Class 11th

Species               B.O

Ne₂                        0

N₂                           3

F₂                 1

O₂                          2

Data contradiction.

$\overrightarrow{a}*\left(\overrightarrow{b}*\overrightarrow{c}\right)=\left(\overrightarrow{a}\cdot \overrightarrow{c}\right)\overrightarrow{b}-\left(\overrightarrow{a}\cdot \overrightarrow{b}\right)\overrightarrow{c}$

Kindly consider the following figure

pva $\left(\sim rvp\right)$ $$

$\sim \left(p\vee q\right)\vee \left(\sim rvp\right)$

its negation as asked in question

$\left(p\vee q\right)\wedge \left(\sim p\wedge r\right)$

= $\left(p\wedge \sim p\wedge r\right)\vee \left(q\wedge r\wedge \sim p\right)$

= $\left(p\wedge r\wedge \sim p\right)\left[asp\wedge \sim pisfalse\right]$

Let base = b

$tan60°=\frac{h}{b}$

$tan30°=\frac{h-20}{b}$

$Mean=\frac{3+12+7+a+\left(43-a\right)}{5}=13$  

Variance = $\frac{3^2+12^2+7^2+a^2+{\left(43-a\right)}^2}{5}-{\left(13\right)}^2$  

$\Rightarrow \frac{2a^2-a+1}{5}\to Naturalnumber$      

Let 2a² – a + 1 = 5x

D = 1 – 4 (2) (1 – 5n)

= 40n – 7, which is not $4\lambda or4\lambda +1from.$  

As each square form is $4\lambda or4\lambda +1$  

Total number of possible relation = $2^{n^2}=2^4=16$ ${}^{{}^{}}{}^{}$

Favourable relations = $?,\left\{\left(x,x\right)\right\},\left\{\left(y,y\right)\right\}$ $$

$\left\{\left(x,x\right),\left(y,y\right)\right\}$

$\left\{\left(x,x\right),\left(y,y\right),\left(x,y\right),\left(y,x\right)\right\}$

Probability = $\frac{5}{16}$

Kindly go through the solution

S > Se > Te > O

$?{\mathcal{l}}_{1}and{\mathcal{l}}_2$ are perpendicular, so

$3*1+\left(-2\right)\left(\frac{\alpha }{2}\right)+0*2=0$

⇒a = 3

Now angle between  ${\mathcal{l}}_{2}and{\mathcal{l}}_3$ ,

$cos\theta =\frac{1\left(-3\right)+\frac{\alpha }{2}\left(-2\right)+2\left(4\right)}{\sqrt[]{1+\frac{{\alpha }^2}{4}+4}.\sqrt[]{9+4+16}}$

$\Rightarrow cos\theta =\frac{\frac{2}{29}}{2}\Rightarrow \theta =co{s}^{-1}\left(\frac{4}{29}\right)=se{c}^{-1}\left(\frac{29}{4}\right)$

Let P (at², 2 at) where

$a=\frac{3}{2}$

T : yt = x + at² so point Q is

$\left(-a,at-\frac{a}{t}\right)$

N : y = -tx + 2at + at³ passes through (5, -8)

$-8=-5t+3t+\frac{3}{2}{t}^3$

$3t^3-4t+16=0$

⇒ t = -2

  So ordinate of point Q is $-\frac{9}{4}$