Question

Using the identity 2nd....find. (4.9)square

Answer 1

$\underline { \blue {By \: Algebraic \:Identity }}$

$\boxed { \pink { (x-y)^{2} = x^{2} - 2xy + y^{2} }}$

$Now , (4.9)^{2} \\= ( 5 - 0.1 )^{2}$

$Here, x = 5 ,\: y = 0.1$

$= 5^{2} - 2\times 5 \times 0.1 + (0.1)^{2} \\= 25 - 1 + 0.01 \\= 24+0.01 \\= 24.01$

Therefore.,

$\red { Value \: of \: (4.9)^{2} }\green {= 24.01}$

•••♪

Answer 2

ByAlgebraicIdentity

\boxed { \pink { (x-y)^{2} = x^{2} - 2xy + y^{2} }}

(x−y)

2

=x

2

−2xy+y

2

\begin{gathered} Now , (4.9)^{2} \\= ( 5 - 0.1 )^{2}\end{gathered}

Now,(4.9)

2

=(5−0.1)

2

Here, x = 5 ,\: y = 0.1Here,x=5,y=0.1

\begin{gathered} = 5^{2} - 2\times 5 \times 0.1 + (0.1)^{2} \\= 25 - 1 + 0.01 \\= 24+0.01 \\= 24.01 \end{gathered}

=5

2

−2×5×0.1+(0.1)

2

=25−1+0.01

=24+0.01

=24.01

Therefore.,

\red { Value \: of \: (4.9)^{2} }\green {= 24.01}Valueof(4.9)

2

=24.01

•••♪

Hope it helps u friend