Melody

Attn:  ImDecentAtMaths

Please stop giving answers only.

It may help a student to cheat but it certainly cannot teach them anything.

f(x)=ax^3+bx^2+cx+d

f(1)=a+b+c+d = 4

a+b+c+d=4

If a = 4 then b,c, and d = 0                                               1  possibility

If a=3 then one of b,c,d is 1 and the others are 0             3  possibility

If a=2 and one of the others is 2                                       3  possibility 

If a=2 and two of the othes are 1                                      3  possibility

If a=1 and the other three are all 1                                   1  possibility

If a=1 and one of the others is 3                                       3  possibility

f a=1 and 2 of the others are 1 and 2 respectively          3!=6  possibility

Total  1+3+3+3+1+3+6 = 20

You need to check that I haven't missed any.

Now you work it our

Hints:

Use Pythagoras's theorem to find the third side of the triangle.

What is the radius of each of the three semicircles?

Find the areas, one at a time 

Add them together.

If you wnat more help show us what you have done for yourself.

hint:

try it with base 10 and see what happens.  :)   The answer will be the same no matter what base you use.

I do not understand what you have done guest.  There are 3 people and 6 cards so they get 2 cards each. 

Why do you have 2 lots of 4?  

6 cards can be ordered in 6! ways      

but ABCDEF  is the same as BACDEF etc

6! / 2^3=  90 ways to distribute the cards. 

3!*3! ways for them to get a read each  = 36 ways

36/90 = 40%

Please don't give just answers like this.  

Incomplete working is better.

Why can't you talk about it? 

You are not posting forbidden questions so I don't see the problem.

a and b are the roots of that equation.  Use the quadratic formula to find them.

Then simplify what you have to find and then do the substitution.  It falls out well.

If you still need help post what you have tried for yourself so that we can see where you get stuck.