Bertie

We are solving the equation f(x)=0. Graphically the roots (solutions) of the equation are the x co-ordinates of the points where the y=f(x) crosses the x axis, i.e. the points where y = 0.

The Newton-Raphson method starts with a point x0 close to a root, takes the tangent to the curve for that value of x, (the tangent at A in the sketch below) and then calculates the x co-ordinate of its intersection with the x axis, (x1 in the sketch below).  (All done algebraically of course). Hopefully x1 will be a better approximation to the root  than was x0. The process is then repeated using x1 as the new starting point to find an x2 then again to find an x3 and so on until converged to whatever degree of accuracy is required.

The method can go awry if the gradient of the tangent is small, it's possible to be kicked a long way (possibly irretrievably) from the original point. In the sketch below, for example, suppose x0 were further to the right so that the tangent to the curve was constructed at B. The next approximation could be way off to the left or the right depending on whether B is to the left or right of the maximima. Either way the next 'approximation' to the root will be useless. On paper, the problem manifests itself with a small value for f '(x) and a correspondingly large correction term f(x)/f '(x).

You missed the & in front of \frac.

Do you really mean 'without a calculator', or are you simply not allowed to use the power ^ button ? The 5th root of 12 is 1.643752 (6dp). How many decimal are you supposed to calculate ? Whatever numerical method you use there is going to be a fair amount of heavyish arithmetic.

Best of the elementary methods is Newton-Raphson, (the formula for which you will find in the first response to this question).

Using that method,

$$f(x)=x^{5}-12, \text{ so that } f'(x)=5x^{4}$$

and so

$$f(x_{n+1})=x_{n}-\frac{x_{n}^{5}-12}{5x_{n}^{4}}=\frac{1}{5x_{n}^{4}}\left(4x_{n}^{5}+12\right).$$

Starting with x₀=1.5, that produces the sequence 1.674, 1.64483, 1.643753, 1.643752, (so convergence to 6dp after 4 iterations).

Good luck if you have to do this on paper !

It's a snooker ball.

The area of the square is 12^2 + 5^2 = 169.

The length of the hypotenuse is the square root of this, 13.

Have you heard of the story of the red indian chief who had three squaws (wives) ?

They were allowed to choose the bedding on which they slept, each according to her own preference. The first chose to sleep on a hippopotamus skin, the second on a lion skin and the third on a tiger skin. (And yes I know what you are going to say, but this was long ago when there were hippopotami and tigers in North America.)

In the fullness of time, the first wife bore triplets, the second wife had twins, but the third could only manage a single baby.

The moral of the story is that "The squaw on the hippopotamus is equal to the sum of the squaws on the other two hides".

There are just about always two methods for dealing with problems like this, a counting method and a probability method, (and often one is more convenient than the other.)

The counting method has been used for both of the correct answers so far,  so here's the probability method.

Suppose that all 17 were injured, some of them critically, and that 4 of them subsequently die, (at intervals of several minutes say). The probability that the first to die is a non-skier is 14/17, that the second third and fourth to die are also non-skiers 13/16, 12/15 and 11/14 respectively. The required probability will be the product of these, 24024/57120 = 143/340.

1 statute mile = 1760 yards = 8 furlongs

1 furlong = 220 yards.

(Five and six furlong races are common distances in English horse racing.)

1 chain = 1/10 of a furlong = 22 yards = the length of a cricket pitch.

1 rod (pole or perch, different names for the same thing) = 1/4 of a chain = 22*3/4 = 16.5 feet.

1 acre = 10 square chains.

(So 1 square rod = 1/16 th of a square chain = 1/160 th of an acre.)

1 square mile = 640 acres.

Something to do with the gravitational force of attraction ?

$$f=g\frac{m_{1}m_{2}}{r^{2}}.$$

We don't have storks in England, (except perhaps a few in the zoos), so that can't be the complete answer.