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No, it is not mandatory to learn Korean for most courses, as they are taught in English. 

However, students who are looking to apply for programs which are Korean taught are required to prove a certain level of Korean proficiency by passing the TOPIK exam. It is recommended to the students to learn basic Korean to interact with the other students or people to explore the place.

$\left|E\right|=\frac{2kP}{r^3}$

$E=-\frac{dv}{dr}$ , $v\propto \frac{1}{r^2}$

$v_1=\sqrt[]{2g10}$

$v_2=\sqrt[]{2g5}$

$\overrightarrow{l}=\Delta \overrightarrow{p}$

$l=0.1\left\{\sqrt[]{2g10}+\sqrt[]{2g5}\right\}$

$=0.1\left\{10\sqrt[]{2}+10\right\}$

$=\left(\sqrt[]{2}+1\right)Ns$

$i_{battery}=\frac{\left(20-5\right)}{200}=\frac{15}{200}A$

$i_{300\Omega }=\frac{5}{300}A$

$i_{zener}=\frac{15}{200}-\frac{5}{300}$  

= 58.33 mA

By conservation of mechanical energy

mg (1) =  $\frac{1}{2}$ mv² + mg (0.5)

v² = 10

v =  $\sqrt[]{10}$ m/s

Georgia Tech acceptance rate stands at about 14%, meaning that out of every 100 students who apply to the university's admission process, only 14 are likely to be offered the opportunity to enroll on the campus grounds. This makes the admission criteria of the university extremely selective.

$i=\frac{7}{3.5k+0.9k\Omega }=\frac{7}{4.4k}$

$V_0=i\times 700\Omega =\frac{7}{4.4k}\times 7k=\frac{9.9}{4.4}=1.1V$

Kinetic energy: Potential energy = 1 : –2

Velocity at maximum height = vcoss30°

L = m (vcos30) H

$=mv\left(\frac{\sqrt[]{3}}{2}\right)*\frac{v^{2}si{n}^{2}30}{2g}$

$=\sqrt[]{3}\frac{m{v}^3}{16g}$

For 2 kg block

T – 2g sin37 = 2a        . (i)

For 4 kg block

4g – 2T =  $\frac{4a}{2}$

2g – T = a                    . (ii)

T = (2g – a)

2g – a – 2g ×  $\frac{3}{5}$  = 2a

3a = 2g ×  $\frac{2}{5}$

$a=\frac{4g}{15}$

x + 2y + 3z = 42

0    x + 2y = 42 ->22 cases

1    x + 2y = 39 ->19 cases

2    x + 2y = 36 ->19 cases

3    x + 2y = 33 ->17 cases

4    x + 2y = 30 ->16 cases

5    x + 2y = 27 ->14 cases

6    x + 2y = 24 ->13 cases

7    x + 2y = 21 ->11 cases

8    x + 2y = 18 ->10 cases

9    x + 2y = 15 ->8 cases

10  x + 2y =12 -> 7 cases

11  x + 2y = 9 -> 5 cases

12  x + 2y = 6 -> 4 cases

13  x + 2y = 3 -> 2 cases

14  x + 2y = 0 -> 1 cases.

R₁ = { (1, 1) (1, 2), (1, 3)., (1, 20), (2, 2), (2, 4). (2, 20), (3, 3), (3, 6), . (3, 18),
(4, 4), (4, 8), . (4, 20), (5, 5), (5, 10), (5, 15), (5, 20), (6, 6), (6, 12), (6, 18), (7. 7),
(7, 14), (8, 8), (8, 16), (9, 9), (9, 18), (10, 10), (10, 20), (11, 11), (12, 12), . (20, 20)}

n (R₁) = 66

R₂ = {a is integral multiple of b}

So n (R₁ – R₂) = 66 – 20 = 46

as R₁ Ç R₂ = { (a, a) : a Î s} = { (1, 1), (2, 2), ., (20, 20)}

RHL $\underset{x\to 0^{+}}{lim}\frac{co{s}^{-1}\left(1-x^2\right)si{n}^{-1}\left(1-x\right)}{x-x^3}$

$\underset{x\to 0^{+}}{lim}\frac{\pi }{2}\cdot \frac{co{s}^{-1}\left(1-x^2\right)}{x}$

$\frac{\pi }{2}\underset{x\to 0^{+}}{lim}\frac{-1}{\sqrt[]{{\left(1-\left(1-x^2\right)\right)}^2}}\left(-2x\right)$

$=\frac{\pi }{2}\underset{x\to 2^{+}}{lim}\frac{2x}{\sqrt[]{2x^2-x^4}}=\pi \underset{x\to 0^{+}}{lim}\frac{x}{x\sqrt[]{2-x^2}}$

$=\frac{\pi }{\sqrt[]{2}}$

LHL $\underset{x\to 0^{+}}{lim}\frac{co{s}^{-1}\left(1-{\left(1+x\right)}^2\right)si{n}^{-1}\left(1-\left(1+x\right)\right)}{1\cdot \left(1-{\left(1+x\right)}^2\right)}$

$=\underset{x\to 0^{-}}{lim}\frac{co{s}^{-1}\left(-x^2-2x\right),si{n}^{-1}\left(-x\right)}{-x^2-2x}$            

$=\frac{\pi }{2}\underset{x\to 0^{-}}{lim}\frac{-si{n}^{-1}x}{-x\left(x+2\right)}=\frac{\pi }{2}\times \frac{1}{2}=\frac{\pi }{2}$            

$tanB\times tanC=\frac{\sqrt[]{x}}{\sqrt[]{x^2+x+1}}\times \frac{1}{\sqrt[]{x\left(x^2+x+1\right)}}$

$=\frac{1}{x^2+x+1}=ta{n}^{2}A$            

tan² A = tan B tan C

It is only possible when A = B = C at x = 1

A = 30°, B = 30°, C = 30° $\left[tanA=tanB=tanC=\frac{1}{\sqrt[]{3}}\right]$                                 

$A+B=\frac{\pi }{2}-C$

$gof\left(x\right)=\{\begin{array}{cc}\hfill f\left(x\right),\hfill & \hfill f\left(x\right)<0\hfill \\ \hfill f\left(x\right),\hfill & \hfill f\left(x\right)>0\hfill \end{array}$

$=\{\begin{array}{lll}{e}^{lnx}=x\hfill & \left(0,1\right)\hfill & e^{-x}\hfill \\ \left(-\infty ,0\right)\hfill & lnx\hfill & \left(1,\infty \right)\hfill \end{array}$

Therefore, gof (x) is many one and into