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Suppose that   g(x) = f^-1(x) .  If   g(-15) = 0 ,  g(0) = 3 ,  g(3) = 9   and   g(9) = 20 ,  what is   f( f(9) ) ?

If     g(x) = f^-1(x)     then we can take  f  of both sides of the equation to get     f( g(x) )  =  x

Since  f( g(x) )  =  x  ,

f( g(3) )  =  3

                        And we know   g(3)  =  9   so we can substitute  9  in for  g(3)

f( 9 )  =  3

Again since  f( g(x) )  =  x  ,

f( g(0) )  =  0

                        And we know   g(0)  =  3   so we can substitute  3  in for  g(0)

f( 3 )  =  0

So we have found....

f( f(9) )   =  f( 3 )  =  0

Given that  a  and  b  are positive integers where  a > b , I do think it is always true that

a mod (a - b)   =   b mod (a - b)

I don't know the best way to explain it, but here is how I convinced myself. In the diagram,

w  is the remainder when  a  is "filled up" with as many blocks of width  a-b  as possible.

x  is the remainder when  b  is "filled up" with as many blocks of width  a-b  as possible.
 

In other words,

w  =  a mod (a - b)

x  =  b mod (a - b)

And we can see that...

w + (a - b)  =  x + (a - b)

w  =  x

Therefore,

a mod (a - b)  =  b mod (a - b)

But I do not think it is always true that if     a mod  x  =  b mod x     then     x  =  a - b

For example,     10 mod 3  =  4 mod 3     but     3  ≠  10 - 4

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NG is correct that   sin(45°)  =  \(\frac{\sqrt{2}}{2}\)

However,   arcsin(45)  ≠  \(\frac{2}{\sqrt{2}}\)

If     a  =  arcsin(45)     then     sin( a )  =  45

But there is no possible angle  a  that can make   sin( a )  =  45

The biggest possible value of  sin( a )  is  1 .

So there is no real answer to this.

\(\ \phantom{=\quad}\frac{6}{\sqrt{36} + \sqrt{27}} + \frac{6}{\sqrt{27} + \sqrt{18}} + \frac{6}{\sqrt{18} + \sqrt{9}}\\~\\~\\ =\quad\Big(\frac{6}{\sqrt{36} + \sqrt{27}}\cdot\frac{\sqrt{36}-\sqrt{27}}{\sqrt{36}-\sqrt{27}}\Big) + \Big(\frac{6}{\sqrt{27} + \sqrt{18}}\cdot\frac{\sqrt{27}-\sqrt{18}}{\sqrt{27}-\sqrt{18}}\Big) + \Big(\frac{6}{\sqrt{18} + \sqrt{9}}\cdot\frac{\sqrt{18}-\sqrt9}{\sqrt{18}-\sqrt9}\Big)\\~\\~\\ =\quad\frac{6\sqrt{36}-6\sqrt{27}}{36-27}+ \frac{6\sqrt{27}-6\sqrt{18}}{27-18} + \frac{6\sqrt{18}-6\sqrt9}{18-9}\\~\\~\\ =\quad\frac{6\sqrt{36}-6\sqrt{27}}{9}+ \frac{6\sqrt{27}-6\sqrt{18}}{9} + \frac{6\sqrt{18}-6\sqrt9}{9}\\~\\~\\ =\quad\frac{6\sqrt{36}-6\sqrt9}{9}\\~\\~\\ =\quad\frac{36-18}{9}\\~\\~\\ =\quad2 \)_

This probably isn't the most sophisticated way, but we can easily solve this just by plugging in different numbers.

Let   f(x)  =  x² + 6x + 9

f(1)  =  1² + 6(1) + 9  =  1 + 6 + 9  =  16

f(2)  =  2² + 6(2) + 9  =  4 + 12 + 9  =  25

f(3)  =  3² + 6(3) + 9  =  9 + 18 + 9  =  36

f(4)  =  4² + 6(4) + 9  =  16 + 24 + 9  =  49

So we can see that there are only  2  positive integers  x  that satisfy  20 < f(x) < 40

They are:     x = 2     and     x = 3

(3^x - 2^y)(3^x + 2^y)  =  9^x - 4^y

                                             We know  3^x - 2^y  =  2   and   9^x - 4^y  =  10

( 2 )( 3^x + 2^y )  =  10

                                             Divide both sides of the equation by  2

3^x + 2^y  =  5

(4x + 2y)²  =  64

                                             Expand the left side of the equation.

(4x + 2y)(4x + 2y)  =  64

16x² + 16xy + 4y²  =  64

                                             Divide both sides of the equation by  4

4x² + 4xy + y²  =  16

                                             Subtract  4xy  from both sides of the equation.

4x² + y²  =  16 - 4xy

                                             Since  xy = 2  we can substitute  2  in for  xy

4x² + y²  =  16 - 4(2)

                                             Simplify the right side of the equation.

4x² + y²  =  8

The list of distinct, positive integers that has a median of 10 and the smallest average is:

1, 2, 3, 4, 9, 11, 12, 13, 14, 15

We know the first four numbers have to be  1, 2, 3, 4  because they are the four smallest distinct, positive integers.

Then the two middle numbers (9 and 11) must sum to 20 to make the median 10. And we can see why we should use 9 and 11 to minimize the average because, for example, if we used 5 and 15, that would force all of the last four numbers to be greater.

Then the last four numbers are the four smallest, distinct positive integers greater than 11.

average  =  (1 + 2 + 3 + 4 + 9 + 11 + 12 + 13 + 14 + 15) / 10  =  84 / 10  =  8.4