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3 years ago

If x+3/3 = y+2/2, then x/3 =

Mathematics

1 answer:

Kisachek [45]3 years ago

8 0

Answer:

y/3

Step-by-step explanation:

x + 3/3 = y + 2/2

x + 1 = y + 1

x = y + 1 - 1

x = y + 0

x = y

so; x/3 = y/3

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12+root of 35 divided by 2 + 12-2 root of 35 divided by 2

nadya68 [22]

So : $12+ \frac{\sqrt{35}}{2} + \frac{12-2 \sqrt{35} }{2}$ ?

First, you want to bring everything together on the top of the fraction.
12=24/2 so you can have $\frac{24+ \sqrt{35}+12-2 \sqrt{35} }{2}$

Add like terms -- 12 +24 = 36 $\sqrt{35}-2 \sqrt{35} = \sqrt{35}$

Now we are at $\frac{24+ \sqrt{35} }{2}$ Which is the lowest possible simplified version while remaining at exact value.

Feel free to check my math! I kinda did this off the top of my head.

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=====================

Work Shown:

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Can I get help with finding the Fourier cosine series of F(x) = x - x^2

trapecia [35]

Assuming you want the cosine series expansion over an arbitrary symmetric interval $[-L,L]$ , $L\neq0$ , the cosine series is given by

$f_C(x)=\dfrac{a_0}2+\displaystyle\sum_{n\ge1}a_n\cos nx$

You have

$a_0=\displaystyle\frac1L\int_{-L}^Lf(x)\,\mathrm dx$
$a_0=\dfrac1L\left(\dfrac{x^2}2-\dfrac{x^3}3\right)\bigg|_{x=-L}^{x=L}$
$a_0=\dfrac1L\left(\left(\dfrac{L^2}2-\dfrac{L^3}3\right)-\left(\dfrac{(-L)^2}2-\dfrac{(-L)^3}3\right)\right)$
$a_0=-\dfrac{2L^2}3$

$a_n=\displaystyle\frac1L\int_{-L}^Lf(x)\cos nx\,\mathrm dx$

Two successive rounds of integration by parts (I leave the details to you) gives an antiderivative of

$\displaystyle\int(x-x^2)\cos nx\,\mathrm dx=\frac{(1-2x)\cos nx}{n^2}-\dfrac{(2+n^2x-n^2x^2)\sin nx}{n^3}$

and so

$a_n=-\dfrac{4L\cos nL}{n^2}+\dfrac{(4-2n^2L^2)\sin nL}{n^3}$

So the cosine series for $f(x)$ periodic over an interval $[-L,L]$ is

$f_C(x)=-\dfrac{L^2}3+\displaystyle\sum_{n\ge1}\left(-\dfrac{4L\cos nL}{n^2L}+\dfrac{(4-2n^2L^2)\sin nL}{n^3L}\right)\cos nx$

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3 years ago

If x+3/3 = y+2/2, then x/3 =​

If x+3/3 = y+2/2, then x/3 =