MC 302 – GRAPH THEORY – HW #5 – 11/17/11 Due Tuesday, 11
... Answer: Since G is not Eulerian k must be odd. If k were even, G would be Eulerian since G is Eulerian IFF every vertex of G has even degree. Thus every vertex of G has odd degree, k. Assume n is odd. Then by the degree sum theorem, the number of edges in G = (n*k)/2. Since n is odd and k is odd, n* ...
... Answer: Since G is not Eulerian k must be odd. If k were even, G would be Eulerian since G is Eulerian IFF every vertex of G has even degree. Thus every vertex of G has odd degree, k. Assume n is odd. Then by the degree sum theorem, the number of edges in G = (n*k)/2. Since n is odd and k is odd, n* ...
Functions and Relations ---
... a correspondence between the set of natural numbers and a second set ( we can list the numbers in a list, 1st, 2nd, 3rd, … ) We can have a finite sequence; there is a beginning term and an ending term a1, a2, a3,… an here an represents the last term and n represents the number of terms in the ...
... a correspondence between the set of natural numbers and a second set ( we can list the numbers in a list, 1st, 2nd, 3rd, … ) We can have a finite sequence; there is a beginning term and an ending term a1, a2, a3,… an here an represents the last term and n represents the number of terms in the ...
5.6: Find Rational Zeros, II
... Assume a has k factors and d has r factors. According to The Rational Zero Theorem, what is the maximum number of possible rational zeros for the polynomial function ...
... Assume a has k factors and d has r factors. According to The Rational Zero Theorem, what is the maximum number of possible rational zeros for the polynomial function ...
