Class 12th

Required probability of obtaining a (sum = 7) = $\frac{13}{96}\left(given\right)$  

    $i.e.2\left(\left(\frac{1}{6}+x\right)\left(\frac{1}{6}-x\right)+\frac{1}{6}*\frac{1}{6}+\frac{1}{6}*\frac{1}{6}\right)=\frac{13}{96}$

$x=\frac{1}{8}$           

Equation of line PQ :

$\frac{x-1}{2}=\frac{y+2}{3}=\frac{z-3}{-6}=k$

So, let Q (2k + 1, 3k – 2, -6k + 3) Q lies on given plane so

2k + 1 – 3k + 2 – 6k + 3 = 5

$-7k=-1ork=\frac{1}{7}$

So, $Q\left(\frac{9}{7},\frac{-11}{7},\frac{15}{7}\right)$

Now required distance = PQ = $\sqrt[]{\frac{4}{49}+\frac{9}{49}+\frac{36}{49}}=1$

The basic unit of energy is a Joule (J), which is equal to one watt of power expended for one second. In the day-to-day scenarios such as household electricity consumption, joule is too small a unit to be convenient. For billing and metering purpose, it is measured in kilowatt-hours (kWh). Electrical energy is measured in kilowatt-hours (kWh) as the commercial unit. 1kWh = 1000 W * 3600 s 

For calculating the units consumed, we will be using the following formula 

Units = [Power (W) x Time (h)]/1000

Kindly consider the following Image.

Common grandchild of M6 and M8 is D

Please find the answer below

Let square is made with piece of length x metre & hexagon with piece of length y metre

x + y = 20 .(i)

$a=\frac{x}{4}\&b=\frac{y}{6}$       

Now let A = area of square + area of hexagon

$A=\frac{x^2}{16}+6*\frac{\sqrt[]{3}}{4}*\frac{y^2}{36}=\frac{x^2}{16}+\frac{\sqrt[]{3}{y}^2}{24}=\frac{x^2}{16}+\frac{\sqrt[]{3}}{24}{\left(20-x\right)}^{2}from\left(i\right)$ ,

for minimum area

$x=\frac{80\sqrt[]{3}}{6+4\sqrt[]{3}}=\frac{80\sqrt[]{3}}{2\sqrt[]{3}\left(\sqrt[]{3}+2\right)}=\frac{40}{2+\sqrt[]{3}}$

$x=40\left(2-\sqrt[]{3}\right)$

=> side of hexagon = $\frac{y}{6}=\frac{20\sqrt[]{3}\left(2-\sqrt[]{3}\right)}{6}=\frac{20\sqrt[]{3}}{6\left(2+\sqrt[]{3}\right)}=\frac{10\sqrt[]{3}}{3\left(2+\sqrt[]{3}\right)}=\frac{10}{2\sqrt[]{3}+3}$

$\left|A\right|=\left|\begin{array}{ccc}\hfill \left[x+1\right]\hfill & \hfill \left[x+2\right]\hfill & \hfill \left[x+3\right]\hfill \\ \hfill \left[x\right]\hfill & \hfill \left[x+3\right]\hfill & \hfill \left[x+3\right]\hfill \\ \hfill \left[x\right]\hfill & \hfill \left[x+2\right]\hfill & \hfill \left[x+4\right]\hfill \end{array}\right|=\left|\begin{array}{ccc}\hfill \left[x\right]+1\hfill & \hfill \left[x\right]+2\hfill & \hfill \left[x\right]+3\hfill \\ \hfill \left[x\right]\hfill & \hfill \left[x\right]+3\hfill & \hfill \left[x\right]+3\hfill \\ \hfill \left[x\right]\hfill & \hfill \left[x\right]+2\hfill & \hfill \left[x\right]+4\hfill \end{array}\right|$

$R_1\to R_1-R_3\&R_2\to R_2-R_3$

$\left|\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill -1\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill -1\hfill \\ \hfill \left[x\right]\hfill & \hfill \left[x\right]+2\hfill & \hfill \left[x\right]+4\hfill \end{array}\right|=192$

$\Rightarrow \left[x\right]=62$

$\Rightarrow x\in [62,63)$              

The SI unit of the electrical power is Watt which is symbolized as W. One watt is equal to one joule per second (J/s). In the context of current and voltage, one watt is the power consumed when one ampere of the current flows through the potential difference of one volt.