Q: How do you write twice the sum of t and 3?

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2 + s(t) -s

3*(t + 7) or 3t + 21

2t/12

(t+3)3 is already factored. Could write it as (t+3)(t+3)(t+3).

Once, Twice, Three Times a Lady (song - the Commodores)

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; }

16 + 49 = 65

10 for t = 1 to 50 20 input a 30 c = c + a 40 next t 50 print c

t(1) = a = 54 t(4) = a*r^3 = 2 t(4)/t(1) = r^3 = 2/54 = 1/27 and so r = 1/3 Then sum to infinity = a/(1 - r) = 54/(1 - 1/3) = 54/(2/3) = 81.

John is 36, Jane is 27 When he was 27, she was 18. The algebraic solution to the equation: Let A = John and B = Jane, and time t = A - B (difference in ages) If t years ago, John was twice as old A = 2 (B - t) and substitute A - B for t A = 2B - 2A + 2B 3A = 4B A = 4/3 B and since A + B = 63 4/3 B + B = 63 7/3 B = 63 B = 21 and A = 28

T

t+31