Alan

These days I tend to use Mathcad mostly, sometimes Matlab and, very occasionally, Maple (though when I was working I used Matlab most).

Difficult to know what to recommend without knowing what you are going to use it for.  The simplest to learn is probably Mathcad, the hardest is Maple (which is very similar to Mathematica).  The most powerful, at least as far as mathematics is concerned, is Maple/Mathematica.  Matlab and Mathcad are probably better for engineering applications.  For secondary school mathematics all of them are more than capable.  They are all pricey!

There are open-source (i.e. free) packages that do a lot of what these commercial packages do (though usually not so slickly).  For example, there is Scilab (a Matlab lookalike); SMath Studio (a Mathcad lookalike); and others.  There are a number of calculators that are mathematically quite powerful also, such as the Texas Instruments TI-Nspire or some of the Casio calculators.  There is also Microsoft Mathematics 4 if you use a Microsoft Windows PC.

As I said, it depends what you intend to use it for!

Like so:

An alternative:

I think you are referring to the remainder when the larger number is divided by the smaller number.  The function for this is usually called mod.    mod(m,n) gives the remainder when m is divided by n.   So mod(1000,12) = 4 and mod(2000,7) = 5.  This function is in Excel.

If you mean exactly two people, then two have probability 0.3 and 4 have probabilty 0.7, and since there are nCr(6,2) = 15 ways of choosing 2 out of 6 the overall probability is 15*0.3^2*0.7^4 = 0.324

If f(x) = x^2 - x then f(x+1)- x = x^2 which means that if x is less than zero f(x) is greater than f(x+1)-x, so your proposition can't be true for all real x.

Let L0 be unstretched length::

Using Hookes law and gravity:

a) 10*9.8 = k*deltaL1            ...(1)      using g = 9.8 m/s^2

     14*9.8 = k*(delta1+0.025)   ...(2)    2.5cm is 0.025m to keep units consistent

Subtract (1) from 2) to get 4*9.8 = 0.025k  so k = 4*9.8/0.025 kg/s^2 or k = 1568 N/m  

b)  Using k in (1) we have 10*9.8 =1568*delta1  so delta1 = 10*9.8/1568 m or delta1 = 0.063m = 6.3cm

 Since L0 + delta1 = 80cm we must have L0 = 80 - 6.3 = 73.7 cm

The force pulling the spring down is F = mass*g  where g is gravitational acceleration (9.8 m/s^2)

The spring force resisting this is given by Hooke's law: F = k*deltax where k is the spring constant and deltax is the extension

so 2.5*deltax = 0.05*9.8

deltax = 0.05*9.8/2.5 = 0.196 m or 19.6 cm

Therefore the final length of the stretched spring is 19.6 + 10 = 29.6 cm

The answer can't be 240N/m - that is not a length!!!

As x → 0 then sin(x) → x and sin(2x) → 2x,  so (x - sin(2x))/(x+sin(x)) → (x - 2x)/(x + x) → -x/2x → -1/2

If x>2 and if x<-2 then x^2>4