Maths

Using the standard equations of a hyperbola:

$9\left({\mathrm{e}}^2+l\right)$ and directrix $focus\mathrm{a}\mathrm{e}=\sqrt[]{10}$

By multiplying both focus and directrix, we get
$\frac{\mathrm{a}}{\mathrm{e}}=\frac{9}{\sqrt[]{10}}$ and $\Rightarrow {\mathrm{a}}^2=9$
Now $\mathrm{e}=\frac{\sqrt[]{10}}{3}$
$\mathrm{}(ae{)}^2=a^2+b^2$

We can convert 50! In terms of prime factor: $\mathrm{}{2}^{\alpha }\cdot 3^{\beta }\cdot 5^{\gamma }$ & using the greatest integer function.

n $=\left. [\frac{50}{3}\right]+\left. [\frac{50}{3^2}\right]+\left. [\frac{50}{3^3}\right]+\left. [\frac{50}{3^4}\right]$
$=16+5+1+0$ The maximum value of n is 22.

$\underset{x\to 7}{lim}\frac{18-\left[1-x\right]}{\left[x-3a\right]}$

exist &   $a\in I.$

$=\underset{x\to 7}{lim}\frac{17-\left[-x\right]}{\left[x\right]-3a}$

exist

RHL =   $\underset{x\to 7^{+}}{lim}\frac{17-\left[-x\right]}{\left[x\right]-3a}=\frac{25}{7-3a}\left[a\ne \frac{7}{3}\right]$

$LHL=\underset{x\to 7^{-}}{lim}\frac{17-\left[-x\right]}{\left[x\right]-3a}=\frac{24}{6-3a}\left[a\ne 2\right]$

LHL = RHL

$\Rightarrow \frac{25}{7-3a}=\frac{8}{2-a}$

$\therefore a=-6$

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

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

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

$a{x}^2+a{y}^2+2ax-2ay=k$

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

Short distance of (11,6)

= $\sqrt[]{12^2+5^2}-5$ $\text{exception 25:}$

= 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

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

Let the capacity of container be 70L (LCM of 5,7 & 10)

Now,

I container, milk = 28L

    Water = 42L

II container, milk = 20L

    Water = 50L

III container, milk = 21L

    Water = 49L

In large container, milk = (28 + 20 + 21)

= 69L

Water = 42 + 50 + 49

= 141L

Milk : Water = 23 : 47

Distance is same. So, speed will be inversely propornonel to time.

Ratio of time taken by car and motorcycle = 28 : 20

= 7 : 5

So, ratio of speed of car and motorcycle = 5 : 7

5/7 = 32/M

M = (32*7)/5 = 44.8 km/hr