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.
First one is 4^2 or just 16.
Second one is using√ So what you want to do is 2√16
Hope this helps
Answer:
Please,I don't get your question & kindly be specific
Answer:
5/12
Step-by-step explanation:
1/4 and 1/6 are fractions.
As we directly can't add fractions whenever denominators are different, so we will have to find out the Least common multiple of 4 and 6.
The LCM Of 4&6 will be 12, so we will write 12 instead of 4&6.
Now, 4*3=12, that is why we will have to multiply the numerator and denominator with 3.
1*3/4*3 = 3/12
Same thing applies to 1/6. In order to get 12 in the denominator we need to multiply 2 with numerator and denominator both.
1*2/6*2= 2/12
However, we have two new fractions with same denominator now, that is 3/12 and 2/12
We can directly add the numerators now, as we have the same denominator. So 3+2/12= 5/12 would be my final answer. (addition of like fractions)
5/12
Answer:
He paid $108.33 down and will owe $216.67
Step-by-step explanation:
:)
