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.
Ok so to solve for percent off
percent off means discount
so therefor we need to find how much was discounted
30-20.10=9.90
so we see that 9.90 was discounted
to find the percent we do
discount/total=percent discount
9.90/30=0.33
percent means parts out of 100
0.33/1 times 100/100=33/100=33%
answer is 33% discout
We need to use Law of sine.
sin A/a = sin C/c
sin A/|CB| = sin C/|AB|
sin A/14 = sin(118⁰)/ 20
sin A = (14*sin(118⁰))/ 20
A=arcsin((14*sin(118⁰))/ 20) ≈ 38⁰
Answer:
Check the attached document for the solutions, cheers
Step-by-step explanation:
I am learning Geometry right now. And then I don’t know can I help you or not.
