Maths

Line L is   $\frac{x-1}{1}=\frac{y-2}{2\sqrt[]{2}}=\frac{z}{0}$

this line L makes an angle of 45° with the plane  $\sqrt[]{2}x+y-z=1$

$\therefore$ Required distance PQ is perpendicular distance of plane from P is i.e.,

$PQ=\frac{\left|\sqrt[]{2}+7-\sqrt[]{2}-1\right|}{\sqrt[]{2+1+1}}=3$

$\left|A-xI\right|=\left|\begin{array}{cc}\hfill 1/2-x\hfill & \hfill 1/2\hfill \\ \hfill a\hfill & \hfill b-x\hfill \end{array}\right|=x^2-\left(b+\frac{1}{2}\right)x+\frac{b-a}{2}$

from Cayley – Hamilton Theorem

$A^2-\left(b+\frac{1}{2}\right)A+\left(\frac{b-a}{2}\right)I=0$

$\Rightarrow A^3=\left({\left(b+\frac{1}{2}\right)}^2-\left(\frac{b-a}{2}\right)\right)A-\left(b+\frac{1}{2}\right)\left(\frac{b-a}{2}\right)I$

$\therefore {\left(b+\frac{1}{2}\right)}^2-\left(\frac{b-a}{2}\right)=1\&\left(b+\frac{1}{2}\right)\left(\frac{b-a}{2}\right)=0$

$\therefore \left(a,b\right)=\left(\frac{1}{2},\frac{1}{2}\right),\left(-\frac{3}{2},-\frac{3}{2}\right),\left(\frac{3}{2},-\frac{1}{2}\right)$

A = {1, 2, 3, 4}. As (4, 4) $\notin$  R so not reflexive

Also not transitive

EHINTZ

words starting with E is 5!

words starting with H is 5!

words starting with I is 5!

words starting with N is 5!

words starting with T is 5!

If the two lines are perpendicular then a = 2

D = 0       $\Rightarrow$    c = -3

(D = abc + 2fgh - af² – bg² - ch²)

$\frac{x}{4}=cost-sint...\left(1\right)$

$\frac{y}{5}=cost+sint$ . . . . (2)

(1)² + (2)² gives

${\left(\frac{x}{4}\right)}^2+{\left(\frac{y}{5}\right)}^2=2$

$xdy+ydx=\frac{xydx}{\sqrt[]{1-x^2}}$

$\frac{d\left(xy\right)}{xy}=\frac{dx}{\sqrt[]{1-x^2}}$

Integrate both sides we get

logexy = sin1 x + C

$y^2=x,a=\frac{1}{4}$

$x^2=-y,b=\frac{1}{4}$

$A=\frac{16\left|ab\right|}{3}=\frac{1}{3}$

$\left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\right]=0$

$\left|\begin{array}{ccc}\hfill 1\hfill & \hfill 6\hfill & \hfill 3\hfill \\ \hfill 3\hfill & \hfill 2\hfill & \hfill 1\hfill \\ \hfill \alpha +1\hfill & \hfill \beta -1\hfill & \hfill 1\hfill \end{array}\right|=0$

$8\beta -24=0\Rightarrow \beta =3$

$\left|\overrightarrow{c}\right|=\sqrt[]{6}$

${\left(\alpha +1\right)}^2+{\left(\beta -1\right)}^2+1=6\Rightarrow {\left(\alpha +1\right)}^2=1$

a = 1 = 1, 1 $\Rightarrow \alpha =-2,0$ $$

+ = 1, 3

Variance of 2001, 2003, 2006, 2007, 2009, 2010 is same as variance of 1, 3, 6, 7, 9, 10

mean =  $\frac{1+3+6+7+9+10}{6}=6$

var (x) =  $\frac{1^2+3^2+6^2+7^2+9^2+10^2}{6}?36$

var (x) = 10