# Simplify

• November 21st 2008, 07:56 AM
turbod15b
Simplify
how do you simplify this(Thinking)

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

and make C the subject

thanks
• November 21st 2008, 09:34 AM
masters
Hello Turbo,

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

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

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

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

Solve for C:

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

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

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

$C(RQ)^2=L$

$\boxed{C=\frac{L}{(RQ)^2}}$
• November 21st 2008, 09:36 AM
turbod15b
thanks alot (Happy)
• November 21st 2008, 09:44 AM
Twig
hi
Hi there

I don´t know what level you´re at, so I don´t know how much I should take
for granted that you know.
Can you see the "most obvious" simplifications you make?
Like shorten out the $y^3$ terms in the first fraction?

When you have factors like this for exampl: $\frac{10x^3y^2}{5x^3y}$
You see that the number ten is actually composed by two prime numbers, namely 2 and 5.
$2 \cdot 5 = 10$
In the same way, $x^3 = x \cdot x \cdot x$

So you´ll end up with(using my example):
$\frac{10x^3y^2}{5x^3y} = 2y$
• November 21st 2008, 10:11 AM
turbod15b
just forgot to mention for the second one

calculate the value of capacitence 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?