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-n=-10 n=10
7*abs(n - 10) which is 7*(10 - n) if n < 10 and 7*(n - 10) if n ≥ 10
int nearPalin(int n){ int temp = n; int count = 0; while(temp>0){ temp /= 10; count++; } if(count%2 == 0){ count = count/2; while(count--) n = n / 10; temp = n; while(n>0){ temp = temp*10 + n%10; n = n/10; } return temp; } else{ count = count/2; while(count--) n = n / 10; temp = n; n = n/10; while(n>0){ temp = temp*10 + n%10; n = n/10; } return temp; } }
10n where n is positive is 10*10*10... n times [or 1 followed by n zeros] 10-n where n is positive is 1/10n = 1/(10*10*10...) n times [or 0.0...01 where there are n-1 zeros between the decimal point and the 1]
If you mean: n-4 = 10 then n =14
(n - 4)*10 = 10*n - 40(n - 4)*10 = 10*n - 40(n - 4)*10 = 10*n - 40(n - 4)*10 = 10*n - 40
-n=-10 n=10
10 x n = 0.06 /10 /10 n = 0.006
7*abs(n - 10) which is 7*(10 - n) if n < 10 and 7*(n - 10) if n ≥ 10
-n(n + 10) or n(-n - 10)
int nearPalin(int n){ int temp = n; int count = 0; while(temp>0){ temp /= 10; count++; } if(count%2 == 0){ count = count/2; while(count--) n = n / 10; temp = n; while(n>0){ temp = temp*10 + n%10; n = n/10; } return temp; } else{ count = count/2; while(count--) n = n / 10; temp = n; n = n/10; while(n>0){ temp = temp*10 + n%10; n = n/10; } return temp; } }
n+5+(3/10)=8+(1/10) n+5-5+(3/10)=8-5+(1/10) n+(3/10)=3+(1/10) n+(3/10)=2+(11/10) n+(3/10)-(3/10)=2+(11/10)-(3/10) n=2+(8/10) n is equal to 2 and 8 tenths
10n where n is positive is 10*10*10... n times [or 1 followed by n zeros] 10-n where n is positive is 1/10n = 1/(10*10*10...) n times [or 0.0...01 where there are n-1 zeros between the decimal point and the 1]
If you mean: n-4 = 10 then n =14
To solve the third term of 10-n², substitute the number 3 for n. 10-n² =10-3² =10-9 =1 The third term of 10-n² is 1.
n - 10
10+n=