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pannaguma......... plz do tel me hwz k=3......... plz na do nudge me..........

_{[ 0]}

^{[ 2 pie]} sin ix sin jx dx=1/2 _{[ 0]}^{[ 2 pie]} [cos(i-j)-cos(i+j)]dx

=1/2[ { sin2pie(i-j)}/(i-j) - {sin2pie(i+j)}/(i+j)]

now wen i , j are any 2 integers s.t. i is nt equal to j

den

i-j vl b an integer n thus _{[ 0]}^{[ 2 pie]} sin ix sin jx dx will b equal to 0

wen i=j then _{[ 0]}^{[ 2 pie]} sin ix sin jx =_{[ 0]}^{[ 2 pie]} sin^2 ix dx

=_[0]^{[ 2 pie]} (1-cos2ix)/2 dx

=pie - _[0]^{[ 2 pie]} cos 2ix dx

= pie -[ sin4pie i ]/2i

= pie which is nt equal to 0

therfore R is symmetric n transitive for i not equal to j

bt R is not reflexive

PLZ DO TEL ME WHETHER EM CORRECT OR NOT.................

Let C(k) b d capacitance wen d dielectric slab is inserted between d 2 plates of d capacitor....... n C b d capacitance widout ne dielectric slab.... den,
dielectric constant ,k= C(k)/ C
= q(k)/q
=100x 10^-6/ 50x 10^-6
=2

w= 100 s^-1
L , R , C r given in d ques
X(L)= wL
X(c)=1/wc
impedance, Z= sq rt[ R^2+(xL-xC)^2]
in dis way cn b obtained
peak emf, Eo= 30 V
thus
peak current, I o= Eo/ Z
resonating frequency, f = 1/ [ 2 pie LC]

PLZ C MY SOL. N TEMME IF DER'S NE FLAW IN DIS
total no. of possible permutations = 15!/5!3!3!2!
Total no. of possible permutations keeping d 3 c's together=13!/5!3!2!
total no. of possible permutations keeping d 2 c's together= 14!/5!3!2!
total no. of possible permutations keeping only 2 c's together at a tym= 14!/5!3!2! - 13!/5!3!2!
therefore
total no. of required permutations = 15!/5!3!3!2! - 13!/5!3!2! - (14!/5!3!2! - 13!/5!3!2!)
= 15!/5!3!3!2!- 14!/5!3!2!

hey........... is d answr 2 d 2nd 1 "n"..............

Na2SO4-> 2 Na+ + SO4 --
t=0 c 0 0
t=T c(1- a) 2ca ca

therfore
i = total no. of moles after dissociation / total no. of moles before dissociation
i= c(1+2a)/ c
i = 1+2a

D<0
9b^2-40ac<0
9b^2<40ac
here c , b^2 are +ve therefore a should also be +ve
rt(40ac)>3b
now by a.m. - g.m. inequality 2a+5c/2>rt(10ac)
i.e. 2a+5b>rt(40ac)>3b
therefore 2a-3b+5c>0

I hope ths ll help.............. followin r d 3 geometrical isomers possible

a=b(mod c) means tht wen a is divided by c thn b is d remainder.........

ya d ans. is 4 bcoz here d no. of tyms is odd.................

actually i gave my ans. 4 d general case...............

32=4(mod7)

32^32=4^32(mod7)

=2^64(mod7)

now v can easily c tht 2^p(mod7) follows d pattern 2,4,1,2,4,1..........

so 32^32=2(mod7)

32^32^32=2^32(mod7)

=4(mod7)

so v can easiliy c tht d remainder will b either 2 or 4 depending on d no. of tyms d term is raised 2 d power of 32.................

m i correct????

d ans. is either 2 or 4........

i think d ans. is 5.........

a=b(mod m ) means tht wen u divide a by m thn b is d remainder..............