@viewpt said:yar samjh nai aa rha hai ye..
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Q: find the are of 5 sided regular polygon of length 10 each?
172.8
are of reguular pentagon 1.72 * s^2
@techsurge said:IFlog(4^x+1)+2x=log15+xlog41) x is a prime number2) 2x is prime number 3)4x is prime number4) 8x is prime
4x is prime....x=.5
@techsurge said:IFlog(4^x+1)+2x=log15+xlog41) x is a prime number2) 2x is prime number 3)4x is prime number4) 8x is prime
4x is a prime no.
@techsurge said:IFlog(4^x+1)+2x=log15+xlog41) x is a prime number2) 2x is prime number 3)4x is prime number4) 8x is prime
4) ??
@pankaj1988 said:Bhai yeh direct formula hai kya?1.72 s^2?
Yes
Read this in book Handa Ka Funda
Cant guarantee its authenticity
@pankaj1988 said:
bhai ans change change kar diya hai...4x is prime ...x=.5
Yes 4x
please share approach
mujhse to ye hila bhi nahi
@viewpt said:wo square hai cube ki jagah..-------------------------------------------Q: A grasshoper at a meter distance froma wall. with everey jump te covers half of the remaining distance. find the no. of jumps required to reach to the wall?
infinite h kya....:O
@techsurge said:IFlog(4^x+1)+2x=log15+xlog41) x is a prime number2) 2x is prime number 3)4x is prime number4) 8x is prime
15*4^x/(4^x+1)=10^2x
x=1/2
then 4x is prime
x=1/2
then 4x is prime
@techsurge said:IFlog(4^x+1)+2x=log15+xlog41) x is a prime number2) 2x is prime number 3)4x is prime number4) 8x is prime
log(4^x + 1)*10^(2x)) = log(15*4^x)
=> (4^x + 1)*(5^(2x)) = 15
x = 1/2 satisfies
=> 4x = 2 => 4x is a prime number ?
@pankaj1988 said:bhai ans change change kar diya hai...4x is prime ...x=.5
if x=.5 --> log3+1 = log15 + log2
= log3+log5+log2
--> log5 = log2+1 which is nt possible as log5??
= log3+log5+log2
--> log5 = log2+1 which is nt possible as log5??
@krum said:15*4^x/(4^x+1)=2x
not getting anywhere, but if it were 1.5
x=1/2
then 4x is prime
u mean to say 1.5 instead of 15
i might have posted wrong
this one somebody gave me
@techsurge said:6/35 is the OA(4c2*(4!/2!)*(4!/2!)*2) / (8!/(2!*2!))
getting 1/70
why (4!/2!) two times ?
@rubikmath said:if x=.5 --> log3+1 = log15 + log2 = log3+log5+log2--> log5 = log2+1 which is nt possible as log5??
When x = 1/2
LHS = log(3) + 1
RHS = log(15) + log(2) = log(30) = log(3) + 1
LHS = RHS ..
@pankaj1988 said:bhai ans change change kar diya hai...4x is prime ...x=.5
pls post approach, did u also change 15 to 1.5 ???
@techsurge said:Yes 4x please share approachmujhse to ye hila bhi nahi
Log (4^x+1) = Log ( 15 * 4^x) - 2x
2x = Log ( (15 * 4^x) / ( 4^x+1) )
=> 10^2x = 15/ ( 1+ 4^x)
Since 4^x is always +ve ...RHS
so I chekd for x= 1=> not satisfied
Again hit at x=.5 => satisfied
x = 4 * .5 =2 prime
@techsurge said:u mean to say 1.5 instead of 15i might have posted wrongthis one somebody gave me
sorry mate, i messed up again :splat:
@jain4444 said:getting 1/70 why (4!/2!) two times ?
in first 4 two country(same) nd 2 plain diff.colour --> 4!/2!
remaining 4 , 2 flag less(same) nd 2 plain diff.colour --> 4!/2!
remaining 4 , 2 flag less(same) nd 2 plain diff.colour --> 4!/2!