how do you simplify this(Thinking)

http://i10.photobucket.com/albums/a1...1/DSC00940.jpg

and make C the subject

thanks

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- Nov 21st 2008, 07:56 AMturbod15bSimplify
how do you simplify this(Thinking)

http://i10.photobucket.com/albums/a1...1/DSC00940.jpg

and make C the subject

thanks - Nov 21st 2008, 09:34 AMmasters
Hello Turbo,

Simplify: $\displaystyle (3x)\left(\frac{10y^3x}{5y^3}-x\right)+\left(\frac{12x^4y^2}{3x^4y}-2y\right)(2y)$

$\displaystyle 3x(2x-x)+(4y-2y)(2y)$

$\displaystyle 3x(x)+2y(2y)$

$\displaystyle \boxed{3x^2+4y^2}$

Solve for**C**:

$\displaystyle Q=\frac{1}{R}\sqrt{\frac{L}{C}}$

$\displaystyle RQ=\sqrt{\frac{L}{C}}$

$\displaystyle (RQ)^2=\frac{L}{C}$

$\displaystyle C(RQ)^2=L$

$\displaystyle \boxed{C=\frac{L}{(RQ)^2}}$ - Nov 21st 2008, 09:36 AMturbod15b
thanks alot (Happy)

- Nov 21st 2008, 09:44 AMTwighi
Hi there

I donŽt know what level youŽre at, so I donŽt know how much I should take

for granted that you know.

Can you see the "most obvious" simplifications you make?

Like shorten out the $\displaystyle y^3$ terms in the first fraction?

When you have factors like this for exampl: $\displaystyle \frac{10x^3y^2}{5x^3y}$

You see that the number ten is actually composed by two prime numbers, namely 2 and 5.

$\displaystyle 2 \cdot 5 = 10$

In the same way, $\displaystyle x^3 = x \cdot x \cdot x $

So youŽll end up with(using my example):

$\displaystyle \frac{10x^3y^2}{5x^3y} = 2y $ - Nov 21st 2008, 10:11 AMturbod15b
just forgot to mention for the second one

calculate the value of capacitence C in farads

C in farads

Indunctance L = 0.2H

Resistance R = 50 ohms

Charge Q = 15C

so if

C = L/(RQ)2

C= 0.2/(50x15)2= 0.000000355

correct?