Two vertical poles are 150m apart and the height of one is three times that of other. If from the middle point of the line joining their feet, an observer finds the angles of elevation of their tops to be complementary, then the height of the shorter pole (in meters) is:

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

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

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

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