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Reil [10]
3 years ago
9

A circle P is circumscribed about a regular hexagon ABCDEF

Mathematics
1 answer:
DanielleElmas [232]3 years ago
5 0

Answer:the answer will be D

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Which expression is equivalent to StartFraction 2 a + 1 Over 10 a minus 5 Endfraction divided by StartFraction 10 a Over 4 a squ
const2013 [10]

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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Answer:

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Step-by-step explanation:

4log(216)=12x

Divide both sides by 12

log(216)/3 = x

Use properties of logs

log(216^(1/3)) = x

Simplify

log(6) = x

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Answer:

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Step-by-step explanation:

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G(x) = 6x + 1

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