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balu736 [363]
3 years ago
15

Write the recurring decimal 0.473737373... as a fraction.​

Mathematics
1 answer:
Semenov [28]3 years ago
4 0

Answer:

\frac{469}{990}

Step-by-step explanation:

We require 2 equations with the repeating digits after the decimal point.

By subtracting the 2 equations eliminates the repeating digits.

let x = 0.47373737.....

Multiply both sides by 100 and 10000, that is

100x = 47.3737... → (1)

10000x = 4737.3737.... → (2)

Subtract (1) from (2)

9900x = 4690 ( divide both sides by 9900 )

x = \frac{4690}{9900} = \frac{460}{990} ← in simplest form

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Power and chain rule (where the power rule kicks in because \sqrt x=x^{1/2}):

\left(\sqrt{\dfrac{\cos(2x)}{1+\sin(2x)}}\right)'=\dfrac1{2\sqrt{\frac{\cos(2x)}{1+\sin(2x)}}}\left(\dfrac{\cos(2x)}{1+\sin(2x)}\right)'

Simplify the leading term as

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Quotient rule:

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(\cos(2x))'=-\sin(2x)(2x)'=-2\sin(2x)

(1+\sin(2x))'=\cos(2x)(2x)'=2\cos(2x)

Put everything together and simplify:

\dfrac{\sqrt{1+\sin(2x)}}{2\sqrt{\cos(2x)}}\dfrac{(1+\sin(2x))(-2\sin(2x))-\cos(2x)(2\cos(2x))}{(1+\sin(2x))^2}

=\dfrac{\sqrt{1+\sin(2x)}}{2\sqrt{\cos(2x)}}\dfrac{-2\sin(2x)-2\sin^2(2x)-2\cos^2(2x)}{(1+\sin(2x))^2}

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=-\dfrac{\sqrt{1+\sin(2x)}}{\sqrt{\cos(2x)}}\dfrac1{1+\sin(2x)}

=-\dfrac1{\sqrt{\cos(2x)}}\dfrac1{\sqrt{1+\sin(2x)}}

=\boxed{-\dfrac1{\sqrt{\cos(2x)(1+\sin(2x))}}}

5 0
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Answer:

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Step-by-step explanation:

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