Melody

Thanks Chris,  

Mathcalc:  Before I answer any more I want you to examine CPhill's answers and tell him whether you understand.

If you do not, try to be specific about what you do not understand.

These are tricky questions I think, and getting the concepts involved is difficult.

If you try very hard to understand and you still do not I shall take a look at them with you.

I think Chris will be ok with that.  We both want you to learn the best you can :)

Thanks Chris . :D

I have answered Shades's  last question  - he must have posted it somewhere else :/

[2nd]  [atan]  number

Thanks Alan I do believe you are correct :) 

( My first answer was rubbish )

I have more ideas:

There are 8 people in a room. They were asked to go in a certain place 2 at a time. In how many ways could the people be paired? Is it an example of permutation or combination problem?

I'll pair them first   [ This is a combination question because order doesn't matter]

choose anyone then there are 7 ways to choose a mate for that person

choose anyone then there are 5 people left to pair him with

choose anyone then there are 3 people left to pair him with

there there is only 2 left so they have to go together

so that is 7*5*3 = 105 possible pairs. [ Just like Alan said ]

Now they leave the room in pairs and order counts so that would be 4! = 2*3*4 = 24   [This bit is a permutation question]

So now I get     7*5*3*4!         105*24 = 2520

So yes I agree that there is 105 ways for them to be paired ( and I think that is what the answer was asking)

 but I think there is 24 orders that those pairs can leave the room.  So there are 105*24 = 2520  ways they can leave the room 

That is a cute icon Solveit :)

Welcome to the forum  :)

@  Injustagod, if you speak to people that way you may find the help you want for yourself stops being offered !

There are 8 people in a room. They were asked to go in a certain place 2 at a time. In how many ways could the people be paired? Is it an example of permutation or combination problem?

Deleted on the grounds that my answer was rubbish   

sin(5x) = sin(3x)

Thanks LambLamb CPhill and guest #4    

I think this answer is probably very similar to  guest     #4      

\(5x=3x+2\pi n\qquad or \qquad 5x=\pi-3x+2\pi n\\ 2x=2\pi n\qquad \qquad or \qquad 8x=\pi+2\pi n\\ x=\pi n\qquad \qquad \quad or \qquad x=\frac{\pi+2\pi n}{8}\qquad where\;\;n\in Z\\\)

(((101-103)^2+(103-107)^2+(107-103)^2+(109-113)^2+(113-127)^2+(127-113)^2+(131-137)^2+(137-131)^2+(139-149)^2+(149-139)^2+(151-157)^2+(157-163)^2+(163-157)^2+(167-179)^2+(179-167)^2+(181-191)^2+(191-181)^2+(193-197)^2+(197-193)^2+(199-197)^2))/19)^0.5

=(((-2)^2+(-4)^2+(4)^2+(-4)^2+(-14)^2+(14)^2+(-6)^2+(6)^2+(-10)^2+(10)^2+(-6)^2+(-6)^2+(6)^2+(-12)^2+(10)^2+(-10)^2+(10)^2+(-4)^2+(4)^2+(2)^2))/19)^0.5

=((4+16+16+16+196+196+36+36+100+100+36+36+36+144+100+100+100+16+16+4))/19)^0.5

          4+16+16+16+196+196+36+36+100+100+36+36+36+144+100+100+100+16+16+4 = 1304

=sqrt(1304/19)

sqrt(1304/19) = 8.284417839979369

((x^4)-20(x^2)) - ((x^4)+2(x^3))

There is no division here but out guest did simplify it correctly.  Thanks guest :)

\(=x^4-20x^2 - 1(x^4+2x^3)\\ =x^4-20x^2 - 1x^4-2x^3\\ =-20x^2 -2x^3\\ =-20x^2 -2x^3\\\)

  K= +4/7     or    K=  - 4/7