Melody

Ways to be all the same flavour = 3

Ways exactly 5 can be the same = 6C5*3*2 = 36

Ways exactly 4 can be the same = 6C4*3*2*2 = 180

Ways exactly 3 can be the same  = 6C3*3*2*2*2 = 480

Add them up = 699

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another thought...

How many ways altogther, no conditions   3^6 = 729

There are no ways to not double up the flavour, I mean they cannot all be differerent.

The only way  NOT to have 3 or more of one flavour is for there to be 2 of each flavour.

6!/(2*2*2) = 90

729-90 = 639.

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If my logic was correct these 2 answers would be the same.  

They are not the same so my logic is flawed.

MAYBE SOMEONE CAN POINT OUT THE LOGIC ERROR THAT I HAVE MADE.    

Another way to do it (which is harder in this instance, is to square both sides

\(|5x+3| = 4x - 1\\ 25x^2+30x+9=16x^2-8x+1\\ 9x^2+38x+8=0\\ x=\frac{-38\pm \sqrt{1156}}{18}\\ x=\frac{-38\pm 34}{18}\\ x=\frac{-19\pm 17}{9}\\ x=\frac{-2}{9}\qquad or \qquad x=\frac{-36}{9}\\ x=\frac{-2}{9}\qquad or \qquad x=-4\\ \)

Check both answers in original equation and you will find that one of them is invalid.

So the other one is the answer.  :)

RR BBBB

lock a R in then there appear to be 3 ways to put the other red in.

So I get 3

So either

RRBBBB

RBRBBB

RBBRBB

Its a typo   x is in the 3rd or 4th quad

Hi Juriemagic,

NICE ANSWER GUEST!

I am about to post this but it is the same as guests.  he/she has explained some things better than I have.

Since I have done it I will post it anyway.

there are not really that many to learn.

You learn a couple and then learn to manipulate them.

the main ones are

\(\boxed{\cos^2 x+sin^2x=1\\ Cos(A\pm B)=cosAcosB\mp sinAsinB\qquad \qquad \text{ The signs are opposite}\\ Sin(A\pm B)=sinAcosB\pm sinBcosA\qquad \qquad \text{ The signs are the same}\\ Tan(A\pm B)=\frac{tanA \pm tanB}{1\mp tanAtanB}}\)

There are some easier ones too like   tanA = sinA/cosA   but they should already be known to you.

\(16Sinx.Cos^3x-8SinxCosx\\ =8sinxcosx(2cos^2x -1)\\ =4(2sinxcosx)*-1(1-cos^2x-cos^2x )\\ =4(sin2x)*-1(sin^2x-cos^2x )\\ =4(sin2x)(cos^2x-sin^2 x )\\ =4(sin2x)(cos2x)\\ =2[2sin2xcos2x]\\ =2sin(4x)\)

You need to memorize

\(sin(A+B) = sinAcosB + cosAsinB\\ so\\ sin(2A)=sinAcosA+cosAsinA=2sinAcosA\)

It can be provben of course but you really need to memorize it.

so Sin24cos24 can be written as   0.5(2sin24cos24) = 0.5*sin48

Note: I have not looked at the whole question.

I do not understand what you are asking ??

Where is the diagram?

Let the answer number be y

the 2 digit number be M

\(10\le M\le99\\ 280\le 28M\le 2772\\ 280\le y^3 \le 2772\\ 7\le y \le 14\\ \)

So y could be  7, 8, 9, 10, 11, 12, 13, or 14

7^3 = 343   and 343 is not divisable by 28 so the answer is not 7

Just try all the other numbers to find the one that works.   

\(7x=\frac{x}{3}=21\\ \)

there is no answer to this.   7x cannot equal  x/3  unless x=0

and

 if x=0 then  x/3 =0 not 21