What are the prime factors of 96prime factors of 144

1 answer:

5 0

96 prime factors (2•2•2•2•2•3)

144 prime factors (2•2•2•2•3•3)

Actual factor of 96 (1,2,34,6,8,12,16,24,32,48,96)

Actual factors of 144 (1,2,3,4,6,8,9,12,16,18,24,36,48,72,144)

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How many arrangements can be made using 2 letters of the word HYPERBOLAS if no letter is to be used more than once?

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Answer:

90 arrangements

Step-by-step explanation:

Since there are no repititions of letters, there are unique 10 letters in total.

THe number of arrangements would be 2 permutation 10. We need the formula for permutation. That is:

$nPr=\frac{n!}{(n-r)!}\\$

Now, n = 10 [total] and r is 2, so we have:

$nPr=\frac{n!}{(n-r)!}\\\\_{10}P_{2}=\frac{10!}{(10-2)!}\\=\frac{10!}{8!}\\=\frac{10*9*8!}{8!}\\=10*9\\=90$

So, there can be 90 arrangements

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What are the prime factors of 96prime factors of 144 ​

What are the prime factors of 96prime factors of 144