The existence of the additive inverse (of ab).
4
I am concerned that this question [ J plus 16-148 plus 4? ] was typed wrong, but I will try to answer it. This appears to be asking: "Simplify J plus 16-148 plus 4." To simplify... J + 16 - 148 + 4 ... it would be J plus the addition of the three numbers 16, (-148), and 4. 16 + (-148) + 4 = ( - 128 ) so the answer is... J - 128
e,j,h,t,c
(b+h)(b+h)= b2+2hb+h2.
* ghytyhthgfjhghjjjjjjjjjjjjjjhh * ]jk * j* j * j * j * j * j jj j j j j j jj h * h * h * h * h * h * h * h * h * h * h * h * h * h * h * h * h * h * h * h * h * h
please tell me why are these letter place in this order and above and below the line a ef hi _______________ bcd g j
#include<stdio.h> typedef struct hole { int id,size,inf,ef,filled,pid,psize; }hole; typedef struct process { int id,size,comp; }process; hole h[50]; process p[50]; int m,n,intfrag,exfrag; void output() { int i; intfrag=exfrag=0; printf("\n\n\nHole Process IF EF\n"); for(i=0;i<m;i++) { printf("\n%d %d %d %d",h[i].size,h[i].psize,h[i].inf,h[i].ef); intfrag=intfrag+h[i].inf; exfrag=exfrag+h[i].ef; } } void final() { int i; for(i=0;i<n;i++) { if(p[i].comp==0) { printf("There is no memory for the process of size %d\n",p[i].size); } } } void input() { int i,max,flag; max=flag=0; printf("\nEnter total no of holes:\n"); scanf("%d",&m); for(i=0;i<m;i++) { h[i].id=i; printf("\nEnter size of hole %d:\n",i+1); scanf("%d",&h[i].size); h[i].inf=0; h[i].ef=h[i].size; h[i].filled=0; h[i].pid=0; h[i].psize=0; } for(i=0;i<m;i++) { if(h[i].size<0) { printf("\nInvalid input as hole size is negative"); flag=1; } if(h[i].size>max) { max=h[i].size; } } printf("\nEnter total no of processes:\n"); scanf("%d",&n); for(i=0;i<n;i++) { p[i].id=i; printf("\nEnter size of process %d:\n",i+1); scanf("%d",&p[i].size); p[i].comp=0; } for(i=0;i<n;i++) { if(p[i].size<0) { printf("\nInvalid input as process size is negative"); flag=1; } if(p[i].size>max) { printf("\nInvalid input as process size is greater than max size of hole\n"); flag=1; } } if(flag==0) { output(); } } void first() { int i,j; for(i=0;i<n;i++) { for(j=0;j<m;j++) { if(p[i].size<=h[j].size && p[i].comp==0 && h[j].filled==0) { p[i].comp=1; h[j].filled=1; h[j].psize=p[i].size; h[j].id=p[i].id; h[j].inf=h[j].size-p[i].size; h[j].ef=0; output(); } } } printf("\n\n\nInternal Fragmentation:%d",intfrag); printf("\n\n\nExternal Fragmentation:%d\n\n",exfrag); final(); } void best() { int min,i,j,sub,var; for(i=0;i<n;i++) { min=999; sub=0; for(j=0;j<m;j++) { if(p[i].size<=h[j].size && p[i].comp==0 && h[j].filled==0) { sub=h[j].size-p[i].size; if(min>sub) { min=sub; var=j; } } } p[i].comp=1; h[var].filled=1; h[var].psize=p[i].size; h[var].id=p[i].id; h[var].inf=h[var].size-p[i].size; h[var].ef=0; output(); } printf("\n\n\nInternal Fragmentation:%d",intfrag); printf("\n\n\nExternal Fragmentation:%d\n\n",exfrag); final(); } void next() { int i,j,a; a=j=0; for(i=0;i<n;i++) { for(j=a;j<m;j++) { if(p[i].size<=h[j].size && p[i].comp==0 && h[j].filled==0) { p[i].comp=1; h[j].filled=1; h[j].psize=p[i].size; h[j].id=p[i].id; h[j].inf=h[j].size-p[i].size; h[j].ef=0; output(); a=(j+1)%m; } } } printf("\n\n\nInternal Fragmentation:%d",intfrag); printf("\n\n\nExternal Fragmentation:%d\n\n",exfrag); final(); } void main() { int ch; printf("\n\n\n**********MEMORY ALLOCATION**********\n\n"); printf("\n1.First Fit Algorithm\n2.Best Fit Algorithm\n3.Next Fit Algorithm"); printf("\nEnter your choice:\n"); scanf("%d",&ch); switch(ch) { case 1: input(); first(); break; case 2: input(); best(); break; case 3: input(); next(); break; } } Written by: Fabianski Benjamin
Given ef is the midsegment of isosceles trapezoid abcd bc equals 17x ef equals 22.5x plus 9 and ad equals 30x plus 12 find ad?
The existence of the additive inverse (of ab).
One triplet (k) and two pairs (h and number). 3j has no like term.
it is symmetric
J. H. H. Coombes died on 1978-02-18.
J. H. H. Coombes was born on 1906-12-28.
J H J M. Meuwissen has written: 'Huwelijksonvruchtbaarheid'
J. H. Haverly died in 1901.
J. H. Müller died in 1830.