Melody

Thanks guest.  You are totally correct.

I failed to recognise that all the even numbers have a factor of 2  

I'll start again

4 numbers are randomly selected from the subset (1,2,3,4,5,6,7,8,9) . What is the probability that the product of the 4 randomly selected integers is a multiple of 14?

One of the numbers has to be 7.  AND at least one has to be even.

Forget about the 7 and there are 4 even and 4 odd numbers from which to choose the other numbers

3 evens                                  4C3                       = 4

2 evens and an odd                4C2*4C1 = 6*4   = 24

1 even and 2 odds                 4C1* 4C2 = 4*6   = 24

  Total  4+24+24 = 52               (just like guest said)

Prob = 52 / 4C9 =  52 / 126

Now we are in agreement.

\(\frac{(y-x)^2}{(y-z)(z-x)} + \frac{(z-y)^2}{(z-x)(x-y)} + \frac{(x-z)^2}{(x-y)(y-z)}\\ =\frac{(y-x)^2(x-y)}{(y-z)(z-x)(x-y)} + \frac{(z-y)^2(y-z)}{(z-x)(x-y)(y-z)} + \frac{(x-z)^2(z-x)}{(x-y)(y-z)(z-x)}\\\\ =\frac{[(y-x)^2(x-y)]\;\;+\;\;[(z-y)^2(y-z)]\;\; +\;\; [(x-z)^2(z-x)]}{(y-z)(z-x)(x-y)} \\ =\frac{[-(y-x)^2(y-x)]\;\;-\;\;[(z-y)^2(z-y)]\;\; -\;\; [(x-z)^2(x-z)]}{(yz-yx-z^2+xz)(x-y)} \\ =\frac{-\{[(y-x)^3]\;\;+\;\;[(z-y)^3]\;\; +\;\; [(x-z)^3]\}}{(yz-yx-z^2+xz)(x-y)} \\ =\frac{-\{[(y-x)^3]\;\;+\;\;[(z-y)^3]\;\; +\;\; [(x-z)^3]\}}{(yzx-yzy-yxx+yxy-z^2x+z^2y+xzx-xzy)} \\ =\frac{-\{[y^3-3y^2x+3yx^2-x^3]\;\;+\;\;[z^3-3z^2y+3zy^2-y^3]\;\; +\;\; [x^3-3x^2z+3xz^2-z^3]\}}{(-y^2z-x^2y+y^2x-z^2x+z^2y+x^2z)} \\ =\frac{-3\{[-y^2x+yx^2]\;\;+\;\;[-z^2y+zy^2]\;\; +\;\; [-x^2z+xz^2]\}}{(-y^2z-x^2y+y^2x-z^2x+z^2y+x^2z)} \\ =\frac{-3(zy^2+yx^2 -y^2x+xz^2-z^2y-x^2z}{(-y^2z-x^2y+y^2x-z^2x+z^2y+x^2z)} \\ =3 \qquad \text{so long as the denominator isn't 0}\)

I'm sure Textot's formula is what you were supposed to use.

LaTex

\frac{(y-x)^2}{(y-z)(z-x)} + \frac{(z-y)^2}{(z-x)(x-y)} + \frac{(x-z)^2}{(x-y)(y-z)}\\
=\frac{(y-x)^2(x-y)}{(y-z)(z-x)(x-y)} + \frac{(z-y)^2(y-z)}{(z-x)(x-y)(y-z)} + \frac{(x-z)^2(z-x)}{(x-y)(y-z)(z-x)}\\\\
=\frac{[(y-x)^2(x-y)]\;\;+\;\;[(z-y)^2(y-z)]\;\;  +\;\;  [(x-z)^2(z-x)]}{(y-z)(z-x)(x-y)} \\
=\frac{[-(y-x)^2(y-x)]\;\;-\;\;[(z-y)^2(z-y)]\;\;  -\;\;  [(x-z)^2(x-z)]}{(yz-yx-z^2+xz)(x-y)} \\
=\frac{-\{[(y-x)^3]\;\;+\;\;[(z-y)^3]\;\;  +\;\;  [(x-z)^3]\}}{(yz-yx-z^2+xz)(x-y)} \\
=\frac{-\{[(y-x)^3]\;\;+\;\;[(z-y)^3]\;\;  +\;\;  [(x-z)^3]\}}{(yzx-yzy-yxx+yxy-z^2x+z^2y+xzx-xzy)} \\
=\frac{-\{[y^3-3y^2x+3yx^2-x^3]\;\;+\;\;[z^3-3z^2y+3zy^2-y^3]\;\;  +\;\;  [x^3-3x^2z+3xz^2-z^3]\}}{(-y^2z-x^2y+y^2x-z^2x+z^2y+x^2z)} \\
=\frac{-3\{[-y^2x+yx^2]\;\;+\;\;[-z^2y+zy^2]\;\;  +\;\;  [-x^2z+xz^2]\}}{(-y^2z-x^2y+y^2x-z^2x+z^2y+x^2z)} \\
=\frac{-3(zy^2+yx^2 -y^2x+xz^2-z^2y-x^2z}{(-y^2z-x^2y+y^2x-z^2x+z^2y+x^2z)} \\
=3  \qquad \text{so long as the denominator isn't 0}

Guest, you made a small mistake.  I like seeing your answers though. :)

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I'm going to try and solve in pairs

m=-1 mod 6 and 1 mod 10

11  works

m=4 mod 7 and 1 mod 10

11 works

now I look, 11 works for all of them

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So the solutions will be of the form    m=(5*7*8*9)k + 11

the first 2 are 11 and  (5*7*8*9) + 11

the 3rd one is         2(5*7*8*9) + 11   and this is bigger than    6*7*8*9

so

There are just 2          they are     11 and  2531

The inverse of funtion x cannot be a function because for one segment of f(x)=-2  The inverse of that is  just -2

      I mean   for the function     For -3<=x<=-1     f(x)=-2

      the inverse would be when x=-2   \(f^{-1}(2)\)   can be any number between -3 and -1   

      Hence there is no unique value of \(f^{-1}(2) \)  Hence \(f^{-1}(x)\)    is not a function.

f(x)         domain [-3,5]    range  [-2,3]

\(f^{-1}(x)\)     domain [-2,3]         range  [-3,5] 

I assume for the last bit

 (x,y)  --> (y,x)     becasue the inverse of a function is its reflection about the line y=x

Here is a mapping

Yes, Good work!   

[-6,-4)  and  (0,6) 

The question is actully asking .  "For what values of of x is the y values of the graph negative."

Well, it is negative between  x= -6 and x= 4      [-6,-4)     

The brackets are curly at x=-4 because at that point y=0 which is not zero.

NOW can you tell me what the other correct intervals are?

You got the right answer but what you have written is not altogether correct.

You have to treat the floor funtion like you would brackets.

So you work out what is inside and THEN take the flor of it befrome moving on to the rest of the expression.

I also desctribed what the floor funtion does rather badly.

The floor funtion simply ROUNDS A NUMBER DOWN to the nearest INTEGER.   

(Thanks for the guidence EP) 

I know you already worked out much of this but I wanted to reiterate the whole lot.  ;)

\(f(1)=\lfloor 1 \rfloor -2 = 1-2=-1\\ f(1.5)=\lfloor 1.5 \rfloor -2 = 1-2 = -1\\ f(3)=\lfloor 3 \rfloor -2 =3-2=1\)

giving    (1, -1)  (1.5, -1) and  (3, 1)

You are just being asked for the equation of this parabola

There is nothing to satisfy.  You are most likely missing an equal sign ...