Question

74
The value of (0.666 ... ... ... ... . . )² is​

Answer 1

let's first convert 0.6666... into p/q form..

let x=0.6666....-->equation 1

as periodicity is 1 , multiply by 10^1=10

10x=6.6666...-->equation 2

subtract equation 2-equation 1

10x=6.666...

(-) x=0.666...

__________

9x=6

__________

x=6/9

x=2/3

so

(2/3)^2

=4/9 (or) 0.444....

Answer 2

To find:

The value of $(0.666\:...\:...)^{2}$

Step-by-step explanation:

First let us express $0.666\:...\:...$ as a fraction.

Let, $x=0.666\:...\:...$

Then $10x=6.666\:...\:...$

Now $10x-x=6.666\:...\:...-0.666\:...\:...$

$\Rightarrow 9x=6$

$\Rightarrow x=\dfrac{2}{3}$

Now, $x^{2}=(\dfrac{2}{3})^{2}$

$\Rightarrow x^{2}=\dfrac{4}{9}$

$\Rightarrow (0.666\:...\:...)^{2}=0.444\:...\:...$

Final answer:

$\boxed{(0.666\:...\:...)^{2}=0.444\:...\:...}$

Read more on Brainly.in

How to convert 146 days into years?

- https://brainly.in/question/8199675

Product of two given proper fractions is always options ...

- https://brainly.in/question/46782322