Answer:
the answer may be this: 8 √3
Answer:
P(3) is true since 2(3) - 1 = 5 < 3! = 6.
Step-by-step explanation:
Let P(n) be the proposition that 2n-1 ≤ n!. for n ≥ 3
Basis: P(3) is true since 2(3) - 1 = 5 < 3! = 6.
Inductive Step: Assume P(k) holds, i.e., 2k - 1 ≤ k! for an arbitrary integer k ≥ 3. To show that P(k + 1) holds:
2(k+1) - 1 = 2k + 2 - 1
≤ 2 + k! (by the inductive hypothesis)
= (k + 1)! Therefore,2n-1 ≤ n! holds, for every integer n ≥ 3.
Answer:
Step-by-step explanation:
range = [-4,6,26]
Step-by-step explanation:
y = 5x - 9.....domain = [1,3,7]....the domain is ur x values, and I am assuming ur looking for the y values that go with the x ones
y = 5x - 9.....when ur domain (ur x) = 1
y = 5(1) - 9
y = 5 - 9
y = -4
y = 5x - 9...when ur domain (x) = 3
y = 5(3) - 9
y = 15 - 9
y = 6
y = 5x - 9...when ur domain(x) is 7
y = 5(7) - 9
y = 35 - 9
y = 26
so ur range (ur y valus) = [-4,6,26]
