Mathematics, 18.03.2020 23:44, BrownieMan123
Let S be the subset of the set of ordered pairs of integers defined recursively by: Basis Step: (0,0)∈S. Recursive Step: If (a, b)∈S, then (a+4,b+5)∈S and (a+5,b+4)∈S. a) List the elements of S produced by the basis step plus the first 3 applications of the recursive step in the definition. b) Use strong induction on the number of applications of the recursive step of the definition of set S given above to show that 9 "divides" (a+b) [or that (a+b) is a multiple of 9] when (a, b)∈S. That is, show that for any (a, b)∈S obtained by n≥0 applications of the recursive step of the definition of S, (a+b) = 9m for some integer m. This is equivalent to proving that set S is a subset of the multiples of 9. Be sure to use strong induction. [Note that the only difference in a proof by induction versus one by strong induction is that the inductive hypothesis is "stronger".]
Answers: 1
Mathematics, 21.06.2019 19:00, asmith219771
What is the expression in factored form? -20x^2 - 12x a. 4x(5x+3) b. -4x(5x-3) c. -4(5x+3) d. -4x(5x+3)
Answers: 2
Mathematics, 21.06.2019 19:30, ashtonsilvers2003
Evaluate the expression for the given value of the variable. ∣-4b-8∣+∣-1-b^2 ∣+2b^3 ; b=-2
Answers: 2
Let S be the subset of the set of ordered pairs of integers defined recursively by: Basis Step: (0,0...
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