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If f(n) o(g(n)), does it imply that g(n) o(f(n))?

No, if f(n) o(g(n)), it does not necessarily imply that g(n) o(f(n)).


How heavy is a frag grenade?

It depends on what kind of grenade it is. The M26 and M26A1 grenades are 21 oz (595 g). The M61 grenade is 16 ounces (453 g). The M67 grenade is 14 ounces (397 g). The MK2 grenade is 21 ounces (595 g). The No. 5 Mark I Mills grenade is 22.5 ounces (638 g).


If f(n) o(g(n)), then how can the relationship between the growth rates of the functions f(n) and g(n) be described?

If f(n) o(g(n)), it means that the growth rate of f(n) is smaller than the growth rate of g(n).


Prove that a graph G is connected if and only if it has a spanning tree?

Proving this is simple. First, you prove that G has a spanning tree, it is connected, which is pretty obvious - a spanning tree itself is already a connected graph on the vertex set V(G), thus G which contains it as a spanning sub graph is obviously also connected. Second, you prove that if G is connected, it has a spanning tree. If G is a tree itself, then it must "contain" a spanning tree. If G is connected and not a tree, then it must have at least one cycle. I don't know if you know this or not, but there is a theorem stating that an edge is a cut-edge if and only if it is on no cycle (a cut-edge is an edge such that if you take it out, the graph becomes disconnected). Thus, you can just keep taking out edges from cycles in G until all that is left are cut-gees. Since you did not take out any cut-edges, the graph is still connected; since all that is left are cut-edges, there are no cycles. A connected graph with no cycles is a tree. Thus, G contains a spanning tree. Therefore, a graph G is connected if and only if it has a spanning tree!


Is the huawei hg520i a g router?

Yes.