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2x-5 is twice a number decreased by 5.
s(8+t)
A series of transformations on quadrilateral S resulted in quadrilateral T. The angles of quadrilateral S and T are congruent but the sides of quadrilateral T are twice as long as quadrilateral S. Which transformation on quadrilateral S must be included to result in quadrilateral T * sorry thats the full question!
Each number in Pascal's triangle is used twice when calculating the row below. Consequently the row total doubles with each successive row. If the row containing a single '1' is row zero, then T = 2r where T is the sum of the numbers in row r. So for r=100 T = 2100 = 1267650600228229401496703205376
8t
2x-5 is twice a number decreased by 5.
2t+3
s(8+t)
Sexually Malested And Raped Twice S M A R T
A series of transformations on quadrilateral S resulted in quadrilateral T. The angles of quadrilateral S and T are congruent but the sides of quadrilateral T are twice as long as quadrilateral S. Which transformation on quadrilateral S must be included to result in quadrilateral T * sorry thats the full question!
int Nodes (Tree *t) { int sum= 0; if (t) { sum+=1; if (t->left) sum += Nodes (t->left); if (t->right) sum += Nodes (t->right); } return sum; }
#include #include void main(){long n,s=0,t,d;clrscr();cout > n;t=n;while(n>0){d=n%10;s=s+d;n=n/10;}cout
#include<stdio.h> #include<conio.h> main() { int n,s,r,t; clrscr(); printf("enter n"); scanf("%d",&n); s=0;t=0; while(n!=0) { r=n%10; { if(r%2!=0) t=t+r; if(r%2==0) s=s+r; } n=n/10; } printf("sum of even position digits%d\n",s); printf("sum of odd position digits%d\n",t); getch(); }
t-j. Happy Christmas!
T
t+31
8t