Rom

\(\left \lfloor \dfrac{2007! + 2004!}{2006!+2005!}\right\rfloor = \\ \left \lfloor \dfrac{2004!(1+2005\cdot 2006 \cdot 2007)}{2004!\cdot 2005(1+2006)}\right\rfloor = \\ \left \lfloor \dfrac{1+2005\cdot 2006 \cdot 2007}{2005(1+2006)}\right\rfloor = \\ \left \lfloor \dfrac{\frac{1}{2005}+2006 \cdot 2007}{1+2006}\right\rfloor = \\ \left \lfloor \dfrac{\frac{1}{2005\cdot 2006}+2007}{\frac{1}{2006}+1}\right\rfloor = 2006\\\)

The last equality comes from the first term in the numerator being so small we can ignore it, 

and the denominator being slightly greater than 1 so that the floor is 2006 rather than 2007.

\(\left|\dfrac 1 2 - c i \right| = \left(\dfrac 1 2 \right)^2 + c^2 = \dfrac 3 4\\ \dfrac 1 4 + c^2 = \dfrac 3 4\\ c^2 = \dfrac 1 2\\ c = \pm \dfrac{1}{\sqrt{2}}\)

\(2z - 3\bar{z} = -2 - 30i\\ \text{let $z=x+iy$}\\ 2x + i2y - 3x + i3y = -2 - 30i\\ -x+i5y=-2-30i\\ x = 2,~y = -6i\\ z = 2-6i\)

\(\text{Suppose you want the numbers of 3 consecutive tiles to sum to $N$}\\ \text{Further let's assume that the set you refer to has $M$ numbers in it}\\ \text{We want the number of 3 element partitions of $N$ from the elements of the set}\\ \text{This is equivalent to sorting $N$ balls into $M$ bins}\\ \text{The number of ways this can be done is given by the stars and bars problem (google) and is}\\ \dbinom{N+M-1}{M-1}\\ \text{Tiles can additionally be colored red or blue. This gives an additional factor of $2^L,\\$ where $L$ is the length of the row of tiles.}\\ \text{So if your row is length $L$ the number of arrangements is }\\ 2^L \dbinom{N+M-1}{M-1}\)

\(v_1 = (1-0.2)v_0 = 0.8v_0\\ v_2 = (1+0.3)v_1 = 1.3(0.8)v_0 =1.04v_0\\ \dfrac{v_2}{v_0} = 1.04 = (1+0.04)\\ \text{Overall there was a 4% increase in the stock value}\)

True

all but one of the prime numbers are odd.

True.

2 is prime.

True.

2 is the only even prime.

The statement is false.

\(a = p b\\ a = \dfrac q c\\ c = \dfrac{q}{pb}\)

\(2=5p \Rightarrow p=\dfrac 2 5\\ 2 = \dfrac q 9 \Rightarrow q = 18\\~\\ c = \dfrac{18}{\frac 2 5 \cdot 3} = 15 \)

well basically you are looking at the set of output from each matrix when fed all possible vectors.

The matrix being invertible should immediately suggest one of the 3 options.

Think about the output of matrix A.  It tells you it maps every vector to the zero vector.

Is the zero vector a point, line, plane, or all of 3D space?

Simlarly with C.  Every output it produces lies along the vector i+j.  That suggests that the output grid covers a .. ?