A variable line passes through a fixed-point P. The algebraic sum of the perpendiculars drawn from the points (2, 0), (0, 2), and (1, 1) on the line is zero. Find the coordinates of the point P.

(Hint: Let the slope of the line be m. Then the equation of the line passing through the fixed-point P (x?, y?) is y – y ? = m (x – x?). Taking the algebraic sum of perpendicular distances equal to zero, we get y – 1 = m (x – 1). Thus (x?, y?) is (1, 1).

  $\left|\frac{1}{2}-2i+1\right|=\alpha \left(\frac{1}{2}-2i\right)+\beta \left(1+i\right)$  

$\sqrt[]{\frac{9}{4}+4}=\alpha \left(\frac{1}{2}-2i\right)+\beta \left(1+i\right)$

$\frac{5}{2}=\alpha \left(\frac{1}{2}\right)+\beta +i\left(-2\alpha +\beta \right)$             

$\frac{\alpha }{2}+\beta =\frac{5}{2}$      ...(1)

 –2α + β = 0                    …(2)

Solving (1) and (2)

$\frac{\alpha }{2}+2\alpha =\frac{5}{2}$

$\frac{5}{2}\alpha =\frac{5}{2}$            

a =

Variance = $\frac{\sum x^2}{n}-{\left(\overline{x}\right)}^2$  

$\frac{60^2+60^2+44^2+58^2+68^2+{\alpha }^2+{\beta }^2+56^2}{8}={\left(58\right)}^2=66.2$            

$\frac{7200+1936+3364+4624+3136+{\alpha }^2+{\beta }^2}{8}=3364=66.2$             

$2532.5+\frac{{\alpha }^2+{\beta }^2}{8}-3364=66.2$            

α² + β² = 897.7 × 8

= 7181.6

Start with

(1) $\overline{E}:\frac{6!}{2!}=360$  

(2)    $\overline{GE}:\frac{5!}{2!},$ $\overline{GN}:\frac{5!}{2!}$  

(3) GTE : 4!, GTN: 4!, GTT : 4!

(4) GTWENTY = 1

⇒ 360 + 60 + 60 + 24 + 24 + 24 + 1 = 553

${\left(1+x\right)}^{11}={}^{11}{C}_0+{}^{11}{C}_{1}x+{}^{11}{C}_2{x}^2+.....{+}^{11}{C}_{11}{x}^{11}$

$=\frac{2^{12}-2-24}{12}$

$=\frac{2^{12}-26}{12}=\frac{4070}{12}=\frac{2035}{6}=\frac{m}{n}$

m + n = 2035 + 6 = 2041

$f\left(x\right)=\{\begin{array}{cc}\hfill 2+2x,\hfill & \hfill x\in \left(-1,0\right)\hfill \\ \hfill 1-\frac{x}{3},\hfill & \hfill x\in [0,3)\hfill \end{array}$

$g\left(x\right)=\{\begin{array}{cc}\hfill x,\hfill & \hfill x\in [0,1)\hfill \\ \hfill -x,\hfill & \hfill x\in \left(-3,0\right)\hfill \end{array}$   ->g(x) = |x|, x Î (–3, 1)

$f\left(g\left(x\right)\right)=\{\begin{array}{cc}\hfill 2+2\left|x\right|,\hfill & \hfill \left|x\right|\in \left(-1,0\right)\Rightarrow x\in ?\hfill \\ \hfill 1-\frac{\left|x\right|}{3},\hfill & \hfill \left|x\right|\in [0,3)\Rightarrow x\in \left(-3,1\right)\hfill \end{array}$            

$f\left(g\left(x\right)\right)=\{\begin{array}{cc}\hfill 1-\frac{x}{3},\hfill & \hfill x\in [0,1)\hfill \\ \hfill 1+\frac{x}{3},\hfill & \hfill x\in \left(-3,0\right)\hfill \end{array}$

Range of fog(x) is [0, 1]

            Range of fog(x) is [0, 1]