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.
I think it is A because i think it is the only one that makes sense
It's letter C .
Hope this help :)
Answer:
1. No solution
2. Infinite solutions
Step-by-step explanation:
1. To solve the system, set the equations equal to each other and solve for x.
2x - 5 = 2x + 7
-5 = 7
This is a false statement. This means there is no solution.
2. To solve the system, graph each equation.
y = -3/4 x - 5/2 has a y-intercept -5/2 and slope -3/4.
3x + 4y = -10 converts to y = -3/4x -5/2.
This graphs as the exact same line. This system has infinite solutions.