If G is a DAG we denote by τ (G) the number of distinct topological sorts of G. Now let G = (V, E) be any DAG. An add-one-vertex operation on G constructs a digraph G 0 = (V 0 , E 0 ) such that • G 0 is also a DAG. • We add exactly one new vertex, call it z that is, V 0 = V ∪ {z}, z ∈ / V . • We add some (possibly 0) new directed edges, that is, E 0 ⊇ E. (a) Prove that for any DAG G = (V, E) there is a way to perform an add-one-vertex operation on G such that the resulting DAG G 0 has the same number of topological sorts, i. e. τ (G 0 ) = τ (G). (b) Prove that for any DAG G there is a way to perform an add-one-

i saw this question before, dont know the asnwer srry

step-by-step explanation:

the degrees would remain the same ( the number to the left of the decimal point.

the multiply the decimal by 60 to get the minutes and use the numbers to the left of the decimal point.

then multiply the decimal portion of that by 60 for the seconds and round off as needed.

39 .956652

the 39 remains the degrees.

the minutes = 0.956652 x 60 = 57.39912

the minutes = 57

the seconds = 0.39912 x 60 = 23.95 = 24 ( rounded to nearest tenth)

so 39.956652 = 39 degrees, 57 minutes and 24 seconds

-75.159809

the -75 remains the degrees

the minutes = 0.159809 x 60 = 9.58854

the minutes = 9

the seconds = 0.58854 x 60 = 35.31, rounded to nearest tenth = 35.3 seconds

so -75.159809 = -75 degrees, 9 minutes and 35.3 seconds.

the first coordinate is north or south, since it it positive it is north.

the second number is east or west, because it is negative it is west.

39 degrees, 57 minutes and 24 seconds n

-75 degrees, 9 minutes and 35.3 seconds w

kaahe btau