Rom

From the diagram about you can see that

$|PX|=\sqrt{5^2+12^2} = 13 \\ \cos\left(\dfrac 1 2 \angle PXS\right)=\dfrac{12}{13} \\ \sin\left(\dfrac 1 2 \angle PXS\right) = \sqrt{1-\dfrac{12}{13}} = \dfrac{5}{13} \\ \cos(2x) = \cos^2(x) - \sin^2(x) \\ \cos\left(\angle PXS\right) = \cos\left(2\cdot \dfrac 1 2 \angle PXS\right) = \dfrac{144}{169}-\dfrac{25}{169} = \dfrac{119}{169}$

a)  $x=p,~y=p+2\\ 3(p) - 4(p+2) + 5 = 0\\ 3p-4p+8 = 0\\ -p+8=0 \\ p=8$

b) There are a few ways to do this.  I'd rewrite the equation for AB as

$3x-4y+5=0 \\ 4y = 3x+5 \\ y = \dfrac 3 4 s + \dfrac 5 4 \\ \text{and you can read the gradient right off as }m=\dfrac 3 4$

c) The key here is that the product of the slopes over perpendicular lines is -1.

$m_{AC} = \dfrac{k-2}{-5-1} = -\dfrac{k-2}{6} \\ \text{we know from (a) that }m_{AB}=\dfrac 3 4 \\ m_{AC} \cdot m_{AB}=-1\\ -\dfrac{k-2}{6} \cdot \dfrac 3 4 = -1 \\ \dfrac{3k-6}{24}=1 \\ 3k=30\\ k=10$

d)

$AB: y=\dfrac 3 4 x+\dfrac 5 4\\ 2x-5y=6 : y = \dfrac 2 5 x - \dfrac 6 5 \\ \dfrac 3 4 x+\dfrac 5 4 = \dfrac 2 5 x - \dfrac 6 5 \\ \dfrac{15-8}{20}x = -\dfrac{24+25}{20}\\ \dfrac{7}{20}x=-\dfrac{49}{20} \\ x=-7 \\ y = \dfrac{2}{5}\cdot (-7)-\dfrac 6 5 = -\dfrac{20} 5 = -4\\ D=(x,y) =(-7,-4)$

I guess they just want 

$A = \pi r^2 = \pi \left(\dfrac{D}{2}\right)^2 \\ D=10 \Rightarrow A = \pi(5)^2 = 25\pi \approx 78.5$

$\text{A point on the line }y=-x+6 \text{ is given by }(x, -x+6)\\ \text{the squared distance between two points u and v is given by}\\ d^2(u,v)=(u_x-v_x)^2+(u_y-v_y)^2 \\ \text{if the distance between two points is equal so will the squared distance be, so}\\ d^2[(10,-10),(x,-x+6)] = d^2[(0,0),(x,-x+6)] \\ (10-x)^2 + (-16+x)^2 = x^2 + (-x+6)^2 \\ 2 x^2-52 x+356=2x^2-12 x+36 \\ 320=40x\\ x=8,~y=(-x+6) = -2\\ (x,y) = (8,-2)$

The mathy way of doing this is noting that for a square the perimeter is a function of s while the area is a function of s² 

So an increase in the perimeter by a factor of K leads to an increase in area by a factor of K²

or in even more mathy terms, the area is directly proportional to the square of the perimeter

Another way of putting this is

$\dfrac{A_1}{A_2} = \left(\dfrac{P_1}{P_2}\right)^2 \\ \dfrac{A_1}{A_2} = \left(\dfrac{12}{24}\right)^2 = \dfrac 1 4$

$\text{Probably the clearest way to do this is to note that}\\ \cos(\angle RPS)=\cos\left(\pi - \angle RPQ\right)=-\cos(\angle RPQ)\\ \cos(\angle RPQ)=\sqrt{1 -\sin^2(\angle RPQ)} = \sqrt{1-\frac{49}{625}} \\ \cos(\angle RPQ) = \sqrt{\dfrac{576}{625}}=\dfrac{24}{25},\text{ and thus}\\ \cos(\angle RPS)=-\dfrac{24}{25}$

I'm showing 841 unique centroids.

I'm going to assume this means please solve for x.

$7^2+x^2 - 7x=x^2\\ 7^2 = 7x \\ x= 7$

$\text{suppose }v=\log(u) \\ \text{then }e^v = u \\ e^{-v} =\dfrac{1}{e^v} = \dfrac 1 u \\ -v = \log\left(\dfrac{1}{u}\right)$