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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(3) = 9

Then the point  (3,9) is on the inverse

Then  because f^-1 (x)  is the inverse of  g

Then the point   (9,3)   is on f(x)

So   f(9)  = 3

Then f(f(9) )   =   f(3)

But g (0)  = 3   

Then the point  (0,3)  is on g(x)  = f^-1(x)

Which means that the point  (3,0)  is  on the graph of f(x)

So

f(f(9))  = f(3)   =  0

How did you solve it?

Is it counting primes? 

oh wait, its fine, after a while i figured out the answer was 400

Good job, CPhill!

8) AC  = 15         BD  = 16      AD  = 19

We can find BC  [ the overlap of  AC and BD ]   thusly

AC + BD  =  15 + 16  = 31

BC =  31 - AD

BC  = 31 - 19

BC  =  12

So   AB  =  3   and  CD  = 4

Proof

A    B        C       D

   3      12      4

AC = 15

BD = 16

AD = 19

Thanks, I understand my mistake. I didnt consider 0 to be a perfect square

7)  IJ  = GJ - GI

     IJ  = (x + 27)  - 6

    IJ =    x + 21

GJ - HJ  = GH

(x + 27) - (x + 24)  = GH

              3  = GH

So....

GH + HI + IJ  = GJ

3 + (2x + 25) + (x + 21)  =  x + 27

3x + 49  =  x + 27

2x  = 27 - 49

2x = -22

x = -11

So

HJ  =  HI + IJ

HJ = ( 2x + 25) + (x + 21)

HJ =  (2(-11)+25) + (-11 + 21)

HJ = (-22 + 25) + (10)

HJ =  3 + 10

HJ  = 13

5)  (2x - 4) + (2x - 7)  =   17

I'll let you finish this one

6) (x + 14) + (x + 14)  = 12

And this one, too

Here's a few

1) Just  add the measures of  angles LMT  and TMN  to get the measure of angle LMN

2) GHQ + QHI   = GHI....so

(29 + x) +( x + 162) =  173

2x + 191  = 173    subtract 191 from each side

2x  = -18        divide both sides by 2

x = -9 

3) TMN + LMT  = LMN

     (x + 72)  + (96 + x)  =  166

        2x  +  168 =  166

         2x  = -2

           x  = -1

So LMT  =  96 - 2  =  94°

4) Not too much to this one

7 + ?  =  8

\(y = A e^{-2x}+Be^{-3x} + 3x^2 - 4x\\ \dfrac{dy}{dx} = -2A e^{-2x} -3Be^{-3x} + 6x-4\\ \left . \dfrac{dy}{dx}\right|_{x=0} = -2A -3B - 4 = 7\\ y(0) = A + B = 4\\ -2A -3B = 11\\ A+B = 4\\ \text{I leave you to solve for $A,~B$}\)

[ ABC]  =  (1/2)AB * Height

10  = (1/2)*6 * Height

10  = 3  * Height

10/3  =  Height

So

[ BCD ]   =  (1/2)BD * Height

[BCD ]  = (1/2)(10) (10/3)

[BCD ]  =  5(10/3)

[ BCD ] =  50 / 3

Area  = (1/2) Base * Height  =  (1/2)AC  *  Height

14 =  (1/2)(AC) * 6

14  = 3 AC      divide both sides by 3

14/3   = AC

Second one  is similar

4(10)^(3t)  =  28        divide both sides by 4

(10)^(3t)  = 7       take the Log of both sides

Log (10)^(3t)  = Log (7)        and we can write

(3t) Log (10)  = Log (7)         Log 10  =  1

3t (1)  = Log (7)

3t  =  Log(7)         divide both sides by 3

t  = (1/3) Log (7)  =   Log (7)^(1/3)   =  Log ( ∛7 )

only on your end

6e^(9t)  =  96     divide both sides by 6

e^(9t)  =  16       take the Ln  of both sides

Ln e^(9t)  = Ln (16)      and by a log property we can write

(9t) * Ln e   = Ln (16)      since Ln e  =  1  we have

(9t)* 1  = Ln (16)

9t  =  Ln (16)     divide both sides by 9

t  =  Ln(16)  / 9   =  (1/9)Ln(16)

Since all of the triangles bordering ABC are equilateral....then the perimeter of the hexagon can be expressed as :

2 (CB) + 2 (CA) + 2 (AB)  =  78

2 ( CB + CA + AB)   =  78      divide both sides by 2

(CB + CA + AB)  =  39

(perimeter of triangle ABC)  = 39

2 can be written  as  2 (a prime)  + 0  (a perfect square)

3 can be written as  2 ( a prime) + 1  (a perfect square)

4  can be written as  3 (prime) + 1 (a perfect square)

5 =   5 (a prime)  + 0 (a perfect square)

6  =   5 (prime) + 1 (a perfect square)

7 = 3 (prime) + 4 (a perfect square)

8 =   7 (prime) + 1 (perfect square)

9 = 5 (prime) + 4 (perfect square)

10

0 (perfect square) + 10 (non-prime)  

1 + 9    sum of two squares

2 (prime) + 8 (non-square)

3 + 7 sum of two primes

4 (perfect square) + 6 (non-prime)

5 + 5  sum of two primes

angle EKJ will span  an arc of  360 - 168  =192

angle EFC  = (4(26) + 24)  =   128°

Thanks, killersteve  !!!!

Thanks, heureka.......I wondered how WolframAlpha  derived that answer....!!!

I want to look at this one again......

1.  B =71°, c  = 8, b  = 16

We have a SSA situation.....so.....we will have either 1 triangle, 2 triangles or 0 triangles

Using the Law of Sines

sin 71 / 16  = sin C / 8

sin C  = .5sin71

arcsin (.5sin 71)  = C ≈  28.2°  or   180 -  28.2  = 151.8°

Note that the second angle value isn't possible since  71 + 151.8  > 180°

So....only one triangle exists

A  = 180 - 71 - 28.2  = 80.8°

And

a/ sin (80.8)  = 16/sin 71

a  = 16sin (80.8)/sin 71  ≈ 16.7

A = 80.8°   a  = 16.7

B = 71°      b  = 16

C = 28.2°   c =  8

x + y  =  sqrt (x^2 + y^2)      square both sides

x^2 + 2xy + y^2  =  x^2 + y^2

2xy  = 0

xy  =  0

Then.....this will be true as long as either x,y  = 0   or both = 0

Similarly

x + y  will not equal sqrt (x^2 + y^2)    whenever x and y are both non-zero

x^2 + 2x  , x  = 3

(3)^2  + 2(3)  =

9 + 6  =

15

First one

Central angles EAJ and EKJ span the same arcs as their measure

So....angle EKJ will span  an arc of  360 - 168  =  192°

Second one

This one is similar to the first

Since arc QRT  =198°

Then arc QAT   = angle QAT  =  360 - 198  = 162°  

Last one

Quadrilateral DEFC  is a cyclic quadrilateral.....opposite angles will sum to 180°

So...The sum of angles  EDC and EFC  = 180°

So

2x  +  (4x + 24)  = 180

6x + 24  = 180       subtract 24 from both sides

6x  = 156   divide both sides by 6

x = 26

So...angle EFC  = (4(26) + 24)  =   128°

In this problem the elementary matrix would be

\(e=\begin{pmatrix}1&0&0&0\\0&1&0&0\\2&0&-1&0\\0&0&0&1\end{pmatrix}\)