-3 = j + 5 => -3 -5 = j + 5 -5 => -8 = j
j = 3
It is also x = 5 8zx + 5 = 2j + 3 → 8zx + 2 = 2j → 4zx + 1 = j Which is the first equation, which has a solution x = 5.
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
f = 54
j=-3 and k=2
4
e must be 0, j can be anything
1/3rd of x
j = 11. cross multiplication
cooking a turkey?
-3j = 44.7 -j = 44.7 ÷ 3 -j = 14.9 j = -14.9
#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; }