5x-10y+15z find the equivalent expression by factoring

Mathematics

1 answer:

Marat540 [252]3 years ago

5 0

$\huge\boxed{5(x-2y+3z)}$

We can divide each term by a factor of $5$ . $5(1x-2y+3z)$

You don't need a coefficient of $1$ , so simplify that. $\boxed{5(x-2y+3z)}$

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Use strong mathematical induction to prove the existence part of the unique factorization of integers theorem (Theorem 4.4.5). I

valentina_108 [34]

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.

5 0

2 years ago

Matches the description in the table below.

tatiyna

I think it is A because i think it is the only one that makes sense

5 0

2 years ago

HELP ME PLZZZ HURRY PLZ

tresset_1 [31]

It's letter C .

Hope this help :)

5 0

2 years ago