Answer:
Lets say that P(n) is true if n is a prime or a product of prime numbers. We want to show that P(n) is true for all n > 1.
The base case is n=2. P(2) is true because 2 is prime.
Now lets use the inductive hypothesis. Lets take a number n > 2, and we will assume that P(k) is true for any integer k such that 1 < k < n. We want to show that P(n) is true. We may assume that n is not prime, otherwise, P(n) would be trivially true. Since n is not prime, there exist positive integers a,b greater than 1 such that a*b = n. Note that 1 < a < n and 1 < b < n, thus P(a) and P(b) are true. Therefore there exists primes p1, ...., pj and pj+1, ..., pl such that
p1*p2*...*pj = a
pj+1*pj+2*...*pl = b
As a result
n = a*b = (p1*......*pj)*(pj+1*....*pl) = p1*....*pj*....pl
Since we could write n as a product of primes, then P(n) is also true. For strong induction, we conclude than P(n) is true for all integers greater than 1.
Answer:
Step-by-step explanation:
In this question, you would solve for "a".
Solve:
K = 4a + 9ab
Since we have our "a" on the same side, we can factor it out from the variables:
K = a(9b + 4)
To get "a" by itself, we would have to divide both sides by 9b + 4:
K/9b+ 4 = a
Your answer would be K/9b+ 4 = a
It would look like this:
Answer:
will give you an explanation if you want tag me on comment.
Where's the graph? If you can give me the graph then I'd be able to answer for you.
The sequence is increasing by the previous number multiplied by negative 3 (x -3). With this information, the correct answer would be A. -648, 1944, 5832
