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artcher [175]
3 years ago
7

Factorize completely 81k^2 - m^2

Mathematics
2 answers:
Genrish500 [490]3 years ago
7 0

81k^2 - m^2 factors to

(9k + m)(9k - m)

Prove this by FOILing.

First - 81k^2

Outer - (-9km)

Inner - 9km

Last - (-m^2)

Put it into equation form

81k^2 - 9km + 9km - m^2

And combine terms to get the original equation,

81k^2 - m^2.

⭐ Answered by Hyperrspace (Ace) ⭐

⭐ Brainliest would be appreciated, I'm trying to reach genius! ⭐

⭐ If you have questions, leave a comment, I'm happy to help! ⭐

PolarNik [594]3 years ago
5 0

(9k+m) X (9k - m)

Step-by-step explanation:

81k^{2} - m^{2}

= (9k)^2 - (m)^2 ( taking the square roots)

=(9k+m) (9k - m)

here we have applied the rules of a^2 - b^2 = ( a+b) ( a -b)

here 9k is taking as a and m is taking as b

thus you get your answer.

hope it helps.

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3 years ago
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2 years ago
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brilliants [131]

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Step-by-step explanation:

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3 years ago
How many 6 digit different locker combinations are possible if no repeat numbers are allowed?
Alex_Xolod [135]

Since we are trying to find the number of sequences can be made <em>without repetition</em>, we are going to use a combination.


The formula for combinations is:

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Since there are 10 total digits, n = 10 in this scenario. Since we are choosing 6 digits of the 10 for our sequence, k = 6 in this scenario. Thus, we are trying to find _{10} C _6. This can be found as shown:

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7 0
3 years ago
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SashulF [63]

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I hope that's help !

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