Melody

EDIT:  this might be the correct answer is the b***s and the  boxes were all INdistinguishable.

for that is the question that I attempted to answer.    LOL

I counted them and I get 7 ways

0,0,6

0,1,5

0,2,4

0,3,3

1,1,4

1,2,3

2,2,2

Since all the b***s and all the boxes are the same, the only difference can be is how many b***s  are in the boxes.

It is good thanks Cemone.

I have always lived in/near Sydney so it suits me :) 

Compared to most places in the world the weather is good.

Australia is a peaceful and relatively wealthy country so there is not much to complain about :)

$$\\\dfrac{sin[2(x-\frac{\pi}{3})]}{cos[2(x-\frac{\pi}{3})]} = \dfrac{\sqrt{3}}{3}\\\\\\
let\;\; x-\frac{\pi}{3}=\theta\\\\
\frac{sin[2\theta]}{cos[2\theta]} = \frac{\sqrt{3}}{3}\\\\
tan(2\theta)=\frac{1}{\sqrt3}
$1st or 3rd quad$\\\\
2\theta=\frac{\pi}{6},\;\;\frac{7\pi}{6},\;\;\frac{13\pi}{6},\;\;\frac{21\pi}{6},\;\;....\\\\
\theta=\frac{\pi}{12},\;\;\frac{7\pi}{12},\;\;\frac{13\pi}{12},\;\;\frac{21\pi}{12},\;\;....\\\\
\theta=\frac{\pi+6n\pi}{12}\qquad where\;\; n\in Z \;\;$(n is an integer)$$$

$$\\y=\sqrt[3]{x+4}-3\\\\
$this is how I think of these$\\\\
(y+3)=\sqrt[3]{(x+4)}$$

now

solve y+3=0  the answer is y=-3    so got down 3 units

solve x+4=0  the answer is x=-4   so go left 4 units

Notice how I used the brackets.  This is important to get the grouping correct.

I do not think everyone thinks of it like this but it works well for me.    

Thanks Chris,

Alan keeps telling me I have to rearrange the equation to make x the subject BEFORE I can swap the x and the y.  (just as Chris has told you here)

I have never understood what difference it makes. Especially not if graphing 'holes' are pre-recognised :) Mmm

If you draw a right angled triangle and put the know things on it you will easily understand what Chris has done.

If you need visual help then ask for it!

arc tan is the same as inverst tan so just use atan( )

Why are you posting this question again.

the answer is still the same it is 100

10*10=100

think of negative numbers as cold

and positive numbers as hot

then

if you add  cold into a room the room gets coder - so  + -    =  -

With PERMUTATIONS   - ORDER COUNTS

There will always be more permutations than combinations (unless the number of elements in the answer is 0 or 1).

with combintations order does not count