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Let & be the angle between the 2 mirrors.
The following table will help U find the no. of images.
n = 360/& Position of object No. of images
even anywhere n-1
odd symmetric n-1
asymmetric n
fraction anywhere [n] where [.] represents GIF
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Consider the example:
The lines whose equations are x+y=1,2x-y=3,2x+3y=6 & 3x+4y=5
(a) form a trapezium
(b) form a rectangle
(c) are concurrent
(d) form a quadrilateral
Since no 2 pairs of lines are parallel,the first 2 options are eliminated.Now,if the point of a pair of lines is the same as the point of intersection of the other 2 lines
then the 4 lines are concurrent,otherwise the lines form a quadrilateral.In this case
U will find that option (d) is the most appropriate.
This is known as process of elimination.
Hope U got it now...
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Just as all the covalent bonds have some partial ionic character,the ionic bonds also have partial covalent character.The partial covalent character of ionic bonds
is given by FAJAN'S rule.
The rule states that :
1.The smaller the size of the cation and the larger the size of the anion,the greater the covalent character of the ionic bond.
2.The greater the charge on the cation,the greater the covalent character
of the ionic bond.
3.For cations of the same size and charge the one with electronic config.
typical of transition elements, is more polarising than the one with a noble
gas config. which is typical of alkali and alkaline earth metal cations.
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Yes the answer is FALSE.
{(TRUE AND FALSE)} AND {TRUE OR (NOT TRUE)}
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The computer exam is on 31st for everyone,irrespective of delhi or outside delhi.
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Here are some of the full forms in communication chapter :
SIM - Subscriber Identification Module
IDEN - Integrated Digital Enhanced Network
WAIS - Wide Area Information System
ISDN - Integrated Services Digital Network
PDA - Personal Digital Assistant
URI - Uniform Resource Identifier
URN - Uniform Resource Name
URL - Uniform Resource Locator
OSI - Open System Interconnections
MIME - Multipurpose Internet Mail Extensions
LD - Laser Diode
VFIR - Very Fast Infra Red
ISP - Internet Service Provider
SNA - Systems Network Architecture
NIC - Network Interface Card
PSTN - Public Switched Telephone Network
DHTML - Dynamic HTML
UMTS - Universal Mobile Telecommunications System
SMSC - Short Message Service Center
HLR - Home Location Register
WCDMA - Wideband CDMA
CDMA - Code Division Multiple Access
HTML - Hyper Text Markup Language
BNC connector - Bayone Neill Concelman connector
AUI connector - Attachment Unit Interface connector
ARPAnet - Advanced Research Projects agency Network
UTP - Unshielded Twisted Pair Cable
STP - Shielded Twisted Pair Cable
DGM - Data Grade Medium
VGM - Voice Grade Medium
TAP - Terminal Access Point
CTS signal - Clear To Send signal
CR signal - Carrier Detect Signal
RTS signal - Request To Send Signal
DSR signal - Data Set Ready Signal
DTR signal - Data Terminal Ready Signal
NSFnet - National Science Foundation network
DCE - Data Communication Equipment
DTE - Data Terminal Equipment
TDMA - Time Division Multiple Access
FTP - File Transfer Protocol
GSM - Global System for Mobile
LAN - Local Area Network
WAN - Wide area Network
MAN - Metropolitan Area Network
XML - eXtensible Markup Language
WLL or WiLL - Wireless in Local Loop
3G - Third Generation mobile communications technology
SMS - Short Message Service
EDGE - Enhanced Data rates for Global Evolution
PPP - point to Point Protocols
SLIP - Serial Line Internet Protocol
NCP - Network Control Protocols
IPCP - IP control Protocol
LCP - Link Control Protocol
TCP/IP - Transmission Control Protocol/Internet Protocol
HTTP - Hypertext Tranfser Protocol
PBX - Private Branch Exchange
GUI - Graphical User Interface
WWW - World Wide Web
bps - bits per second
Bps - Bytes per second
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Let f(x) = integral of (ln t)/(1+t) dt from 1 to x.
Then find f(e)+f(1/e).
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You will have to use complex numbers to solve this integral.
The integral can be represented as the real part of ecos&{cos(sin&)+isin(sin&)} d&
=real part of ecos&.eisin& d&
=real part of ecos&+isin& d&
=real part of e power ei& d&
=real part of {1+ ei&/1! + e2i&/2! + ...... } d&
={1+ cos&/1! + cos2&/2! + ......} d& [considering the real part of the above expr.]
=& + sin& + (1/2!)[sin2&/2] + ...... -----> 1
Now applying the proper limits we get the first term of 1 to be 2 pi and all other terms 0.
Cheers!!!!!!!
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Ok friends, the answer is 2 pi.
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P.Bahadur is the best for physical chemistry.
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Hey where are you all guys???????
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Integrand : ecos& cos(sin&)
Limits : Lower limit- 0
Upper limit- 2 pi
Variable of integration : &
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Let X denote the figure which does not recur and assume they are l in number.
Let Y denote recurring period of consisting of m figures.
Let R denote the value of the recurring decimal.
Then, R=0.XYYY
Therefore, 10lR=X.YYY --- 1
And, 10l+m.R=XY.YY --- 2
Subtracting 1 from 2 we get, R=(XY-X)/(10l+m - 10l).
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