Rom

\(f(x) = \dfrac{3x^2-4x+1}{x-1}=\\ \dfrac{(3x-1 )(x-1 )}{x-1}=3x-1\\ \text{but with a "hole" at $x=1$ so $3(1)-1=2$ is not in the range of $f(x)$}\)

\(\text{$x = 1$ is known as a removable singularity}\)

\(\text{$[-16,16]$ is 32 units wide}\)

The maximum, 1, will occur when one of the variable is 1 and the rest 0

\((a+b+c+d-1)=0\\ (a+b+c+d-1)^2=0\\ a^2+b^2+c^2+d^2 - 2(a+b+c+d)+2(ab+ac+ad+bc+bd+cd)+1 = 0\\ a^2+b^2+c^2+d^2+2(ab+ac+ad+bc+bd+cd)= 1\\ \text{It should be clear that to maximize the sum of the squares we want the second set of terms to be 0}\\ \text{This occurs, as noted above when only 1 element is non-zero}\)

If you look at how many possible 2nd cards satisfy the condtion for a given first card

you'll see there are 2 inital cards each that have 10-19 cards that satisfy it,

and then 30 initial cards that have 20 cards that will satisfy it.

That leads to 

\(p = \dfrac{1}{25}\dfrac{1}{49}\sum \limits_{k=10}^{19}k + \dfrac{30}{50}\dfrac{20}{49}=\dfrac{89}{245}\)

\(a) \text{ The partitions of 10 are }\\ (2,2,6), ~(2,3,5),~(2,4,4),~(3,3,4),\\ \text{ and all arrangements of these among the colors}\\ \text{there will be $\dbinom{3}{1}+3! + \dbinom{3}{1}+\dbinom{3}{1} = 15$ arrangements}\)

\(b) \text{ Is basically the same deal but now we find the partition of 20 rather than 10}\\ (2,9,9),~(3,8,9),~(4,7,9),~(4,8,8),~(5,6,9),~(5,7,8),~(6,6,8),~(6,7,7),~\text{and all arrangements by color}\\ \text{as before this will be}\\ 3 + 3! + 3! + 3 + 3!+3! +3+3= 36\)

There is no matrix A that satisfies the 3 equations.

(1,1,1) = (1,0,0)+(0,1,1)

A(1,1,1) = A((1,0,0)+(0,1,1)) = A(1,0,0) + A(0,1,1) = (-3,4,0) + (1,2,3) = (-2, 6, 3) != (3,2,1)

I think what you are after are known as the Peano Axioms

\(\text{we have essentially 4 entities given: ma, ka, la, and a}\\ \text{having placed the 3 pairs there are 4 slots in which the a can be placed}\\ \text{the 3 pairs can be arranged in $3!$ ways and then there are 4 ways to place the a, for a total of}\\ N=3!\cdot 4 = 6\cdot 4 = 24\)

\(|p|=\dfrac{1500}{4}=375\\ |g|=\dfrac{1500}{5}=300\\ |b|=\dfrac{1500}{3}=500\\~\\ \text{let $X$ be the set with none of the above}\\ |X| = 1500 - |p \cup g \cup b|\\~\\ \text{The smallest value of $|X|$ is obtained if the 3 sets above don't intersect}\\ \text{In this case $|X|=1500-375-300-500 = 325$}\)

\(x^{2001}+\left(\dfrac 1 2 - x \right)^{2001} = \\ x^{2001} -x^{2001} + \sum \limits_{k=1}^{2001}\left(\dfrac 1 2\right)^k (-x)^{2001-k} = \\ \sum \limits_{k=1}^{2001}\left(\dfrac 1 2\right)^k (-x)^{2001-k} = \\ \dfrac 1 2 \sum \limits_{k=0}^{2000}\left(\dfrac 1 2\right)^k (-x)^{2000-k} = \\ \dfrac 1 2\left(\dfrac 1 2 - x\right)^{2000} = 0\)

\(\text{$x=\dfrac 1 2$ is the only solution and it is a root of order 2000}\)