Melody

Graph y=(3x+4)+(x^2 - 5 + 6) + (|3-x|-8) + (((x+4)^3) - 8) + ((sqrt(5-x))-3) + (log_2(x) - 4) - (1/x) + (x^2-8+15)/(x-3)(x+4) - 3^x + 9

Did you have fun making that up ?

Here is you diagram, can you work it out now.

\(3^0=1 \;\;(base10)\;\;\equiv 1 \;\;(base16)\;\; \\ 3^1=3 \;\;(base10)\;\;\equiv 3 \;\;(base16)\;\; \\ 3^2= \;\;(base10)\;\;\equiv 9 \;\;(base16)\;\; \\ 3^3=27 \;\;(base10)\;\;\equiv B \;\;(base16)\;\; \\ 3^4=81 \;\;(base10)\;\;\equiv 51 \;\;(base16)\;\; \\ \)

The pattern for the last digit is now set.

\(3^{0+4n} \;\;ends \;\;in\;\; 1 \;\;(base16)\;\; \\ 3^{1+4n} \;\;ends \;\;in\;\; 3 \;\;(base16)\;\; \\ 3^{2+4n}\;\;\;\;ends \;\;in\;\; 9 \;\;(base16)\;\; \\ 3^{3+4n} \;\;\;\;ends \;\;in\;\;B \;\;(base16)\;\; \\ \)

2019 = 4*504+3

So 

\(3^{2019}\quad \text{has a unit digit of WHAT in base 16.}\)

** Please give a response

continued here 

Yes it is 13/25     (Max I think you made a small error)

Here is my contour probability  graph

Thanks for the equine dental lessen Ginger :)

Compute the value of 

1-2+3-4+...+2019-2020+2021

(1+3+5+ ..... +2021) - ( 2+4+6+ ..... +2020)

these are just arithmetic progresstions

The sum of an AP = \(\frac{n}{2}(a+L)\)

There are 2021 terms in total  half that 1010.5   so there are 1010 even numbers and 1011 odd numbers

(1+3+5+ ..... +2021) - ( 2+4+6+ ..... +2020)

\(=[\frac{1011}{2}(1+2021)]-[\frac{1010}{2}(\color{red}{2} \color{black}+2020)]\\ =[1011*(1011)]-[505*(2022)]\\ =1022121 - 1021110\\ =1011\)

I found my careless error, the corrected number is in red.

Consider

\(y=3-x   \;\;where\;\; x\le3\\ when \;\;x\le3,\;\;y\ge0\)

\(y=3-x   \;\;where\;\; x\le3\\ -x=y-3\\ x=3-y\\ \text{the inverse will be}\quad y=3-x\;\;where\;\;x\ge 0 \)

since I only am asked for the inverse values in this domain I do not need to concern myself with the original function when x>3 as this would give an inverse function domain of x<0

To show you,   if 

f(x)=-x^3+2x^2+3x then

f(3)=-27+18+9 = 0    for values bigger than 3 the value will be negative.   

\(f^{-1}(0)=3-0=3\\ f^{-1}(6)=3-6=-3\\ \)

Any question?  ask away :)

 Here is the actual graph

It's a pleasure :)

Your 5C1 gives you   5 choices    DLRIE   

If you chose  L  then your 4C1 will give you the 4 coices   DRIE   say you then chose D

you now have     COL    COD    RIE

alternatively you could have first chosen  D   and then chosen  L   

then you would have  COD   COL   RIE

But these are the same, you have counted this outcome twice.