Hi, this is probably dead easy for most of you but can you help me re-arrange this formula so that CºC¹ are the subject.

The formula is:

F= 1/(2π√(Rº R¹ Cº C¹))

F equals the inverse of two pi times the square root of Rº x R¹ x Cº x C¹

Thanks

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- Aug 10th 2009, 10:06 PMJ4553Rearrange Formula
Hi, this is probably dead easy for most of you but can you help me re-arrange this formula so that CºC¹ are the subject.

The formula is:

F= 1/(2π√(Rº R¹ Cº C¹))

F equals the inverse of two pi times the square root of Rº x R¹ x Cº x C¹

Thanks - Aug 10th 2009, 10:58 PMMatt Westwood
Square it, multiply by $\displaystyle C_0 C_1$ and then divide by $\displaystyle F^2$.

- Aug 10th 2009, 11:33 PMJ4553
Thanks but my maths sucks.

So if someone could just tell me what http://www.mathhelpforum.com/math-he...0dfe0e89-1.gif equals, that'd be great.

This is for an extensive project building a graphic equalizer and I'd really not waste time stuck on something like this (Happy) - Aug 11th 2009, 02:25 AMsongoku
Hi J4553

I think it's better if you try it by yourself. Matt Westwood's advice is excellent !

$\displaystyle F=\frac{1}{2\pi\sqrt{R_0 R_1 C_0 C_1}}$

Steps (as suggested by Matt Westwood) :

1. Square both sides

2. Multiply both sides by $\displaystyle C_0 C_1$

3. Divide both sides by $\displaystyle F^2$ - Aug 11th 2009, 04:59 AMJ4553
http://www.mathhelpforum.com/math-he...79ea93a6-1.gif

Steps (as suggested by Matt Westwood) :

1. Square both sides

2. Multiply both sides by http://www.mathhelpforum.com/math-he...0dfe0e89-1.gif

3. Divide both sides by http://www.mathhelpforum.com/math-he...8ee19dbf-1.gif

Ok I got http://www.mathhelpforum.com/math-he...0dfe0e89-1.gif=http://www.mathhelpforum.com/math-he...8ee19dbf-1.gif/(2 π x Rº x R¹)

Yes / No / Way Off?

- Aug 11th 2009, 05:13 AMsongoku
Hi J4553

Your answer is wrong ^^

$\displaystyle F=\frac{1}{2\pi\sqrt{R_0 R_1 C_0 C_1}}$

If you square both sides, you'll get :

$\displaystyle F^2=\frac{1}{4\pi^2{R_0 R_1 C_0 C_1}}$

Then, multiply both sides by $\displaystyle C_0 C_1$ will give you :

$\displaystyle F^2C_0 C_1=\frac{1}{4\pi^2{R_0 R_1}}$

Finally, Divide both sides by $\displaystyle F^2$ will give you....^^ - Aug 11th 2009, 05:19 AMJ4553
Second attempt...

http://www.mathhelpforum.com/math-he...0dfe0e89-1.gif=http://www.mathhelpforum.com/math-he...8ee19dbf-1.gif/(4 π ² x Rº x R¹)

- Aug 11th 2009, 05:23 AMsongoku
Hi J4553

Still wrong ^^

$\displaystyle F^2C_0 C_1=\frac{1}{4\pi^2{R_0 R_1}}$

Divide both sides by F^2 will give you :

$\displaystyle C_0 C_1 = \frac{1}{4\pi^2{R_0 R_1}F^2}$

:) - Aug 11th 2009, 05:25 AMJ4553
Well I tried, thanks for your help! (Rock)

- Aug 11th 2009, 05:27 AMsongoku