4x+2y=100
x+y=30
x=y-30
4(y-30)+2y=100
120-2y=100
y=10
x+10=30
x=20
So 20 cows and 10 hens
-1 = √1
-1 = 1
Why does this happen?
Its takes time to understand if u don't get the logic but very simple if u are through
1!= 1
2!= 2
3!= 6
4!= 24
5!= 120
From 5! every no. ends with a zero...
Any no ending with zero raised to any power gives a number which has the last digit as zero.
So all the other no.s from 5! onwards will end in zero so we should calculate only the last digit for the following
(1!)^(1!) + (2!)^(2!) + (3!)^(3!) +(4!)^(4!)
(1!)^(1!) = 1
(2!)^(2!) = 2^2 = 4
(3!)^(3!) = 6^6 =... ends with 6 (since 6 raised to any power gives a no. whose last digit is always 6
(4!)^(4!) = 24^24 = 4^24 * 6^24 =....ends with 6
{since 6^24 ends with 6 and
4 raised to the odd power gives no. ending wit 4
4 raised to the even power gives no. ending with 6
eg:
4^1 = 4
4^2 = 16
4^3 = 64
4^4 = 256 and so on...
4^24 ends with 6
6*6 ends with 6}
hence the last digit will be the addition of these last digits
1+4+6+6 = 17 hence ends with 7
Prove that 0.999999....... = 1
Nope..it's true.totally wrong question...
This is wrong.-1 = √-1 * √-1
-1 = √(-1*-1)
This is correct√-1 * √-1 = -1
First we set:
x=0.999999999…… (infinitely recurring)Multiplying both sides by 10, we have,
10x=9.999999999….. (infinitely recurring)subtracting the first equation from the second one,
10x - x = 9.999999999…… - 0.999999999…….Therefore,
9x = 9We divide both sides by 9 to get,
x = 1so do we have, from the first statement,
1 = .999999999….. ?
Which 2 rupees you twat?Three friends went to a restaurant and had a bill of Rs. 30 ...each decided to pay Rs.10...the owner was impressed by the waiter who served this friend and return Rs. 10 to the waiter telling him to return Rs. 2 to each person (2*3=6) and to keep Rs. 4 with u (waiter).. (6+4=10)...As Rs. 2 is returned so each person paid Rs.8 (10-2) so 8*3=24 (all three together) and Rs. 4 from waiter make 24+4=28 ..Where is this Rs. 2 gone ?
Hah easy.....Any 8th standard guy who has learnt to change Recurring decimals to fractions would have solved this.Prove that 0.999999....... = 1
Check this out -
i = √-1.
i² = -1.
i² = i * i
-1 = √-1 * √-1
-1 = √(-1*-1)
-1 = √1
-1 = 1
Why does this happen?
That's the reason whyIt can be i*i or even -i * -i
and √1 can be -1 or 1.
Consider both cases then.
Check this out -
i = √-1.
i² = -1.
i² = i * i
-1 = √-1 * √-1
-1 = √(-1*-1)
-1 = √1
-1 = 1
Why does this happen?
Nope...If there are, say 30 people in room, then 1+2+3+4.....+(30-1)/365 = 1.917.... and probability of anything can't exceed 1.Hmmm...
Let there be N people in the room,
The probability of the person 1 having the same birthday as any other is 0/365.
The probability of the person 2 having the same birthday as any other is 1/365.
The probability of the person 3 having the same birthday as any other is 2/365.
Going this way, the probability of any on the N persons having a birthday as any other is
1+2+3...+(N-1)/365?
Is it right?
This does not happen!!
Actually, there's a rule dat
√(a*b)=√a * √b , if and only if one or both of a & b are +ve (i.e. >0)!!!
So,
√-1 * √-1 is not equal to √(-1*-1)