Answer:i guess yes
Step-by-step explanation:
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:
domain: (-∞, 5)∪(5, ∞)
domain in interval form: {x|x ≠ 5}
Step-by-step explanation:
slndkndjakabssj
pasensya na god bless!!!!
Answer:
D.2
Step-by-step explanation:
First, you need to divide 34 and 60 by A.4, when you divide that you see that 4 dose not go in to 34 evenly and go"s in to 60 evenly, so this is incorect.
Now we will divide 34 and 60 by B.3, when you divide, you see that 3 dose not go into 34 evenly and go"s in to 60 evenly, so this is also incorect.
Then we divide 34 and 60 by C.1, when we divide this one you find that 1 go"s in to 34 evenly and 60 evenly, so this is a posible answer.
finally we divide 34 and 60 by D.2, when we do this we find that 2 go"s in to 34 evenly and 60 evenly, so this is also a posible answer.
All though C. and D. both go in to 34 evenly and 60 evenly, we are looking for the GCF (Greatest Common Factor). So the biggest number of C.1 and D.2, is D.2
So the answer is D.2.
Hoped this helped, Have a grate day!
