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
4
It's an algebraic expression that can be simplified to: j-128
There are no brackets!
(j^3 + 3k^4)(j^6 - 3j^3k^4 + 9k^8)
2j+1/j+5 = 1 Multiply all terms by j+5 to eliminate the fraction: 2j+1 = j+5 2j-j = 5-1 j = 4
4
It's an algebraic expression that can be simplified to: j-128
There are no brackets!
(j^3 + 3k^4)(j^6 - 3j^3k^4 + 9k^8)
j
2j+1/j+5 = 1 Multiply all terms by j+5 to eliminate the fraction: 2j+1 = j+5 2j-j = 5-1 j = 4
-19+j
#include <iostream> int main() { int i,j; for(i=1;i<6;i++) // Change 6 to a higher number to get more lines of output { for(j=i;j<i+4;j++) // Change 4 to a higher number to get more numbers in each line cout<<j; cout<<"\n"; } return 0; }
-4 (9j+5) = -36 j - 20 or - (36j + 20)
Ctrl plus J is used to justify text or a paragraph.
e must be 0, j can be anything
//to multiply two matrices... #include<stdio.h> main() { int a[4][4]; int b[4][4]; int c[4][4]; int i,j,k; printf("\n Enter elements into the first matix....\n"); for(i=0;i<4;i++) { for(j=0;j>4;j++) { scanf("%d",&a[i][j]); } } printf("\n Enter elements into the second matrix.....\n"); for(i=0;i<4;i++) { for(j=0;j<4;j++) { scanf("%d",&b[i][j]); } } printf("\n 1st Matrix.....\n"); for(i=0;i<4;i++) { for(j=0;j<4;j++) { printf("\t%d",a[i][j]); } printf("\n"); } printf("\n 2nd Matrix......\n); for(i=0;i<4;i++) { for(j=0;j<4;j++) { printf("\t%d",b[i][j]); } printf("\n"); } for(i=0;i<4;i++) { for(j=0;j<4;j++) { for(k=0;k<4;k++) { c[i][j]+=(a[i][k]*b[k][j]); } } } printf("\n Resultant Matrix...\n"); for(i=0;i<4;i++) { for(j=0;j<4;j++) { printf("\t%d",c[i][j]); } printf("\n"); } }