Melody

n+q=63

5n+25q=1075

solve simultaneously

I think the no accounts is just a temporary problem.

I really do not like the way this question is presented.  

Lets see if I can do better... 

\(g(x)=x+sinx\\ f(x)=x^4\\ y(x)=(f\circ g)(x)=[g(x)]^4=(x+sinx)^4 \)

now I have just referred to g(x) as g  etc but you have to remember that g, f and y are all functions of x.  

I have also stopped talking about f becasue I find it confusing.

This is probably not the most mathematically correct presentation but I think it is easier to understand.

\(y=(x+sinx)^4\\ let \;\; g=x+sinx\quad and\quad y=g^4\\ so\\ \frac{dg}{dx}=1+cosx \quad and\quad \frac{dy}{dg}=4g^3=4(x+sinx)^3\\ \frac{dy}{dx}=\frac{dy}{dg}*\frac{dg}{dx}\\ \qquad \text{notice that the dg's cancel each other out - There is a chain of cancelling}\\ \frac{dy}{dx}=4(x+sinx)^3*(1+cosx)\\ \frac{dy}{dx}=4(x+sinx)^3*(1+cosx)\\\)

Now look at the pattern of what has happened.  That is an easier way to use this.

LaTex:

y=(x+sinx)^4\\
let \;\; g=x+sinx\quad and\quad  y=g^4\\
so\\
\frac{dg}{dx}=1+cosx \quad and\quad  \frac{dy}{dg}=4g^3=4(x+sinx)^3\\
\frac{dy}{dx}=\frac{dy}{dg}*\frac{dg}{dx}\\
\qquad \text{notice that the dg's cancel each other out}\\
\frac{dy}{dx}=4(x+sinx)^3*(1+cosx)\\
\frac{dy}{dx}=4(x+sinx)^3*(1+cosx)\\

You are very welcome :)

WEll I have very little idea but this is where my play took me.

How many subsets of S are there altogether.    This is the number of ways that A can be chosen

\(\displaystyle \sum_{k=0}^n\;\;\binom{n}{k}\)

subsets b and C can be shosen the same way so

number of ways subsets A,B and C can be chosen is       \(\left[\displaystyle \sum_{k=0}^n\;\;\binom{n}{k}\right]^3\)

But what if  A is a subset of B and B is a subset of C ...

Ways to chose C is    \(\displaystyle \sum_{k=0}^n\;\;\binom{n}{k}\)

then

ways to chose B is     \(\displaystyle \sum_{m=0}^k\;\;\binom{k}{m} \qquad m\le k\)

ways to choose A is    \(\displaystyle \sum_{t=0}^m\;\;\binom{m}{t}\qquad t\le m\)

Put thse toghether and I get the number of ways   \(A \subseteq B \subseteq C\)

is    \( \displaystyle \left[ \sum_{k=0}^n\;\;\; \binom{n}{k}\left[\sum_{m=0}^k\;\;\binom{k}{m}\left[\sum_{t=0}^m\; \binom{m}{t}\right]\right]\right]\\~\\ \)      that is probably not displayed in the right order.   

So prob would be

\( \frac{\displaystyle \left[ \sum_{k=0}^n\;\;\; \binom{n}{k}\left[\sum_{m=0}^k\;\;\binom{k}{m}\left[\sum_{t=0}^m\; \binom{m}{t}\right]\right]\right]\\~\\ }{\left[\displaystyle \sum_{k=0}^n\;\;\binom{n}{k}\right]^3} \)

I don't know how to display that properly, the order and the brackets are most likely wrong.......

It is written  as (square root of 3) times y    ---  which is correct.

\(\sqrt3y \qquad not\quad \sqrt{3y}\)

Hi,

I am sorry that you are having this problem.  We always encourage our guests to become members,

This is an ongoing problem.  I wasn't sure if it had been fixed but obviously it has not.  

I will let admin know but if it was easily fixed I am sure he would have already done so.

All I can suggest is that you sign your posts with the username that you choose for yourself. 

That way you will become known even before you manage to make an account.   

There are 360 degrees in a whole circle, you have 60 degrees.  SO you have   \(\frac{60}{360}\)   of a circel

The area of a whole cirlce is  \(A=\pi r^2\)

If this was the whole circle the area would be    \(\pi* 30^2 = 900\pi \;feet\;squared\)

SO the area of just the sector will be     \(\frac{60}{360}*\pi*900 \;\;feet squared\)

and you can do the calculation.

Thanks Bilderboi and EP

You did make a mistake builder boi which EP picked up on. 

Here is my answer just to stop any confusion.  

\(\frac{\frac{x}{\sqrt3}}{y}\\ =\frac{x}{\sqrt3}\div y\\ =\frac{x}{\sqrt3}\times \frac{1}{y} \\ =\frac{x}{y\sqrt3}\qquad \text{I know that you already got this far} \\ \text{now you just need to reationalize the denominator}\\ =\frac{x}{y\sqrt3} \times \frac{\sqrt3}{\sqrt3}\\ =\frac{\sqrt3\; x}{3y} \)