Math Problems
#Problems #Quest22
#Solutions | #Quest22
e^x=t
dt=e^xdx
1/(x-1)(x³-1)=1/(x-1)²(x²+x+1)=(x+1)/(3x²+3x+3) +(2-x)/3(x-1)²=(x+1)/3(x²+x+1) +1/3(x-1)²-1/3(x-1)
Author:Anar
e^x=t
dt=e^xdx
1/(x-1)(x³-1)=1/(x-1)²(x²+x+1)=(x+1)/(3x²+3x+3) +(2-x)/3(x-1)²=(x+1)/3(x²+x+1) +1/3(x-1)²-1/3(x-1)
Author:
Math Problems
#Problems #Quest1 (repost)
#Solutions | #Quest1
sum of the n first positive odds is n².
So just divide the sixth positive odd = 11, by the twenty-first positive odd = 41.
Answer: 11/41.
Author:@Sidneifs
sum of the n first positive odds is n².
So just divide the sixth positive odd = 11, by the twenty-first positive odd = 41.
Answer: 11/41.
Author:
Math Problems
#Problems #Quest28
#Solutions | #Quest28
Prove that f(x) = cx3^x thats it ig, nothing more to be done, then u can easily compute that thing to be n/In3
Author:@IvaLobachevsky
Prove that f(x) = cx3^x thats it ig, nothing more to be done, then u can easily compute that thing to be n/In3
Author:
Math Problems
#Problems #Quest37
#Solutions | #Quest37
(1+x)ⁿ=C(n;0)*1+C(n;1)*1*x+C(n;2)*1*x²+...+C(n;n-1)*x^(n-1) +C(n;n)*x^n
n(1+x)^(n-1)=C(n;1)+2C(n;2)*x+3*C(n;3)x²+...+(n-1)*C(n;n-1)*x^(n-2)+n*C(n;n)*x^(n-1)
x=1
C(n;1)+2*C(n;2)+3*C(n;3)+...+n*C(n;n)=n*2^(n-1)
Author:Anar
(1+x)ⁿ=C(n;0)*1+C(n;1)*1*x+C(n;2)*1*x²+...+C(n;n-1)*x^(n-1) +C(n;n)*x^n
n(1+x)^(n-1)=C(n;1)+2C(n;2)*x+3*C(n;3)x²+...+(n-1)*C(n;n-1)*x^(n-2)+n*C(n;n)*x^(n-1)
x=1
C(n;1)+2*C(n;2)+3*C(n;3)+...+n*C(n;n)=n*2^(n-1)
Author:
Math Problems
#Problems #Quest6
#Solutions | #Quest6
So here there are only 7 people, where the 7th person is both a lawyer and CA. We will split the problem into cases: when the "both" person is part of the committee and when he isn't.
Case 1: For this committee, the "both" person must be part of it, so we really are choosing 2 people from the remaining 6. So we have 1⋅C6,2
Case 2: Here, the "both" person isn't part of the committee, so we have 3 Lawyers we can choose for our necessary lawyer, 3 for our necessary CA, and finally 4 for the remaining person who can be anything. This leaves us with 3⋅3⋅4 committees. However, we don't care about distinguishing between when the person is chosen, so we divide by the permutations of all committees, leaving us with 3⋅3⋅4/3!=6.
Adding the two cases, we get 21.
Author:@UltimateTOE
So here there are only 7 people, where the 7th person is both a lawyer and CA. We will split the problem into cases: when the "both" person is part of the committee and when he isn't.
Case 1: For this committee, the "both" person must be part of it, so we really are choosing 2 people from the remaining 6. So we have 1⋅C6,2
Case 2: Here, the "both" person isn't part of the committee, so we have 3 Lawyers we can choose for our necessary lawyer, 3 for our necessary CA, and finally 4 for the remaining person who can be anything. This leaves us with 3⋅3⋅4 committees. However, we don't care about distinguishing between when the person is chosen, so we divide by the permutations of all committees, leaving us with 3⋅3⋅4/3!=6.
Adding the two cases, we get 21.
Author:
Math Problems
#Problem #Quest44
#Solutions | #Quest44
This is basically the number of ways to select a committee with at least k people, have k people be the leaders of the committee, then choose a person out of the other people not in the committee(but in the n people). Therefore, we can instead just choose k people as the leaders first, then the person out of the other people, then everyone else can be in or not in the committee, so the answer is binom (n)(k) . (n-k). 2^(n-k-1).
Author: @MathMrIO
This is basically the number of ways to select a committee with at least k people, have k people be the leaders of the committee, then choose a person out of the other people not in the committee(but in the n people). Therefore, we can instead just choose k people as the leaders first, then the person out of the other people, then everyone else can be in or not in the committee, so the answer is binom (n)(k) . (n-k). 2^(n-k-1).
Author: @MathMrIO