Class 11th

We cannot call the trailing zeroes as significant numbers. If you remember from the rules, it needs to have a decimal point somewhere to be considered as one. What is recommended is that one can use the measure as to the power of 10.  

Resonance energy is the energy difference between most stable resonating structure andactual structure.

HBr adds to alkene in accordance with Markovnikov's rule.

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

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

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

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

b = 2

-> a + b = 3

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]