8t
s(8+t)
4t = 24 Therefore t = 24/4 t = 6
62 + t
There are infinitely many common multiples of 24 and 4. Those are all numbers of form t = 24n, where n = 1, 2, 3, ..., because 24 itself is divisible by 4. The least common multiple of 4 and 24 is 24.
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; }
26
t+31
8t
Let x and y be positive numbers such that xy=13. Thus, y=13/x. We want to minimize t=x+y=x+13/x. Differentiate t=x+13/x. t'=1-13x-2 Set this equal to zero. 1-13x-2=0 13x-2=1 x2=13 x=sqrt(13) Now solve for y: y=13/x=13/sqrt(13)=sqrt(13)
s(8+t)
t w e n t y
24 = t + 624 - 6 = t +6 - 618 = t
The sum is 2.05 and you don;t need to estimate such a simple sum!
4t = 24 Therefore t = 24/4 t = 6
62 + t
24 = T + 6 take 6 from each side and you get: 24 - 6 = T therefore 18 = T