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3 years ago

5

in one week a kiln removes 1/3 of the moisture from a stack of wood. What fraction of the moisture remains in the lumber after 5

weeks in a kiln?

1 answer:

Evgesh-ka [11]3 years ago

7 0

Answer:

-1 2/3 that's the best i could do sorry if its wrong

Step-by-step explanation:

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Subtract the following polynomials, then place the answer in the proper location on the grid. Write the answer in descending pow

Vadim26 [7]

Hi! :)

12 a^2 + 60 a - 8
- (16 a^2 - 32 a + 9)

= -4 a^2 + 92 a - 17

3 0

3 years ago

Solve for x does anyone know how to do this?

riadik2000 [5.3K]

Answer:

x = 9

Step-by-step explanation:

The sum of the 4 angles in a quadrilateral = 360°

Sum the given angles and equate to 360

90 + 109 + 9x - 5 + 85 = 360, that is

9x + 279 = 360 ( subtract 279 from both sides )

9x = 81 ( divide both sides by 9 )

x = 9

6 0

3 years ago

Is this correct help plzz

const2013 [10]

None of the offered choices is correct.

The radical cannot be simplified. If rewritten, it must be written as something like
$x \sqrt[3]{ \frac{6}{x}}$
or
$\frac{6x}{ \sqrt[3]{36x}}$

3 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

F(v)=6+v^2 how do I solve

Nostrana [21]

Step-by-step explanation:

Well, this is a function. In order to solve this, you need a v value. Once you get a v value, you can plug into into 6+v^2 and wala- you get your answer.

Example, lets say

v is 3

then

F(3) = 6 + 3^2

f(3) = 6 + 9

f(3) =15

Hope this helped

6 0

3 years ago