Alan

Yes, I think so.

Yes, in part 1 that's quite right.  But in part 2, you have pink1_glued_to_pink2 as different from  pink2_glued_to_pink1.  

Here's another way of thinking about part 2.

Imagine the two pink b***s are together in positions 1 and 2.  There are 8!/(3!*5!) = 56 ways of arranging the others.  Now move the pink b***s to positions 2 and 3.  There are another 56 ways of arranging the others.  Now move the pink b***s ...etc.  There are 9 positions where you can have the two pink b***s together in this way, so in total there are 9*56 = 504 ways the two pink b***s can be together (this is half your number).

All this assumes the positions are ordered of course.  For example it assumes that ppooowwwww is different from wwwwwooopp (you mustn't look at the b***s from the other side!!).

Melody, in part 1 you didn't differentiate between the 2 pink b***s (.i.e. you treated pink1/pink2 as the same as pink2/pink1).  So why did you differentiate in part 2 (at least, it looks to me like you did, unless I've misinterpreted what you mean by being glued together in 2 different ways)?

Chris has reminded me that L'Hopital's rule may also be used here.

This says differentiate the numerator and denominator separately, and form the ratio of the resulting two expressions, then take the limit of this as x tends to zero.

So, if you are familiar with L'Hopital's rule, this would be the better approach.

Your reply is good for linear simultaneous equations, Pokemonfan88, but the question asked about non-linear simultaneous equations.  These are often more difficult to deal with.  There aren't really any general rules, especially if they are very complicated, but sometimes rearranging one to find one unknown in terms of another and substituting into another equation will give an equation that can be solved.  Often one needs to resort to numerical solution methods though.

If you have a specific set of non-linear simultaneous equations in mind Anonymous, post them here and we'll see what we can do.

Perhaps this just means find Sn, in which case multiply both sides by 350.

$${\mathtt{Sn}} = {\mathtt{350}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}{\left({\mathtt{30}}^\circ\right)} \Rightarrow {\mathtt{Sn}} = {\mathtt{303.108\: \!891\: \!324\: \!4}}$$

Yes. just type in the fractions and then press Enter (not the = sign!).  The calculator display shows the result as a decimal, but above the calculator it is shown in both fraction and decimal form.  See below:

As follows?

I suspect sally1 is expecting this to be taken just a little further, so taking up where Melody left off:

(f(a+h) - f(a))/h = (7/(a+h) - 7/a)/h

                        = (7a - 7(a+h))/(a*(a+h)*h) 

                        = -7h/(a*(a+h)*h)

                        = -7/(a*(a+h))

In purely numerical terms the left and right sides are not equal; but when interpreted as rotations about a point they give rise to the same net rotation as Chris has explained.  Here is a picture that might help to clarify this.