Melody

The easiest way to start is to multiply both sides by 30   (that will get rid of all the fractions.)

Let f(x)=2x-3 For what value of x is (f(f(x)) = x?

\(f(x)=2x-3\\ so\\ f(f(x))=2(2x-3)-3\\ \text{so when does}\\ 2(2x-3)-3=x \)

well you will have to solve it to find out ;)

You are more likely to get a good answer if you present your question in a readable format.

\(T=11\sqrt{M}\\ \frac{T}{11}=\sqrt{M}\)

Now square both sides.

The second one is easy.  The last one is not so bad either.

I assume you are being asked to make x the subject.

I'll start half way through your working.

\(y - 3x = 50x^2 - 60x + 26\\ y = 50x^2 - 57x + 26\\ 2y-52 = 100x^2 - 114x \\ \frac{2y-52}{100} = x^2 - 1.14x \\ \text{Complete the square}\\ \frac{2y-52}{100} +(0.57)^2= x^2 - 1.14x+(0.57)^2 \\ \frac{2y-52}{100} +0.3249= (x-0.57)^2 \\ \frac{2y-52+32.49}{100} = (x-0.57)^2 \\ \frac{2y-19.51}{100} = (x-0.57)^2 \\ \pm\frac{\sqrt{2y-19.51}}{10} = (x-0.57) \\ x=0.57\pm\frac{\sqrt{2y-19.51}}{10} \\ x=\frac{5.7\pm\sqrt{2y-19.51}}{10} \\\)

You will need to check for careless mistakes.

Convert these to base 10.

the base 10 equation is    A=6B-22

it was base 6 so A and B must both be 0,1,2,3,4,or5

so solve this.

\(0\le6B-22\le5\)

there is only one possible answer.

Then solve for A

OR alternatively ....

rationalize the denominator first by multiplying top and bottom by sqrt5

\(\sqrt{\frac{\frac{5}{\sqrt{80}}*\sqrt5+\frac{\sqrt{845}}{9}*\sqrt{5}+\sqrt{45}*\sqrt5}{\sqrt5*\sqrt5}}\\ \sqrt{\frac{\frac{5}{\sqrt{16}}+\frac{\sqrt{5*169}}{9}*\sqrt{5}+\sqrt{5*9}*\sqrt5}{5}}\\\)

etc

Ignore the biggest square root till the end

Simplify each term on the numerator separately like this.

\(\frac{5}{\sqrt{80}}=\frac{5}{\sqrt{16*5}}=\frac{5}{4\sqrt{5}}\)

etc

Now you need to rationalize the denominator so multiply every term, top and bottom by   sqrt5

Keep going from there.   It all falls out quite nicely.

Try comparing them in pairs from the top down.

A = 100^500 = (100^400) * (100^100)  

B = 200^400 = (100^400) * (2^400) = (100^400) * (16^100) < A

So B is less than A

B = (100^400) * (16^100) = (100^300) *(100^100) * (16^100) = (100^300) *(1600^100) 

C =  300^300 = (100^300) * (3^300) =  (100^300) * (27^100) < B

So C is less than B

And you can do the others.

It might be easier to compare them from the bottom up but it is still the same basic method.