"10 n on a p" likely refers to "10 newtons on a pound." This phrase could relate to the conversion of units of force, where newtons (N) are a metric unit and pounds (lb) are used in the imperial system. Specifically, 1 pound is approximately equal to 4.448 newtons, so 10 newtons would be roughly equivalent to 2.25 pounds. This conversion helps in comparing forces measured in different unit systems.
With 10 straight cuts, you can create a maximum of 56 pieces of pie. This is based on the formula for the maximum number of pieces ( P(n) = \frac{n(n + 1)}{2} + 1 ), where ( n ) is the number of cuts. For 10 cuts, substituting into the formula gives ( P(10) = \frac{10 \times 11}{2} + 1 = 56 ).
The answer is negative four BECAUSE... 20/5 is POSITIVE four 20/-5 is a NEGATIVE four because a positive divided by a negative is a negative. Easy way to remember negs/pos: n * p = n p* n = n p * p = p n * n = p n / p = n p / n = n p / p = p n / n = p There are always two ways to get a positive, and two ways to get a negative. Very simple.
Multiplying by a negative changes the sign of the original number. P x N = N N x N = P Multiplying by a positive keeps the sign of the original number. P x P = P N x P = N
Proof: P{T>n+m/T>n}=P{T>n+m,T>n}/P{T>n} (Bayes theorem) =P{T>n+m}/P{T>n} =((1-p)^(n+m))/(1-p)^n = (1-p)^(n+m-n) = (1-p)^m (1-p)^m = {T>m} So T>m has the same probability as T>m+n given that T>n, which means it doesn't care (or don't remember) that n phases had passed.
E= P x r x (1 + r)n / ((1+r)n -1) Here p=principal amount r = interesr rate per month(ex: if interest rate per annum is 10% then 10/(12*100)) n= tenure in months
#include<stdio.h> void main() { int *p,n,s=0; printf("Enter Number :"); scanf("%d",&n); for(p=&n;*p>0;p++) { s=s*10+*p%10; *p=*p/10; } printf("Reverse Number=%d",s); }
#include<stdio.h> #include<conio.h> void main() { clrscr(); int r=0,d,m,n; printf("Enter a value="); scanf("%d",&n); m=n; do { d=m%10; m=m/10; r=(r*10)+d; } while(m!=0); printf("%d is the reverse",r); getch(); }
Try 151,600! Permutations & Combinations. P(n,r)=n!(n−r)! not P(n,r)=n!/(n!-r!)r! ?
I'll use these symbols for each coin: P = Penny; D = Dime; N = Nickel 12 P 7 P & 1 N (7 + 5) 2 P & 2 N (2 + 10) 1 D & 2 P (10 + 2)
P= positive N=negative P x N = N N x P = N P x P = P N x N = P Hope that helps!?!?!
N - p% = N - p% of N = N*(1 - p%) = N*(1 - p/100) or N*(100 - p)/100
BJT is nothing but the addition of two PN junction diodes. There are two types of BJT= P-N-P or N-P-N P-N N-P + or + N-P = P-N-P P-N =N-P-N SCR is a thyristor which is made adding two BJTs. Of course they are made of sillicon. Exempli gratia: P-N-P + + N-P-N = P-N-P-N comparison between scr bjt and mosfet Check the related link for further information.
With 10 straight cuts, you can create a maximum of 56 pieces of pie. This is based on the formula for the maximum number of pieces ( P(n) = \frac{n(n + 1)}{2} + 1 ), where ( n ) is the number of cuts. For 10 cuts, substituting into the formula gives ( P(10) = \frac{10 \times 11}{2} + 1 = 56 ).
#include<stdio.h> #include<conio.h> main() { int a[10][10],b[10][10],c[10][10],m,n,i,j,p,q,op; printf("enter the order of matrix a:n"); scanf("%d",&m,&n); printf("enter the %d elements of a\n",m*); for(i=0;i<m;i++) for(j=0;j<n;j++) scanf("%d",&a[i][j]); printf("enter the order of matrix b:n"); scanf("%d",&p,&q); printf("enter the %d elements of b\n",p*q); for(i=0;i<p;i++) for(j=0;j<q;j++) scanf("%d",&b[i][j]); printf("enter the option\n"); scanf("%d",&option); switch(op) { case '+' : if(m==p&&n==q) printf("the resultant matrix c is:\n"); for(i=0;i<m;i++) for(j=0;j<n;j++) c[i][j]=a[i][j]+b[i][j]; printf("%d",c[i][j]); printf("\n"); break; case '/' : if(n==p) { for(i=0;i<m;;i++); {for(j=0;j<q;j++) {printf("%d",c[i][j]); } } c[i][j]=0; for(p=0;p<n;p++) c[i][j]=c[i][j]+a[i][p]*b[p][j]: } printf("resultant matrix is:\n"); for(i=0;i<m;;i++); {for(j=0;j<q;j++) {printf("%d\t",c[i][j]); } } printf("\n"); getch(); }
The answer is negative four BECAUSE... 20/5 is POSITIVE four 20/-5 is a NEGATIVE four because a positive divided by a negative is a negative. Easy way to remember negs/pos: n * p = n p* n = n p * p = p n * n = p n / p = n p / n = n p / p = p n / n = p There are always two ways to get a positive, and two ways to get a negative. Very simple.
normal variable stores a value of the given datatype where as the pointer variable stores the address of a variable. for example int n=10; int *p; p=&n; here p is a pointer variable and n is a normal variable.p stores the address of n where as n stores an integer value. *p prints the value of n,p prints the address of n.
Positive. p*p=p p*n=n n*n=p