1 is the loneliest number per Randy Nguyen
Emi = l * r * ((1 + r)^n / (1 + r)^n - 1) * 1/12 where l = loan amt r = rate of interest n = no of terms
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The answer: you probably don't want to know. Here it is: P = L[c(1 + c)n]/[(1 + c)n- 1] where P equals payment, L equals loan amount, n equals number of months in the loan term and c equals monthly interest rate. If you are just wanting to calculate what a monthly mortgage payment would be, there are several mortgage calculators available online. Here's a link to a good one: http://www.bankrate.com/calculators/mortgages/mortgage-calculator.aspx
Continuing Education Loan (private)
R.D.=P*n+P(n+1)/2400
This browser is totally bloody useless for mathematical display but...The probability function of the binomial distribution is P(X = r) = (nCr)*p^r*(1-p)^(n-r) where nCr =n!/[r!(n-r)!]Let n -> infinity while np = L, a constant, so that p = L/nthenP(X = r) = lim as n -> infinity of n*(n-1)*...*(n-k+1)/r! * (L/n)^r * (1 - L/n)^(n-r)= lim as n -> infinity of {n^r - O[(n)^(k-1)]}/r! * (L^r/n^r) * (1 - L/n)^(n-r)= lim as n -> infinity of 1/r! * (L^r) * (1 - L/n)^(n-r) (cancelling out n^r and removing O(n)^(r-1) as being insignificantly smaller than the denominator, n^r)= lim as n -> infinity of (L^r) / r! * (1 - L/n)^(n-r)Now lim n -> infinity of (1 - L/n)^n = e^(-L)and lim n -> infinity of (1 - L/n)^r = lim (1 - 0)^r = 1lim as n -> infinity of (1 - L/n)^(n-r) = e^(-L)So P(X = r) = L^r * e^(-L)/r! which is the probability function of the Poisson distribution with parameter L.
n(1-R)L is an expression: it is not a formula.
n : 2 l : 1 ml : -1, 0, or 1
(N-1)=(4-1)= N=3 l=0,1,2,3
If the sequence (n) converges to a limit L then, by definition, for any eps>0 there exists a number N such |n-L|N. However if eps=0.5 then whatever value of N we chose we find that whenever n>max{N,L}+1, |n-L|=n-L>1>eps. Proving the first statement false by contradiction.
Quantum numbers consist of four values: the principal quantum number (n), the azimuthal quantum number (l), the magnetic quantum number (m_l), and the spin quantum number (m_s). For a valid set, n must be a positive integer (n = 1, 2, 3,...), l must be an integer from 0 to n-1, m_l must range from -l to +l, and m_s can be either +1/2 or -1/2. For example, the set (n=2, l=1, m_l=0, m_s=+1/2) is valid, while (n=2, l=2, m_l=0, m_s=+1/2) is not, because l cannot equal n.
Emi = l * r * ((1 + r)^n / (1 + r)^n - 1) * 1/12 where l = loan amt r = rate of interest n = no of terms
#include<stdio.h> #include<string.h> int max(int a,int b) { return a>b?a:b; }//end max() int main() { char a[]="xyxxzxyzxy"; char b[]="zxzyyzxxyxxz"; int n = strlen(a); int m = strlen(b); int i,j; for(i=n;i>=1;i--) a[i] = a[i-1]; for(i=m;i>=1;i--) b[i] = b[i-1]; int l[n+1][m+1]; printf("\n\t"); for(i=0;i<=n;i++) { for(j=0;j<=m;j++) { if(i==0 j==0) l[i][j]=0; else if(a[i] == b[j] ) l[i][j] = l[i-1][j-1] + 1; else l[i][j] = max(l[i][j-1],l[i-1][j]); printf("%d |",l[i][j]); } printf("\n\t"); } printf("Length of Longest Common Subsequence = %d\n",l[n][m]); return 0; }
For the principal quantum number ( n = 2 ), the possible values of the azimuthal quantum number ( l ) are 0 and 1 (since ( l ) can take on values from 0 to ( n-1 )). For each value of ( l ), the magnetic quantum number ( m_l ) can take values from (-l) to (+l). Therefore, for ( l = 0 ), ( m_l = 0 ) (1 combination), and for ( l = 1), ( m_l ) can be (-1, 0, +1) (3 combinations). In total, there are ( 1 + 3 = 4 ) possible combinations of ( l ) and ( m_l ) for ( n = 2 ).
The secondary quantum number, l, represents the shape of an orbital and can have values ranging from 0 to n-1, where n is the principal quantum number. Therefore, l can have values from 0 to (n-1).
For an electron with n=5, the possible values of l range from 0 to 4 (l=0, 1, 2, 3, 4). The value of l depends on the principal quantum number (n) according to the rule that l can be any integer value from 0 to n-1.
Potassium has an electronic configuration of 1s2, 2s2, 2p6, 3s2, 3p6, 4s1 The azimuthal quantum number of 0corresonds to an s subshell, 1 p subshell You can see that you have n=1, l=0 n=2, l=0; n=2, l=1 n=3, l=0; n=3, l=1, n=4, l=0