Does prime factorization work for every number explain why or why not

Prime factorization only applies to integers greater than 2:

If you only consider these numbers, then yes: you can factor all of them.

Starting from 2, every other number is either prime, or divisibile by a prime we already found. Let's see a little bit of steps.

We begin with 2. It's divisible only by 1 and itself, so it's prime.

Then we have 3. It's divisible only by 1 and itself, so it's prime.

Then we have 4. It's divisibile by 2, so we can break into . The "remaining" 2 is prime, and we're done with the factorization.

Then we have 5. It's divisible only by 1 and itself, so it's prime.

Then we have 6. It's divisibile by 2, so we can break into . The "remaining" 3 is prime, and we're done with the factorization.

And you go on like this. The idea is that you either find a new prime or you can break the number into smaller pieces that you have already identified and factored.

720

step-by-step explanation:

first i did work 2nd i did work and 3rd i did work

so firstly, we need to isolate the y variables to be able to solve these inequalities. firstly, add 0.5x on both sides of the first inequality and subtract 1.5x on both sides of the second inequality:

now since the slope is positive for the first inequality, this means that the line going upwards belongs to the first inequality, and the line going downwards belongs to the second inequality.

next, since y ≥ in the first inequality, this means that the shaded region will be above the first inequality's line, thus shading regions a and b.

next, since y ≤ in the second inequality, this means that the shaded region is going to be below this line, thus shading regions a and c.

since both lines shade region a, this means that region a is the solution.

step-by-step explanation:

mark brainliest!