Rom

\(\text{Think about it a moment and it should be clear that the range of }f(x) = \lfloor x\rfloor - x \text{ is }\\ f(x) \in (-1,0]\)

\(f(x,y) = \arctan\left(\dfrac x y\right)\\ f_x = \dfrac{1}{1+\left(\dfrac x y\right)^2}\cdot \dfrac 1 y = \dfrac{y}{x^2+y^2}\)

\(f_y = \dfrac{1}{1+\left(\dfrac x y \right)^2}\cdot \dfrac{-x}{y^2} = -\dfrac{x}{x^2+y^2}\)

\(f_{xx} = -\dfrac{y}{(x^2+y^2)^2}\cdot 2x = -\dfrac{2xy}{(x^2+y^2)^2}\)

\(f_{yy} = \dfrac{2xy}{(x^2+y^2)^2}\)

\(f_{xy}=f_{yx} =\dfrac{ 1(x^2+y^2)-2y(y)}{(x^2+y^2)^2} = \dfrac{x^2-y^2}{(x^2+y^2)^2}\)

52 x 51 = 2652

You multiplied an even number by an integer and obtained an odd number... Impressive!

\(f(x)=\dfrac{1}{\lfloor x^2-7x+13\rfloor}\)

\(\text{The floor function will map }[0,1) \to 0\\ \text{which results in undefined division by 0}\\ \text{Thus the interval }[0,1) \text{ must be restricted from the floor function}\\ \text{thus }x^2-7x+13<0 \text{ OR } x^2-7x+13 \geq 1\)

\(x^2 - 7x+13 < 0\\ \left(x-\dfrac 7 2\right)^2 +\dfrac 3 4 < 0\\ \text{this clearly cannot occur with real numbers}\)

\(x^2 - 7x+13 \geq 1\\ x^2 - 7x + 12 \geq 0\\ \left(x-\dfrac 7 2\right)^2 - \dfrac 1 4 \geq 0\\ \left(x-\dfrac 7 2\right)^2 \geq \dfrac 1 4\\ \left(x-\dfrac 7 2\right) \geq \dfrac 1 2 \text{ OR }\left(x-\dfrac 7 2\right) \leq -\dfrac 1 2\\ x \geq 4 \text{ OR }x \leq 3\\ x \in (-\infty, 3]\cup [4,\infty)\)

\(\text{1000 is not divisible by 9}\\ \dfrac{999}{9}=111 \text{ numbers in 1-999 that are divisible by 9}\\ 1000-111 = 889 \text{ numbers that are not divisible by 9}\)

for the second question do you mean any length arrangement or 9 letter arrangements

\(\text{I assume you mean the chairs are distinct and thus the chair choice matters}\\ \text{There are 5 ways to choose 2 adjacent chairs}\\ \text{There are 2 ways to orient the 2 boys in those two adjacent chairs} \text{That gives us 10 ways to place the boys}\\ \text{then there are }3!=6 \text{ ways to arrange the girls in the 3 remaining chairs}\\ \text{That gives us }10\cdot 6 = 60 \text{ ways to arrange the 5 people as described}\)

The remaining questions have been answered recently here.  Search for them.

\(f(x) = \dfrac 1 3 x^2 - 3x + 12 = \\ \dfrac 1 3(x^2 -9x+36 )= \\ \dfrac 1 3\left(\left(x-\dfrac 9 2\right)^2-\dfrac{81}{4}+36\right) =\\ \dfrac 1 3\left(x-\dfrac 9 2\right)^2- \dfrac{27}{4}+12 = \\ \dfrac 1 3\left(x-\dfrac 9 2\right)^2+\dfrac{21}{4}\)

\(f(x)=0\\ \dfrac 1 3\left(x-\dfrac 9 2\right)^2 = -\dfrac{21}{4}\\ \left(x-\dfrac 9 2\right)^2= -\dfrac{63}{4}\\ x-\dfrac 9 2 = \pm i\sqrt{\dfrac{63}{4}}\\ x = \dfrac 9 2 \pm i\dfrac 3 2 \sqrt{7}\\ x = \dfrac 3 2\left(3\pm i \sqrt{7}\right) \)

\(\dfrac{1}{|m|}\geq \dfrac 1 8\\ \dfrac 1 m \geq \dfrac 1 8 \bigcup \dfrac 1 m \leq - \dfrac 1 8\\ m \leq 8 \bigcup m \geq -8\\ m \in [-8,8],~m \in \mathbb{Z}-\{0\}\\ \text{16 total integers}\)

\(b=t-a\\ a^2+b^2 = a^2 + (t-a)^2 = \\ 2a^2 - 2at + t^2 = \\ 2\left(a-\dfrac t 2 \right)^2 + \dfrac{t^2}{2}\\ \text{This clearly has a minimum value of }\dfrac{t^2}{2} \text{ at }a = \dfrac t 2\)

your answer was correct, and guessing has it's place in solving problems,

but there was a concrete way to solve this one.

\(\text{For this problem the simplest route is just completing the square}\\ y=x^2+8x = \\ x^2 + 8x + 16 - 16 = \\ (x+4)^2 - 16\\ \text{This is a minimum at }x=-4 \text{ and the minimum is }y=-16\)