Melody

What is the average of all positive integers that have four digits when written in base 3, but three digits when written in base 6? Write your answer in base 10.

\(3^3\le x \le3^4-1\qquad \cup \qquad 6^2\le x\le6^3-1\\~\\ 27\le x \le 80 \qquad \cup \qquad 36 \le x\le 215\\~\\ 36\le x \le 80\)

\((36+37+ ....+80)  /  (80-36+1)\\~\\ =(36+37+ ....+80)  /  45\\~\\ =\frac{45}{2}(36+80) / 45\\~\\ =\frac{1}{2}(116) \\~\\ =58\)

Or I could have just said  36+  (80-36) / 2 = 36+22 = 58

thank you for showing us your anwers.  

The wording in the question is not completely clear.     (spelling edit, it was 'working' and maybe that was confusing)

It is hard to make the wording exact for questions like this.

But yes a and b are correct.  

c) there are 6 different  2 coloured squares.  But lots of each.

So he has 6 to choose from for his first and 6 to choose from for his second = 36 if he can choose 2 the same.

If they both have to be different then he has   6*5 = 30 possible choices

So tell us what you have done.

asinus is right.  I did the last bit wrong.

x= pi/3      or  x = 5pi/3

5!=120

so it looks like you want the middle one.

so it will start with a 3  

How many are less than 30000?    That would be 2*4!=48

how many start with 31?    That would be 3! = 6

So another 6 will start with 32

So that means there are 60 that are less than 34000

So the 60th one must be  32541

Have you tried doing any of this for yourself?

cos(2x)-5cos(x)+3=0

\(cos(2x)-5cos(x)+3=0\\ cos^2x-sin^2x-5cosx+3=0\\ cos^2x-(1-cos^2x)-5cosx+3=0\\ 2cos^2x-1-5cosx+3=0\\ 2cos^2x-5cosx+2=0\\ let\;\;y=cosx\\ 2y^2-5y+2=0\\ y=\frac{5\pm\sqrt{25-16}}{4}\\ y=\frac{5\pm 3}{4}\\ y=\frac{8}{4}\qquad y=\frac{2}{4}\\ cosx=2\qquad or \qquad cosx=0.5\\ cosx=0.5\\ x=2\pi n \pm \frac{\pi}{6}\qquad n\in Z\\ x=\frac{\pi}{6}\quad or \quad x=\frac{11\pi}{6}\qquad 0

The LaTex is not rendering.   I have taken a pic of what it is supposed to be looking like.

end in

04, 08, 12, ..... 96

96/4 = 24

Both of those absolute values will be great or equal to zero.

The only time the denominator will be zero is when both absolute expresssions are = 0

so   

\(x^2+3x-4 =0\qquad and \qquad   x^2+9x-10 = 0\\ (x+4)(x-1)=0 \qquad and \qquad   (x+10)(x-1) = 0\\ so \qquad x=1 \)

the only real value of x are not in the domain of  f(x) is 1

Here is the graph:

Have you even attempted to work out what the dimensions of all the different bits are?

Have you done anything at all towards answering this?