All Answers 358156 Answers

i*(6+7i)

$${i}{\mathtt{\,\times\,}}\left({\mathtt{6}}{\mathtt{\,\small\textbf+\,}}{\mathtt{7}}{i}\right) = {\mathtt{\,-\,}}{\mathtt{7}}{\mathtt{\,\small\textbf+\,}}{\mathtt{6}}{i}$$

I just entered it - the calc automaticals takes i to be   $${\sqrt{-{\mathtt{1}}}}$$

I am assuming you are the same person who has been asking all the combination/permutation questions lately. 

I would like to see you attempt these questions and then we can talk about whether your answers are likely to be correct.  How about working with us.  You will learn a lot better if you do.   :)

total number of commities - no restrictions         $${\left({\frac{{\mathtt{10}}{!}}{{\mathtt{3}}{!}{\mathtt{\,\times\,}}({\mathtt{10}}{\mathtt{\,-\,}}{\mathtt{3}}){!}}}\right)} = {\mathtt{120}}$$

total number that Fred is in         $${\left({\frac{{\mathtt{9}}{!}}{{\mathtt{2}}{!}{\mathtt{\,\times\,}}({\mathtt{9}}{\mathtt{\,-\,}}{\mathtt{2}}){!}}}\right)} = {\mathtt{36}}$$

total number that Fred is not ine       $${\left({\frac{{\mathtt{9}}{!}}{{\mathtt{3}}{!}{\mathtt{\,\times\,}}({\mathtt{9}}{\mathtt{\,-\,}}{\mathtt{3}}){!}}}\right)} = {\mathtt{84}}$$

$${\mathtt{84}}{\mathtt{\,\small\textbf+\,}}{\mathtt{36}} = {\mathtt{120}}$$

These 2 are complementary events,

either one happens OR the other happens. 

Hi TG

Here is a video clip that i enjoyed - I just pulled it out of our Reference thread (that is a sticky note)

It is on fractals.  I think that you will like it :)

How many ways are there to put 6 b***s in 3 boxes if three b***s are indistinguishably white, three are indistinguishably black, and the boxes are distinguishable?

AGAIN i HAVE ANSWERED A DIFFERENT QUESTION HERE.

THIS IS   BOXES ALL THE SAME,   3 BLACK AND 3 WHITE B***S

Again I am basing this one my original answer - if it its wrong then they are all wrong

Thanks for that link, Anonymous.....that really helped me see the logic behind the formula!!!

If you graph it you can see immediately that y=0 at 0 and-0.1846

By substitution then

y=4(6)-8

y=24-8

y=16

hi Albert,   

My answers keep disappearing - I shall try again  

I did not answer this in the first place because I did not understand the question.

DO YOU  want to find the area between the curves    $$y=x \;\;and\;\; y=\frac{x^3}{9}$$     for x=0 to x=3   ?

(I think that this was CPhill's assumption)

[You were logged on when I published this :/  ]

I want the centroid location x bar and y bar of the uncommon area. I don't think it has been answered right.

Well, I got 7 ways if the boxes are all  the same so I am going to start with those solutions. (I hope I got them all. I think i did)

I ANSWERED THE WRONG QUESTION HERE

THIS IS THE ANSWER TO

6 IDENTICAL B***S AND 3 DIFFERENT BOXES  :  

I HAVE ANSWERED THE ACTUAL QUESTION IN MY NEXT POST

0,0,6     the 6 could go in any box so that is  1*3 = 3

0,1,5         3!    permutations here                       =  6

0,2,4         3!    permutations here                       =  6

0,3,3       the 0 could go in any box so that is  1*3 = 3

1,1,4        the 4 could go in any box so that is  1*3 = 3

1,2,3         3!    permutations here                       =  6

2,2,2       there is only one possibility here            =  1

 3+6+6+3+3+6+ 1 =  28 ways      

Also was his answer showing the equation for numerator and denominator or just the numerator? Seems like the later.

Where does y2+((y1-y2)/2) dy = dA come into the workings shown by CPhill? 

Yep - I see that :)

If all the b***s are different and all the boxes are different then

The first ball can go into any of 3 boxes

the second ball can go into any of 3 boxes

.....

so there will be    

$${{\mathtt{3}}}^{{\mathtt{6}}} = {\mathtt{729}}$$                 options

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.

here in california its hot and since its spring i feel great knowing that shorts are always there for me. does anybody on this site live there?

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  + -    =  -