Let ABC be a right triangle (∠C=90°). Let CH be the altitude of the triangle ABC.

Which of the following statements best describes the solution below ?

stira [4]

Answer:

D

Step-by-step explanation:

8 0

3 years ago

Algebra isn’t my best subject, please help

kogti [31]

The Distributive property states that when we have the statement a(b + c), we can "distribute" the a by multiplying a times both terms that are inside the parenthses.

So in this problem, 2(5 + 7) can be simplified by distributing the

2 through both terms that are inside the set of parenthses.

So we have (2 · 5) + (2 · 7).

8 0

3 years ago

Jim and Jackie are married with three children at home and a mortgage. Jim’s net pay per year is $67,000 and Jackie does not hav

Darya [45]

The answer is A. they should work on their plan for managing income.

4 0

3 years ago

Read 2 more answers

Omar has 2 3/4 cup of Joe to make dumplings. If he uses 3/16 cup of Joe for each dumpling how many whole dumplings can Omar make

Fiesta28 [93]

Answer:

44/3 or 14 and 2/3

Step-by-step explanation:

First convert 2 and 3/4 in a improper fraction which gives you 11/4.

Next, you divide 11/4 and 3/16.

It will give you 44/3 which can be 14 and 2/3.

6 0

3 years ago

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

3 years ago