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Natasha2012 [34]
3 years ago
14

M. Score: 0 of 1 pt

Mathematics
1 answer:
beks73 [17]3 years ago
5 0

Answer:

  3×5×53

Step-by-step explanation:

You can use divisibility rules to find the small prime factors.

The number ends in 5, so is divisible by 5.

  795/5 = 159

The sum of digits is 1+5+9 = 15; 1+5 = 6, a number divisible by 3, so 3 is a factor.

  159/3 = 53 . . . . . a prime number,* so we're done.

795 = 3×5×53

_____

* If this were not prime, it would be divisible by a prime less than its square root. √53 ≈ 7.3. We know it is not divisible by 2, 3, or 5. We also know the closest multiples of 7 are 49 and 56, so it is not divisible by 7. Hence 53 is prime.

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Step-by-step explanation:

Writing the description in algebraic translation

\frac{b^{-2}}{ab^{-3}}

so we have to find the expression which will be equal to \frac{b^{-2}}{ab^{-3}}.

Considering the expression

\frac{b^{-2}}{ab^{-3}}

\mathrm{Apply\:exponent\:rule}:\quad \frac{x^a}{x^b}=x^{a-b}

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so the expression becomes

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Therefore,

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