Maths

As per the problem, we have Length of the first train = 120 m

Length of the second train = 180 m

As per the problem,

$\frac{(120+180)}{(8+s)}=5$

8 + s = 60

s = 52 m/sec

6X + 5Y = 218 . (1)

5X - 3Y = 24 . (2)

Solve to get X = 18, Y = 22.

  $S_{10}=530\Rightarrow 5\left[2a+9d\right]=530$

2a + 9d = 106 . (i)

$S_5=140\Rightarrow \frac{5}{2}\left[2a+4d\right]=140$              

a + 2d = 28 . (ii)

Solving (i) & (ii) a = 88 d = 10

$\therefore S_{20}-S_6=14a+175d=1862$            

$\overrightarrow{a}=\alpha \overrightarrow{b}+\beta \overrightarrow{c}=\left(2\alpha +\beta \right)i+\left(\alpha -\beta \right)J+\left(\alpha +\beta \right)k$

is perpendicular to   $\overrightarrow{d}=3\widehat{i}+2\widehat{j}+6\widehat{k}$

$\Rightarrow 3\left(2\alpha +\beta \right)+2\left(\alpha +\beta \right)+6\left(\alpha +\beta \right)=0$              

->14a + 7b = 0 Þ b = -2a .(i)

$\Rightarrow \overrightarrow{a}=3\alpha j-\alpha k$

$\left|\overrightarrow{a}\right|=\sqrt[]{9{\alpha }^2+{\alpha }^2}=\sqrt[]{10}\Rightarrow \left|\alpha \right|=1\Rightarrow \alpha \pm 1,for\alpha =1,\overrightarrow{a}=3\widehat{j}-\widehat{k}$       

for $\alpha =-1,\overrightarrow{a}=-3\widehat{j}+\widehat{k}$  

$\overrightarrow{d}*\overrightarrow{a}=\pm \left|\begin{array}{ccc}\hfill i\hfill & \hfill j\hfill & \hfill k\hfill \\ \hfill 3\hfill & \hfill 2\hfill & \hfill 6\hfill \\ \hfill 0\hfill & \hfill 3\hfill & \hfill -1\hfill \end{array}\right|$

$=\pm 42$              

Let x be the required percent.

Using the successive percentage change rule, we have

150 + x + $\frac{150x}{100}$ $\frac{}{}$ = 0

$\frac{}{}$ $\frac{250x}{100}$  = – 150

x = – 60%

so 60%

Required time t =

$\frac{1760}{2\pi r}=\frac{1760}{\left[2*\left(\frac{22}{7}\right)*\frac{3.5}{2}\right]}=160min.$

  $f\left(x\right)=[\begin{array}{l}-\frac{4}{3}{x}^3+2x^2+3x,x>0\\ 3x{e}^x,x\le 0\end{array}$

  $f\text{'}\left(x\right)>0forx<0e^x\left(x+1\right)>0\Rightarrow X>-1\Rightarrow x\in \left(-1,0\right)$            

for  x > 0  $-4x^2+4x+3>0$   

$\Rightarrow x\in \left(0,\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)$  

Again    at  x = 0  f'(0) = 3 > 0

$\therefore f\left(x\right)isincreasingin\left(-1,\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)$            

CP of 378 = SP of 525

$\begin{array}{l}\frac{378}{525}=\frac{SP}{CP}\\ \frac{SP}{CP}=0.72\end{array}$

So, 28% loss

$cose{c}^{2}xdy+2dx=\left(1+ycos2x\right)cose{c}^{2}xdx.$

$\Rightarrow \frac{dy}{dx}+2sin2x=1+ycos2x.$

$I.F.=e^{-{\displaystyle \int cos2xdx}}=e^{-\frac{sin2x}{2}}$

$\therefore Solutiony{e}^{-\frac{sin2x}{2}}={\displaystyle \int e^{-\frac{sin2x}{2}}.cos2xdx.Put\frac{sin2x}{2}=t\Rightarrow cos2xdx=dt}$

$y\left(0\right)=-1+e^{-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\Rightarrow {\left(y\left(0\right)+1\right)}^2={\left(e^{-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right)}^2=e^{-1}$

$\frac{{(27+5)}^{332}}{9}=\frac{{(5)}^{332}}{9}=\frac{5^2*{(125)}^{110}}{9}=25*{(-1)}^{110}=7$