Вычислить значение

функции f (t) = t+7:t+3 при t=2. Вычислите значения функции для всех t=0,1,2, ...,9.

F (t) = t+7:t+3

t=0 f (0) = 0+7:0+3=бесконечность

t=1 f (t) = 1+7:1+3=1-7-3=11

t=2 f (2) = 2+7:2+3=2+3.5+3=8.5

t=3 f (3) = 3+7:3+3=3+2. (3) + 3=8. (3)

t=4 f (4) = 4+7:4+3=4+1.75+3=8.75

t=5 f (5) = 5+7:5+3=5+1.4+3=9.4

t=6 f (6) = 6+7:6+3=6+1.1 (6) + 3=10.1 (6)

t=7 f (7) = 7+7:7+3=7+1+3=11

t=8 f (8) = 8+7:8+3=8+0.875+3=11.875

t=9 f (9) = 9+7:9+3=7+0. (7) + 3=10. (7)

T=0 = > f (t) = 7/3

t=1 = > f (t) = 2

t=2 = > f (t) = 9/5

t=3 = > f (t) = 5/3

t=4 = > f (t) = 11/7

t=5 = > f (t) = 3/2

t=6 = > f (t) = 13/9

t=7 = > f (t) = 7/5

t=8 = > f (t) = 15/11

t=9 = > f (t) = 4/3