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lana [24]
3 years ago
9

Which expression is equivalent to StartFraction 2 a + 1 Over 10 a minus 5 Endfraction divided by StartFraction 10 a Over 4 a squ

ared minus 1 EndFraction?
Mathematics
1 answer:
const2013 [10]3 years ago
4 0

Answer:

\frac{(2a + 1)^2}{50a}

Step-by-step explanation:

Given

\frac{2a + 1}{10a - 5} / \frac{10a}{4a^2 - 1}

Required

Find the equivalent

We start by changing the / to *

\frac{2a + 1}{10a - 5} / \frac{10a}{4a^2 - 1}

\frac{2a + 1}{10a - 5} * \frac{4a^2 - 1}{10a}

Factorize 10a - 5

\frac{2a + 1}{5(2a - 1)} * \frac{4a^2 - 1}{10a}

Expand 4a² - 1

\frac{2a + 1}{5(2a - 1)} * \frac{(2a)^2 - 1}{10a}

\frac{2a + 1}{5(2a - 1)} * \frac{(2a)^2 - 1^2}{10a}

Express (2a)² - 1² as a difference of two squares

Difference of two squares is such that: a^2- b^2= (a+b)(a-b)

The expression becomes

\frac{2a + 1}{5(2a - 1)} * \frac{(2a - 1)(2a + 1)}{10a}

Combine both fractions to form a single fraction

\frac{(2a + 1)(2a - 1)(2a + 1)}{5(2a - 1)10a}

Divide the numerator and denominator by 2a - 1

\frac{(2a + 1)((2a + 1)}{5*10a}

Simplify the numerator

\frac{(2a + 1)^2}{5*10a}

\frac{(2a + 1)^2}{50a}

Hence,

\frac{2a + 1}{10a - 5} / \frac{10a}{4a^2 - 1} = \frac{(2a + 1)^2}{50a}

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stich3 [128]
So we can start with the full of possibilities and eliminate them one by one.

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Now we know that any prime greater than 2 is odd as otherwise it would have 2 as a factor, so we can eliminate all of these digits that would be an even number, leaving:

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We know that 11,13,17 and 19 are all primes, so we cannot eliminate any more of these, leaving the set:

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Answer:

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Step-by-step explanation:

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