the rectangle

21 mm = Length of the rectangle.

8 0

3 years ago

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

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