Question

If :

x = $\huge\boxed{{\bf{3.\bar{3}\:-0.\bar{9}}}}$
y = $\huge\boxed{{\bf{4.\bar{1}\:-0.\bar{5}}}}$

Find value of x² + y² - 2xy

$\Rightarrow \bf{Question \:\:for \:\:only \:\:stars \:\:and \:\:mods}$

Solve as soon as possible

Answer 1

Step-by-step:

The Value of x:

$\sf{=3+0.\overline{3}-0.\overline{9}}$

$\sf{=3+\dfrac{3}{9}-\dfrac{9}{9}}$

$\sf{=3-\dfrac{6}{9}}$

$\sf{=\dfrac{7}{3} }$

The Value of y:

$\sf{=4+0.\overline{1}-0.\overline{5}}$

$\sf{=4+\dfrac{1}{9}-\dfrac{5}{9}}$

$\sf{=4-\dfrac{4}{9}}$

$\sf{=\dfrac{32}{9} }$

$\boxed{\sf{x^2-2xy+y^2=(x-y)^2\;[Algebraic\;Identity]}}$

This means we should look for the value of $\sf{(x-y)^2}$.

The values of x, y are given.

→ $\sf{x-y=\dfrac{7\times 3}{3\times 3}-\dfrac{32}{9} }$

→ $\sf{x-y=\dfrac{21-32}{9} }$

→ $\sf{x-y=-\dfrac{11}{9} }$

→ $\sf{(x-y)^2=(-\dfrac{11}{9} )^2}$

$\sf{\therefore{(x-y)^2=\dfrac{121}{81} }}$

For more:

Finding the value without actually dividing(non-terminating)

How to find the value of $\sf{\dfrac{121}{81} }$ without actually dividing

when we know that $\sf{999999999=3^4\times 37\times 333667}$?

The fraction:

$\sf{=\dfrac{121}{81} }$

$\sf{=1+\dfrac{40}{3^4} }$

$\sf{=1+\dfrac{40\times 37\times 333667}{3^4\times 37\times 333667}}$

$\sf{=1+\dfrac{493827160}{999999999}}$

$\sf{=1+493827160\times 0.\overline{000000001}}$

$\sf{=1.\overline{493827160}}$

Therefore, its decimal value is $\sf{1.\overline{493827160}}$.

Answer 2

Answer:

yes bro the 1st is correct

Step-by-step explanation:

The Value of x:

\sf{=3+0.\overline{3}-0.\overline{9}}=3+0.3−0.9

\sf{=3+\dfrac{3}{9}-\dfrac{9}{9}}=3+93−99

\sf{=3-\dfrac{6}{9}}=3−96

\sf{=\dfrac{7}{3} }=37

The Value of y:

\sf{=4+0.\overline{1}-0.\overline{5}}=4+0.1−0.5

\sf{=4+\dfrac{1}{9}-\dfrac{5}{9}}=4+91−95

\sf{=4-\dfrac{4}{9}}=4−94

\sf{=\dfrac{32}{9} }=932

\boxed{\sf{x^2-2xy+y^2=(x-y)^2\;[Algebraic\;Identity]}}x2−2xy+y2=(x−y)2[AlgebraicIdentity]

This means we should look for the value of \sf{(x-y)^2}(x−y)2 .

The values of x, y are given.

→ \sf{x-y=\dfrac{7\times 3}{3\times 3}-\dfrac{32}{9} }x−y=3×37×3−932

→ \sf{x-y=\dfrac{21-32}{9} }x−y=921−32

→ \sf{x-y=-\dfrac{11}{9} }x−y=−911

→ \sf{(x-y)^2=(-\dfrac{11}{9} )^2}(x−y)2=(−911)2

\sf{\therefore{(x-y)^2=\dfrac{121}{81} }}∴(x−y)2=81121

For more:

Finding the value without actually dividing(non-terminating)

How to find the value of \sf{\dfrac{121}{81} }81121 without actually dividing

when we know that \sf{999999999=3^4\times 37\times 333667}999999999=34×37×333667 ?

The fraction:

\sf{=\dfrac{121}{81} }=81121

\sf{=1+\dfrac{40}{3^4} }=1+3440

\sf{=1+\dfrac{40\times 37\times 333667}{3^4\times 37\times 333667}}=1+34×37×33366740×37×333667

\sf{=1+\dfrac{493827160}{999999999}}=1+999999999493827160

\sf{=1+493827160\times 0.\overline{000000001}}=1+493827160×0.000000001

\sf{=1.\overline{493827160}}=1.493827160

Therefore, its decimal value is \sf{1.\overline{493827160}}1.493827160