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Just to confirm the above, here is a short computer code that spits out EQUAL values for x, y and z, no matter what initial values you assign to them.

a=1; b=1;d=1;c=a^b; e=b^d;f=d^a;if(c==e and e==f, goto7, goto10);print"x=", a,", ",; print"y=",b,", ",; print"z=",d,", ",; a++;if(a<100, goto3, 0);a=1;b++;if(b<100, goto3, discard=0; a=1;b=1;d++;if(d<100, goto3, 0)

x= 1 , y= 1 , z= 1 , x= 2 , y= 2 , z= 2 , x= 3 , y= 3 , z= 3 , x= 4 , y= 4 , z= 4 , x= 5 , y= 5 , z= 5 , x= 6 , y= 6 , z= 6 , x= 7 , y= 7 , z= 7 , x= 8 , y= 8 , z= 8 , x= 9 , y= 9 , z= 9 , x= 10 , y= 10 , z= 10 .............and so on.

Any  base angle  of either of the two triangles will have a measure of

[180 - 30] / 2  =    150/2   = 75°  =  x

[The two triangles are isosceles  with their base angles equal  because the sides opposite these angles are equal]

C  = 92

So.....Angles  A  and B must be equal

A + B + C  = 180

B + B + 92  = 180

2B + 92  = 180   subtract 92 from each side

2B  = 88   divide both sides by 2

B = 44°

First Part - The first 20 digits. This is very simple.
13^13 = 302,875,106,592,253 
Multiply the above number by log 13(base 10). Carry the result to about 100 decimal places. You should get:
302,875,106,592,253 x log 13 =337,385,711,567, 664.8232318987826741897646811857360282268728608288726576980008833486046624631780203460135940132237440066.
From the above result, keep the fractional part ONLY.
Raise the fractional to the power 10:
10^0.8232318987826741897646811857360282268728608288726576980008833486046624631780203460135940132237440066 =6.656284847586597803523004328401344362443297464463201858579506080321635285934050922391456377579985062630734182573941. Remove the decimal point and count the first 20 digits from the left =66,562,848,475,865,978,035 - This is the first 20 digits of your number.

Second Part - The last 20 digits.
This part is very difficult to do without a programming language. For this purpose, I wrote a short code in Python which gives you the last 20 digits:
1- %%time
2 - # Fast python code that first calls pow()  
3 - # then applies % operator
4 - import math as m
5 - a =13
6 - b =302875106592253 
7 - p =100000000000000000000  
8 - # Using direct fast method to compute  
9 - # (a ^ b) % p. 
10 - d = pow(a, b, p) 
11 - print( f"{d :,d}")  
The last 20 digits are = 91,055,086,363,688,549,053

Thanks Tiggsy,

I used GeoGebra.

It is free. 

I have a downloaded version but I think you can use it online as well.

I do most of my diagrams with it.     

 BAC  = 28°  = DCP     [ a consequence of parallels AB and CD  cut by transversal AC ]

And since DP = PC then angle DCP  = angle PDC

Therefore angle DPC  =  180 - 2(28)  =  180 - 56  = 124

And by the exterior angle theorem

Angle QPD + Angle QDP  = Angle DPC

    90  + Angle QDP  =  124

Angle QDP =  124 - 90 =  34°

In order for x^y = y^z = z^x to hold true, it must mean that: x = y = z. Otherwise, if they are not equal to each other, x^y will never equal to y^z or to z^x. Therefore, x must equal y and must equal z. Or, x = y. 

See the image here :

Let ABC  be an equilateral triangle  of side 6

The area of this triangle  is   (√3/4) *side^2  = (√3/4)*6^2  = √3 (36/4) = 9√3

Any point  falling into  [ ILMNKJ]  will be nearer to the center O  than to any vertex

Looking at triangle  ICL....

IL  = 2     and this triangle  is similar to triangle ABC

So.....the scale factor  of ICL to ABC  = (2/6) =   (1/3)

So....the area of triangle  ICL  =  area of triangle ABC * (scale factor)^2  = 9√3 * (1/3)^2  =  √3

And triangles AJK and BNM  are congruent to ICL  so they have the same areas

So  [ ILMNKJ] has an area  of [ABC]  - [ ICL] - [ AJK] - [ BNM]  =

9√3 - √3 - √3 - √3   =  

9√3 - 3√3  =

6√3

So.....the probability that P is closer to the center O than to  any vertex  =

[ ILMNKJ]         6√3          6            2

________  =   _____  =  ___ =    ___

[ ABC ]             9√3          9            3

The second one is easier if we identify what the parts of each martix represent

The first matrix

   S          M        L

[ 1.50    2.50    6.50   ]    prices of the three sizes at store 1

[ 2.00    3.50    8.00   ]    prices of the three sizes at store  2

The second matrix  

No. on shelf     No. in stockroom

 of each size     of each size

  10                    24              S

   18                   48              M

   6                     12              L

The first row of the first matrix x the first column of the second gives the total  price of the stock on the shelves in store 1....so  this is  entry a11  in the resulting matrix  = 

(1.50)(10 ) + (2.50)*(18)  + (6.50)(6)  =  99

The first row of th first matrix x the secolumn of the second matrix gives the total price of the stock in the stockroom at store 1....this   is entry a12  in the resulting matrix =

(1.50)(24 ) + (2.50)(48) + (6.50)(12)  = 234

The second row of the first matrix x the first column of the second matrix  gives  the total price of the stock on the shelves at store 2......this is entry a21  in the resulting matrix =

(2.00)(10) + (3.50)(18) + (8.00)(6)  = 131

Finally.....second row of the first matrix x second column of the second matrix gives the total price of the stock in the stockroom of store 2.....this is entry a22 in the resulting matrix  =

(2.00)(24) + (3.50)(48) + (8.00)(12)  = 312

So     we have the following resultant (product) matrix :

Shelves   Stockroom

99            234       Store 1

131          312       Store 2

The rows represent  the  different stores

And the columns represent the location of the stock within each store

First one  

-9    is correct

Second one

-3R2 + 4R1  .....as before......I assume that the result goes in R1

-3R2   =  -24     3    -9

4R1  =    -12    20    8

              ___________

              -36   23    -1   

So.....the resulting matrix  is

-36   23   -1

   8     1    3

if you could check this ne as well, wasnt sure what to put at the bottom, if you could help with that ?thank you again, this is the last three 

Not quite sure how to solve this either, but after writing out a couple of steps, there seem to be some patterns in  the number of mangoes different people have... Not sure about a proof yet though. Would really appreciate it if someone more knowledgeable could look at this! 

Yeah that's what I thought too thanks for answering 

Yeah that's what I thought too thanks for answering 

Was a bit confusing but I get how you got the answer, desmos helped my understand it more as well. Thanks

Ths one is really easy! Wow, thanks for yuor help. I will take notes

If you mean by "have the same graph" that you can pick up the graph of one polynomial and place it exactly on top of the other one, 

i.e. that p1(x) = p2(x) then yes they have the same roots.

As the other poster noted the converse is not true.  Polynomials with identical roots can have different graphs.

BEFORE her 6th try is 5 tries

The probability that she DOESN"T solve it daily is 3/4

1- (3/4)⁵ = 781/1024 = 76.3%      ???

75c-300=25c+200   subtract 25c from both sides, add 300 to both sides

50c  =  500     divide both sides by 50

c =  10

9  = -6   + x / 3       add 6 to both sides

15  =  x  / 3           multiply both sides by 3

45  =  x

Here is another way:
1 - Start with 1
2 - Divide 1 by 4 = 1/4
3 - Take one of the remaining 1/4
4 - Subdivide that second 1/4 into 3 parts, or:
5 - (1/4) / 3 = (1/4) x (1/3) = 1 / 12
6 - Add 1/4 in (2) above to 1 / 12 in (5) above, or:
7 - 1 / 4 + 1 / 12 = 1 / 3 - which is what you want. 

The second one requires a little thinking

If the center of the circle lies on the line  y = 6x + 2  and the circle is tangent to the x axis  at (-2, 0)  then the the x coordinate of the center of the circle  is at  x = -2

So.....the y cordinate of the center  must be y = 6(-2) + 2  =  -10

So....the center  is  ( -2, -10)

And since this circle is tangent to the x axis......the radius must = 10

So.....the equation is   (x + 2)^2 + (y + 10)^2  = 100

See the graph here :   https://www.desmos.com/calculator/8abh7hhozi

The first  isn't too difficult 

We have the form   (x - h)^2  + (y - h)^2  = r^2

The center  is ( x , h)    and the radius  =  sqrt (r^2) = r

So....we have

(x - 2)^2 + (y + 3)^2  = 80   writing this in a slightly different form we have

(x - 2)^2  + ( y - (-3) )^2  = 80

So.....the center  is   (2, -3)    and the radius  = sqrt (80)  =  sqrt (16 *5)  = 4sqrt (5)

Egyptian Fractions have the form :  1 / positive integer 

1/3 =  1/a + 1/b

Let z  = 3

a,b  must be > 3

So let a  =  z + m

And let b =  z + n

So we have

1/ z  =    1/ [z + m]  +  1/[;z + n]

1/z  =  [ 2z + m + n] / [(z + m) (z + n)]      cross-multiply

(z + m) (z + n)  =  z (2z + m + n ]

z^2 + mz + nz + mn =  2z^2 + mz + nz

z^2 + mn = 2z^2

z^2  = mn       so

3*2 = mn

9 = mn

So the possibilites for m,n  are

m   n

1    9

3    3

So....the possible fractions are

1/ [3 + 1]   +  1/ [ 3 + 9]  =     1/4 + 1/12

1/[3 + 3] + 1/[3 + 3]  =  1/6 + 1/6

Only the first is what we need

So

1/4  + 1/12