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Wittaler [7]
3 years ago
11

Write the recurring decimal 0.5 as a fraction in its simplest form.​

Mathematics
2 answers:
sergejj [24]3 years ago
8 0

Answer:

0.5×2/2=1/2 is the required fraction.

emmasim [6.3K]3 years ago
3 0

Answer:

Step-by-step explanation:

Let x = 0.555...

10x = 5.555...

10x-x = 5.555... - 0.555...

9x = 5

x = 5/9

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We split [2, 4] into n subintervals of length \dfrac{4-2}n=\dfrac2n,

[2,4]=\left[2,2+\dfrac2n\right]\cup\left[2+\dfrac2n,2+\dfrac4n\right]\cup\left[2+\dfrac4n,2+\dfrac6n\right]\cup\cdots\cup\left[2+\dfrac{2(n-1)}n,4\right]

so that the right endpoints are given by the sequence

x_i=2+\dfrac{2i}n=\dfrac{2(n+i)}n

for 1\le i\le n. Then the Riemann sum approximating

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is

\displaystyle\sum_{i=1}^nf(x_i)\dfrac{4-2}n=\frac8{n^2}\sum_{i=1}^n(n+i)=\frac8{n^2}\left(n^2+\frac{n(n+1)}2\right)=\frac{12n+4}n

The integral is given exactly as n\to\infty, for which we get

\displaystyle\int_2^42x\,\mathrm dx=\lim_{n\to\infty}\frac{12n+4}n=12

To check: we have

\displaystyle\int_2^42x\,\mathrm dx=x^2\bigg|_2^4=4^2-2^2=16-4=12

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