@saurav.kgp said:what is the ans of this question? Find the value of x^3 if x^2+x+1=0.
x^2 +x +1 =0 ==> multiply by x on both sides ==> x^3 + x^2 +x = 0 we know x^2 +x = -1
==> x^3 - 1 =0 ==. x= 1
==> x^3 - 1 =0 ==. x= 1
@saurav.kgp said:what is the ans of this question? Find the value of x^3 if x^2+x+1=0.
x = (-1)^(2/3)
so x^3 = (-1)^2 or 1....
But Discriminant is coming out to be negative......
@Logrhythm and @ravi.theja what if we write x^2+x+1=0 as (x+1/2)^2+(3/4)=0,
such that (x+1/2)^2= a negative number and hence imaginary roots? what's wrong in this approach?
such that (x+1/2)^2= a negative number and hence imaginary roots? what's wrong in this approach?
Please solve it .
@CrookDinu said:@bs0409 :hw 11?? my ans is coming 10..share ur approach
n will be of length 11 like 1xxxxxxxxxx
Now for sum to be 2, one of the x's have to be 1.
Also one more case is 20000000000.
So total=11
@deepkarg said:@Logrhythm and @ravi.theja what if we write x^2+x+1=0 as (x+1/2)^2+(3/4)=0, such that (x+1/2)^2= a negative number and hence imaginary roots? what's wrong in this approach?
x^2+x+1=0 has imaginary roots only. No real roots....
a,b,c,d,e-prime
a
a+b+c+d+e=350
Find a
@uutkarshsingh said:@mailtoankit@LogrhythmThe way you people have solved is correct... but i am failing to understand why my method is wrong... ... MY APPROACH : first lets select 48 out of 52..... i.e 52c48.... then distribute 48 cards in 3 sets ... 16 in each... i.e (48)!/((16!)^3*(3!))...then multiply both... i.e 52c48 * (48)!/((16!)^3*(3!)) ....... = (52!)/(4*(16!)^3*3) .... divisor main mera extra 4 aa raha hai.. i dont knw wh its wrng ..aprch orrct lag rahi hai....
The bold and underlined part will not be there...........
@meenu05 said:Please solve it .
log6 (1/2) +2logx(_/2) = log36 (1/2)=log6 (1/ _/2)
logx (2) = log6( _/2)=(1/2)log6 (2)=log36 (2)
so x=36
@bs0409 said:x^2+x+1=0 has imaginary roots only. No real roots....a,b,c,d,e-primea is min and e is maxa+b+c+d+e=350Find a
2??
there are two containers A and B filled with oil with different prices and their volumes are 140litres and 60 litres resp. equal quantities are drawn from both A and Bin such a manner that the oil drawn from A is oured into B and the oil drawn from B is poured in A. the price per litre becomes equal in both A and B. how much oil is drawn from each A and B?
42 , 30 , 32 , CBD
@jain4444
Shld be 42..
Let CP of Oil1 be x per kg and that of Oil2 be y per kg..Let n L of Oil be drawn from each container A n B..
Now, Total price of Oil in A = (140-n)x + ny..Thus, Price per L of A = [(140-n)x + ny]/140
Again, Total Price of Oil in B= (60-n)y + nx..Thus, Price per L of B= [(60-n)y + nx]/60
Given, [(140-n)x + ny]/140 = [(60-n)y + nx]/60
=> 42(x-y) = n(x-y)
=>n = 42..
Please solve it .
@jain4444 said:there are two containers A and B filled with oil with different prices and their volumes are 140litres and 60 litres resp. equal quantities are drawn from both A and Bin such a manner that the oil drawn from A is oured into B and the oil drawn from B is poured in A. the price per litre becomes equal in both A and B. how much oil is drawn from each A and B? 42 , 30 , 32 , CBD
42
@jain4444 said:there are two containers A and B filled with oil with different prices and their volumes are 140litres and 60 litres resp. equal quantities are drawn from both A and Bin such a manner that the oil drawn from A is oured into B and the oil drawn from B is poured in A. the price per litre becomes equal in both A and B. how much oil is drawn from each A and B? 42 , 30 , 32 , CBD
42
@naveenkrs
Log(base 10)[1 - {1- (1-x^2)^-1 }^-1]^(-1/2) = 1
=> [1 - {1- (1-x^2)^-1 }^-1]^(-1/2) = 10
=> [1 - {1- (1-x^2)^-1 }^-1]^-1 = 100[Squaring Both Sides]
=> [1 - {1- 1/(1-x^2) }^-1]^-1 = 100
=>[1 - {1- (x^2-1)/x^2}^-1]^-1 = 100
=>[1 - (x^2 - 1)/x^2]^-1 = 100
=>[1/x^2]^-1 = 100
=> x^2 = 100
=>x = +/- 10..