Ask question

Ask question

All categories

3 years ago

11

By which angle must AB turn about point A in the clockwise direction so that it coincides with AE ? A. 90° B. 180° C. 270° D. 36

0°

1 answer:

dmitriy555 [2]3 years ago

5 0

Answer:

B. 180°

Step-by-step explanation:

Rotate AB 180 degrees in the clockwise direction.

You might be interested in

I need help please :(

mixer [17]

Answer:

x = 7

Step-by-step explanation:

Given:

∠DEF = 117°

∠DEG = (12x + 1)°

∠GEF = (5x - 3)°

Find:

value of x

Computation:

∠DEF = ∠DEG + ∠GEF

117° = (12x + 1)° + (5x - 3)°

117° = 17 x - 2

x = 7

7 0

2 years ago

Read 2 more answers

The coordinates A(2, 1), B(7, 1), C(2, 4) form what type of polygon?

Digiron [165]

Step-by-step explanation:

$Slope \: of \: AB = \frac{1 - 1}{7 - 2} = \frac{0}{5} = 0 \\ \\ Slope \: of \: BC = \frac{4 - 1}{2 - 7} = \frac{3}{ - 5} = - \frac{3}{ 5} \\ \\ \because \: Slope \: of \: AB \neq \: Slope \: of \: BC \\ \\ \therefore \: given \: points \: are \: non - collinear. \\$

Hence, coordinates A(2, 1), B(7, 1), C(2, 4) form a triangle.

6 0

3 years ago

Read 2 more answers

6 1/2+n=12. please explain for me I dunno how to do this.

Temka [501]

Hey there!

Let's first define our goal. Our goal here is to be able to isolate the variable, n, and solve for it by moving everything to the other side. We see that we have two sides of the equation, and by using the property of equality which states that whatever you do to one side you do to the other and the expression remains equal, we can solve this equation.

It looks like the only thing we have on the left side with n is that 6 1/2. In order to get rid of it, we must subtract it from both sides, because the property of equality states you must do the inverse operation.

We have:

n = 12 - 6 1/2

n = 5 1/2

Hope this helps!

7 0

2 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

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

3 0

2 years ago

Please help me, picture below.

Lesechka [4]

The answer is the last one, 8 square root 6 because when factoring out 486 you will get 9 square root 6. When factoring out 24, you will get 2 square root 6 and the 6 is already factored out. After that you add and subtract and you will get the last answer 8 square root 6. Hope that helped.

8 0

3 years ago