Melody

Please put only one question per post.

a)  I would prove this using induction.     I will outline the method here.

Step 1:

Show true for n=0

Step 2:

Assume true for n=k where k is some integer.

then prove that it will be true for n=k+1   (I've done this myself so I know it works)

Step 3

Assume true for n=k

Prove it is true for n=k-1

Step 4.  (Summary)

Since it is true for n=0 it must be true for n=1, n=2 .......etc

also, since it is true for n=0 it must be true for n=-1, n=-2 ..... etc

Therefore it must be true for all integer values of n.

Now try filling in the blanks.

Believe it or not initial guest answerer, this is Ginger's way of saying that she likes you.

OR at least that she does not dislike you yet.     LOL

Here is an answer that you can learn from.

The probability that 1 person has it is     1/3000

The prob that a person does not have it is  2999/3000

If there are 6 people in a room the prob that no one has it is    (2999/3000)^6

so

the prob that at least 1 of the 6 people has it is     1-  (2999/3000)^6   which is approx    0.002

\(6x^3y^4+24x^6y^2-36x^3y^3\)

Hint:

What is the greatest number that goes into 6,24 and 36?

You can also factor out x^3,  that is the biggest one for x

And what is the biggest factor of y that you can factor out?

Now, what is the answer?

I have not heard these terms used although most of the time I am sure they are self-explanatory.

I'm not sure about the first one, is it infinite or is it a jump?  Looks like both .....

the others I would check using desmos.

I have said this to you before.  Don't you know what I mean?

I'll do 3 just to show you.

answered.

I think I have thought this through but I have not completed all the maths.

I let XYZ be circumscribed by a circle of radius 3  (I find it easier to use actual numbers)

area Triangle XGY = 0.5*3*3sin120 =(9sqrt3)/4

area Triangle X'GY' = 2/3 of that = (3sqrt3)/2

Now find the area of one little triangle.

the equation of XZ is  y=sqrt3x+3

point A has a y value of 1 so the x value is (-2sqrt3)/3    

Y' is the intersection of   x^2+y^2=4  and  y=1   so I think it is  (-sqrt3,1)

Distance AY'=  sqrt3/3

So from that you can work out the area of each of the little triangles.

Then you can work out the rest.

(maybe there is a better way, but that will work)

You will need to check all my logic and arithmetic.

Nice work asinus and guest :)

Oh dear ilorty, you have made a couple of basic errors   

sqrt 50 is   only a half of the diagonal of the square.  So it is in fact the radius of the big circle

Also half of sqrt(50) is definitely not sqrt(25)

Never mind, it happens to us all.    

Hi Jenny,

Use desmos graphing calculator and you can check your answers for yourself  

If you are a smart man you can work this out for yourself.  Just draw a rough number line.