Answer

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Hint: First of all let 1.1 m = a and 0.4 = b. Now to get the value of $\left( 1.1m-0.4 \right)\left( 1.1m+0.4 \right)$, use the identity $\left( a-b \right)\left( a+b \right)={{a}^{2}}-{{b}^{2}}$. Put ${{a}^{2}}={{\left( 1.1m \right)}^{2}}\ and\ {{b}^{2}}={{\left( 0.4 \right)}^{2}}$

Complete step-by-step answer:

Here we have to solve $\left( 1.1m-0.4 \right)\left( 1.1m+0.4 \right)$

First of all let us assume the given expression to be,

\[E=\left( 1.1m-0.4 \right)\left( 1.1m+0.4 \right)\]

Let us take 1.1 m to be ‘a’.

Also, let us take 0.4 to be ‘b’.

By putting the assumed values of 1.1 m and 0.4, we get,

E = (a –b) (a + b)

By solving above expression, we get,

E = a (a +b) – b (a + b)

By further solving the above expression, we get,

$E={{a}^{2}}+ab-ba-{{b}^{2}}$

As we know that ab = ba, therefore by applying this in above expression, we get,

$E={{a}^{2}}+ab-ab-{{b}^{2}}$

We know that x – x = 0. By using this in above expression we get,

\[E={{a}^{2}}-{{b}^{2}}\]

Hence we get, \[E=\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}\]

Therefore, we have got an identity that is

$\left( a+b \right)\left( a-b \right)=\left( {{a}^{2}}-{{b}^{2}} \right)$

As we have assumed that a = 1.1 m and b = 0.4,

By putting values in above identity, we get,

$\left( 0.1m+0.4 \right)\left( 0.1m-0.4 \right)={{\left( 0.1m \right)}^{2}}-{{\left( 0.4 \right)}^{2}}$

As we know that ${{\left( a.b \right)}^{n}}={{a}^{n}}.{{b}^{n}}$. By applying this in RHS of above equation, we get,

$\begin{align}

& \left( 0.1m+0.4 \right)\left( 0.1m-0.4 \right)={{\left( 0.1m \right)}^{2}}-{{\left( 0.4 \right)}^{2}} \\

& =\left( 0.01 \right){{m}^{2}}-\left( 0.16 \right) \\

\end{align}$

Hence, we get $\left( 0.1m+0.4 \right)\left( 0.1m-0.4 \right)=\left( 0.01{{m}^{2}} \right)-\left( 0.16 \right)$

By using the suitable identity that is $\left( a-b \right)\left( a+b \right)={{a}^{2}}-{{b}^{2}}$

Note: Students must remember this identity because this identity is very useful in mathematics. Also students should take special care while multiplying, squaring etc. terms which contain decimal.

For example, students often make this mistake of writing ${{\left( 0.4 \right)}^{2}}=1.6$ which is wrong. Actually${{\left( 0.4 \right)}^{2}}=\left( 0.16 \right)$. So this mistake must be avoided.

Complete step-by-step answer:

Here we have to solve $\left( 1.1m-0.4 \right)\left( 1.1m+0.4 \right)$

First of all let us assume the given expression to be,

\[E=\left( 1.1m-0.4 \right)\left( 1.1m+0.4 \right)\]

Let us take 1.1 m to be ‘a’.

Also, let us take 0.4 to be ‘b’.

By putting the assumed values of 1.1 m and 0.4, we get,

E = (a –b) (a + b)

By solving above expression, we get,

E = a (a +b) – b (a + b)

By further solving the above expression, we get,

$E={{a}^{2}}+ab-ba-{{b}^{2}}$

As we know that ab = ba, therefore by applying this in above expression, we get,

$E={{a}^{2}}+ab-ab-{{b}^{2}}$

We know that x – x = 0. By using this in above expression we get,

\[E={{a}^{2}}-{{b}^{2}}\]

Hence we get, \[E=\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}\]

Therefore, we have got an identity that is

$\left( a+b \right)\left( a-b \right)=\left( {{a}^{2}}-{{b}^{2}} \right)$

As we have assumed that a = 1.1 m and b = 0.4,

By putting values in above identity, we get,

$\left( 0.1m+0.4 \right)\left( 0.1m-0.4 \right)={{\left( 0.1m \right)}^{2}}-{{\left( 0.4 \right)}^{2}}$

As we know that ${{\left( a.b \right)}^{n}}={{a}^{n}}.{{b}^{n}}$. By applying this in RHS of above equation, we get,

$\begin{align}

& \left( 0.1m+0.4 \right)\left( 0.1m-0.4 \right)={{\left( 0.1m \right)}^{2}}-{{\left( 0.4 \right)}^{2}} \\

& =\left( 0.01 \right){{m}^{2}}-\left( 0.16 \right) \\

\end{align}$

Hence, we get $\left( 0.1m+0.4 \right)\left( 0.1m-0.4 \right)=\left( 0.01{{m}^{2}} \right)-\left( 0.16 \right)$

By using the suitable identity that is $\left( a-b \right)\left( a+b \right)={{a}^{2}}-{{b}^{2}}$

Note: Students must remember this identity because this identity is very useful in mathematics. Also students should take special care while multiplying, squaring etc. terms which contain decimal.

For example, students often make this mistake of writing ${{\left( 0.4 \right)}^{2}}=1.6$ which is wrong. Actually${{\left( 0.4 \right)}^{2}}=\left( 0.16 \right)$. So this mistake must be avoided.

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