All categories

2 years ago

Use the image from problem 11 to answer this problem.

Mathematics

1 answer:

guajiro [1.7K]2 years ago

4 0

Answer:

34.

Step-by-step explanation:

All angles of a triangle combined equal 180, so if 1 angle measures 146, subtract 146 from 180. 180-146=34

You might be interested in

The area of the rectangle is 6 square feet. If its width is 2 1/7 feet, find its length.

densk [106]

Answer:

L = 2 4/5 feet

Step-by-step explanation:

A = L x W

6 = L x 2 1/7

divide both sides by 2 1/7

6 divided by 2 1/7 = L (is the same as multiplying by the reciprocal)

6 x 7/15 = L

L = 14/5 or 2 4/5

5 0

3 years ago

What is the perimeter of this polygon? (2, 4) (-1,2) (3,-2) (-2,-3)

Ber [7]

Answer:

I think the answer is 19

Step-by-step explanation:

7 0

3 years ago

Tyree and four friends go to the movies. Each person buys a movie ticket for $7, a snack for $5, and a drink for $2. Write an ex

Mandarinka [93]

Answer:

Tyree and all of his friends get 1 ticket for $7. Next they all get 1 snack for $5. Then they all get a drink for $2. So:

1=7

1=5

1=2

Then you take each person (which there are five of then) and you times all five people by 7, 5, 2.

5 x 7=35

5 x 5=25

5 x 2=10

After that you add it all together:

35 plus 25=60

60 plus 10=70

Step-by-step explanation:

your welcome

4 0

3 years ago

Read 2 more answers

Solve using Fourier series.

Olin [163]

With $2L=\pi$ , the Fourier series expansion of $f(x)$ is

$\displaystyle f(x)\sim\frac{a_0}2+\sum_{n\ge1}a_n\cos\dfrac{n\pi x}L+\sum_{n\ge1}b_n\sin\dfrac{n\pi x}L$
$\displaystyle f(x)\sim\frac{a_0}2+\sum_{n\ge1}a_n\cos2nx+\sum_{n\ge1}b_n\sin2nx$

where the coefficients are obtained by computing

$\displaystyle a_0=\frac1L\int_0^{2L}f(x)\,\mathrm dx$
$\displaystyle a_0=\frac2\pi\int_0^\pi f(x)\,\mathrm dx$

$\displaystyle a_n=\frac1L\int_0^{2L}f(x)\cos\dfrac{n\pi x}L\,\mathrm dx$
$\displaystyle a_n=\frac2\pi\int_0^\pi f(x)\cos2nx\,\mathrm dx$

$\displaystyle b_n=\frac1L\int_0^{2L}f(x)\sin\dfrac{n\pi x}L\,\mathrm dx$
$\displaystyle b_n=\frac2\pi\int_0^\pi f(x)\sin2nx\,\mathrm dx$

You should end up with

$a_0=0$
$a_n=0$
(both due to the fact that $f(x)$ is odd)
$b_n=\dfrac1{3n}\left(2-\cos\dfrac{2n\pi}3-\cos\dfrac{4n\pi}3\right)$

Now the problem is that this expansion does not match the given one. As a matter of fact, since $f(x)$ is odd, there is no cosine series. So I'm starting to think this question is missing some initial details.

One possibility is that you're actually supposed to use the even extension of $f(x)$ , which is to say we're actually considering the function

$\varphi(x)=\begin{cases}\frac\pi3&\text{for }|x|\le\frac\pi3\\0&\text{for }\frac\pi3$

and enforcing a period of $2L=2\pi$ . Now, you should find that

$\varphi(x)\sim\dfrac2{\sqrt3}\left(\cos x-\dfrac{\cos5x}5+\dfrac{\cos7x}7-\dfrac{\cos11x}{11}+\cdots\right)$

The value of the sum can then be verified by choosing $x=0$ , which gives

$\varphi(0)=\dfrac\pi3=\dfrac2{\sqrt3}\left(1-\dfrac15+\dfrac17-\dfrac1{11}+\cdots\right)$
$\implies\dfrac\pi{2\sqrt3}=1-\dfrac15+\dfrac17-\dfrac1{11}+\cdots$

as required.

5 0

2 years ago

- What is g(x)= -4(x - 5)2 + 8 in standard form

Ulleksa [173]

Answer:

g(x) = -8x + 48

4 0

1 year ago