Melody

Ok  I hope that you are feeling all better now. 

It is good to see you back Rosala,

Is that a picture of me? Blue always was my colour  

Yes you got just a little scolding.  Not enough of a one to stay away for days though.

Zegroes was feeling bad.  I was saying things that he didn't want to hear.  I didn't want you to say anything to make him him feel even worse.  That is all.

You over reacted.  But I am really glad that you are back.  I like having you here.  You are fun.   

Ok Alan, I just worked out your answer.   

I didn't know I could do this

$$\begin{array}{rll}
\frac{}{}dp&=&-2dt\\\\
\int \frac{1}{p}dp&=&\int-2 dt
\end{array}$$

You are integrating both sides but one is being integrated with respect to p and the other with respect to t.

I didn't know you could do this.  Can you try to think of a simple explanation of why this is allowed?

Thank you.  

Alan's understanding of calculus is much better than mine.  I am sure his method is the best one, however

I am having trouble following his logic.    I will come back to that.

This is how I would do it:

$$\begin{array}{rlll}
\frac{dp}{dt}&=&-2p\qquad \qquad &p(0)=20\\\\
\frac{dt}{dp}&=&\frac{1}{-2p}\qquad \qquad &p(0)=20\\\\
\int{\frac{dt}{dp}}dp&=&\int \frac{1}{-2p}dp\\\\
t&=&\frac{1}{-2}ln(p)+c &when\; t=0\;p=20\;\;substitute\; to\; find\; c
\\\\
0&=&\frac{1}{-2}ln(20)+c\\\\
c&=&\frac{ln(20)}{2}\\\\
t&=&\frac{-ln(p)}{2}+\frac{ln(20)}{2}\\\\
2t&=&-ln(p)+ln(20)\\\\
2t&=&ln(\frac{20}{p})\\\\
e^{2t}&=&e^{ln(\frac{20}{p})}\\\\
p&=&\frac{20}{e^{2t}}\\\\

\end{array}$$

You really don't need algebra for this.

$5 is spent on to toys that are equal in price so each toy must be $5/2 = $2.50

I'll take your word for it Aziz. lol 

Why would you ever want to do this?  The units in Aziz's answer would be Kg².  What would that be good for?

I'd like to take Alan's explanation and expand upon it a little.

NOTE: THIS IS ONLY TRUE AS THE ANGLE TENDS TO 0

   $$The circumference of a circle is $2\pi r$\\
If the radius is 1 unit then the circumference is $2\pi$\\
So if the angle is measured in radians then the arc length will be equal to the angle that it is subtended from. This is for a unit circle.\\\\$$

You can do this by a normal expansion if you want but it is a difference of 2 squares and you should commit this to memory.

$$\boxed{(a+b)(a-b)=a^2-b^2}$$

So for your question the simplification will be 

$$\\(3 \sqrt2 + \sqrt5) (3 \sqrt2 - \sqrt5)\\\\
=(3 \sqrt2)^2-(\sqrt5)^2\\\\
=3^2 (\sqrt2)^2-(\sqrt5)^2\\\\
=9*2-5\\\\
=18-5\\\\
=13$$

That is the right answer anonymous but it is not quite finished.    

I can see just by looking that this will be the difference of 2 squares.  This is something that you should recognise.

$$\boxed{(a+b)(a-b)=a^2-b^2}$$

So

$$\\(2x+3)(2x-3) \\
=(2x)^2-3^2\\
=2x^2-9$$