What are the prime factorization of 25

Mathematics

1 answer:

forsale [732]4 years ago

6 0

Answer:

25 = 5^2

Step-by-step explanation:

Factor the following integer:

Hint: | Is 25 divisible by 2?

25 is not divisible by 2 since 25 is odd and 2 is even:

25 = 25 (25 is not divisible by 2)

Hint: | Is 25 divisible by 3?

The sum of the digits of 25 is 2 + 5 = 7, which is not divisible by 3. This means 25 is not divisible by 3:

25 = 25 (25 is not divisible by 2 or 3)

Hint: | Is 25 divisible by 5?

The last digit of 25 is 5, which means 25 is divisible by 5:

25 = 5 5:

25 = 5×5 (5 is not divisible by 2 or 3 since 25 is not)

Hint: | Now try to factor 5. Is 5 prime?

No primes less than 5 divide into it. Therefore 5 is prime:

25 = 5×5

Hint: | Express 25 as a prime power.

There are 2 copies of 5 in the product:

Answer: 25 = 5^2

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Please help with this problem 32n^4 _____= 4n^2

sertanlavr [38]

Answer:

4*n^2 and n cannot equal 0

Step-by-step explanation:

First, since you have a variable int he denominator you need to write on the side that n cannot = 0, you may write it as n=/= 0, so a slash through an equal sign. You need to include this with your answer.

I'm assuming this is a fraction.The concept is actually pretty simple. What happens if you have 2/4? You can simplify it into 1/2. The concept is basically saying 2/(2*2) and here 2/2 = 1 so the 2 ont he top cancels out with one of the 2s on the bottom. This canceling only works for multiplication and division.

You can also do this with variables. x^3/x^7 can be rewritten as (x*x*x)/(x*x*x*x*x*x*x). here the three on top cancel out three ont he bottom leaving 1/(x*x*x*x) = 1/x^4. This also works if there are more on top. x^8/x^2 = x^6.

We do both in your problem. Let's work with one at a time. First the numbers. 32/4 = 8 Now the variables n^4/n^2 = n^2 because two of the ns int he numerator canceled the two in the denominator. Now put them together to have your answer. 4n^2 and you have to add that n=/=0

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4 years ago

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vovikov84 [41]

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