Maths

  $\frac{dy}{dx}=\frac{ax-by+a}{bx+cy+a}$

$=bxdy+cydy+ady=axdx-bydx+adx$                

$=\frac{c{y}^2}{2}+ay-\frac{a{x}^2}{2}-ax+bxy=k$              

$a{x}^2+a{y}^2+2ax-2ay=k$            

$\Rightarrow x^2+y^2+2x-2y=\lambda$              

Short distance of (11,6)

= $\sqrt[]{12^2+5^2}-5$

= 13 – 5

= 8

  $x={\displaystyle \sum _{n=0}^{\infty }{a}^n}=\frac{1}{1-a}y={\displaystyle \sum _{n=0}^{\infty }{b}^n=\frac{1}{1-b}{\displaystyle \sum _{n=0}^{\infty }{c}^n=\frac{1}{1-c}}}$               

Now,

a, b, c -> AP

1 – a, 1 – b, 1 – c -> AP

$\frac{1}{1-a},\frac{1}{1-b},\frac{1}{1-c}\to HP$

x, y, z -> HP

x + 2y + z = 2

$\alpha x+3y-z=\alpha$

$-\alpha x+y+2z=-\alpha$            

$\Delta =\left|\begin{array}{ccc}\hfill 1\hfill & \hfill 2\hfill & \hfill 1\hfill \\ \hfill \alpha \hfill & \hfill 3\hfill & \hfill -1\hfill \\ \hfill -\alpha \hfill & \hfill 1\hfill & \hfill 2\hfill \end{array}\right|=1\left(6+1\right)-2\left(2\alpha -\alpha \right)+1\left(\alpha +3\alpha \right)=7+2a$            

$\alpha =-\frac{7}{2}$                

  $\overline{z}=i{z}^2$

Let z = x + iy

x – iy = I (x² – y² + 2xiy)

Case-I

x = 0

-y² = -y

y = 0, 1

Case – II

$y=-\frac{1}{2}$

$\Rightarrow x^2-\frac{1}{4}=\frac{1}{2}\Rightarrow x=\pm \frac{\sqrt[]{3}}{2}$

Area of polygon

$=\frac{1}{2}\left|\begin{array}{ccc}\hfill 0\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill \frac{\sqrt[]{3}}{2}\hfill & \hfill \frac{-1}{2}\hfill & \hfill 1\hfill \\ \hfill \frac{-\sqrt[]{3}}{2}\hfill & \hfill \frac{-1}{2}\hfill & \hfill 1\hfill \end{array}\right|=\frac{1}{2}\left|-\sqrt[]{3}-\frac{\sqrt[]{3}}{2}\right|=\frac{3\sqrt[]{3}}{4}$  

Z distribution and Chi-Squared are some of the most popular distribution patterns of probability, and it is vital to recognise the variations between them and when to use the distribution pattern. A Z table is of no use when the operation revolves around a smaller sample size. On the other hand, the distribution of a sum of independent regular k squares in standard normal variables is the chi-square distribution of k degrees of freedom. The tests are used for the independence of two variables in an incident table and to assess the observable data for 

P : Ramu is intelligent

Q : Ramu is rich

R : Ramu is not honest

Give statement, “Ramu is intelligent and honest if any only if Ramu is not rich”

$=\left(P\wedge \sim R\right)\iff \sim Q$

So, negation of the statement is

$\sim \left[\left(P\wedge \sim R\right)\Rightarrow \sim Q\right]$

$=\left(\left(P\wedge \sim R\right)\wedge Q\right)\vee \left(\sim Q\wedge \left(\sim P\vee R\right)\right)$

$2sin\left(\frac{\pi }{22}\right)sin\left(\frac{3\pi }{22}\right)sin\left(\frac{5\pi }{22}\right)sin\left(\frac{7\pi }{22}\right)sin\left(\frac{9\pi }{22}\right)$

$=\frac{2sin\frac{32\pi }{11}}{2^{5}sin\frac{\pi }{11}}=\frac{1}{16}$

Median = $\frac{2k+12}{2}=k+6$

Mean deviation = $\sum \frac{\left|x_i-M\right|}{n}=6$

$\Rightarrow \frac{\left(k+3\right)+\left(k+1\right)+\left(k-1\right)+\left(6-k\right)+\left(6-k\right)+\left(10-k\right)+\left(15-k\right)+\left(18-k\right)}{8}$

$\therefore \frac{58-2k}{8}=6$

k = 5

$Median=\frac{2*5+12}{2}=11$

$\overrightarrow{a}=\widehat{i}-\widehat{j}+2\widehat{k}$

$\overrightarrow{a}*\overrightarrow{b}=2\widehat{i}-\widehat{k}$

$\overrightarrow{a}.\overrightarrow{b}=3$

${\left|\overrightarrow{a}*\overrightarrow{b}\right|}^2+{\left|\overrightarrow{a}.\overrightarrow{b}\right|}^2={\left|\overrightarrow{a}\right|}^2.{\left|\overrightarrow{b}\right|}^2$$$

$=\frac{\overrightarrow{b}.\overrightarrow{a}-{\left|\overrightarrow{b}\right|}^2}{\left|\overrightarrow{a}-\overrightarrow{b}\right|}=\frac{3-\frac{7}{3}}{\sqrt[]{\frac{7}{3}}}=\frac{2}{\sqrt[]{21}}$