How do you simplify ${(0.1x + 0.4y)^2}$?

Hint: To simplify ${(0.1x + 0.4y)^2}$, we first convert the coefficients of $x$ and $y$. These coefficients are decimal numbers and are to be converted into its simplest form of fractions. Then, we can apply the formula ${(a + b)^2} = {a^2} + {b^2} + 2ab$ to evaluate the value of ${(0.1x + 0.4y)^2}$.

Complete step-by-step solution:
We first convert the coefficients of $x$ and $y$ into fractions.
We have the coefficient of $x$ as $0.1$ and the coefficient of $y$ as $0.4$.
To convert $0.1$ into fractions, we write it as the decimal number is written in the numerator and the denominator is $1$. Then, we multiply both the numerator and the denominator by $10$.
$\dfrac{{0.1}}{1} = \dfrac{{0.1}}{1} \times \dfrac{{10}}{{10}} = \dfrac{1}{{10}}$
So, $\dfrac{1}{{10}}$ is the fractional form of $0.1$
Similarly, $0.4$ can be written as
$\dfrac{{0.4}}{1} = \dfrac{{0.4}}{1} \times \dfrac{{10}}{{10}} = \dfrac{4}{{10}}$
So, $\dfrac{4}{{10}}$ is the fractional form of $0.4$
Therefore, ${(0.1x + 0.4y)^2}$can be written as
$ = {\left( {\dfrac{x}{{10}} + \dfrac{{4y}}{{10}}} \right)^2}$
Here, we use the formula ${(a + b)^2} = {a^2} + {b^2} + 2ab$ where $a = \dfrac{1}{{10}}$ and $b = \dfrac{4}{{10}}$
$ = {\left( {\dfrac{x}{{10}}} \right)^2} + {\left( {\dfrac{{4y}}{{10}}} \right)^2} + 2 \times \dfrac{x}{{10}} \times \dfrac{{4y}}{{10}}$
Now, we know that ${\left( {\dfrac{m}{n}} \right)^2} = \dfrac{{{m^2}}}{{{n^2}}}$ . Applying the same to the above equation, we get:
\[ = \left( {\dfrac{{{x^2}}}{{{{10}^2}}}} \right) + \left( {\dfrac{{{{\left( {4y} \right)}^2}}}{{{{10}^2}}}} \right) + 2 \times \dfrac{x}{{10}} \times \dfrac{{4y}}{{10}}\]
We also know that ${(mn)^2} = {m^2}{n^2}$. Applying the same to the above equation, we get:
\[
   = \left( {\dfrac{{{x^2}}}{{100}}} \right) + \left( {\dfrac{{{4^2}{y^2}}}{{100}}} \right) + \dfrac{{8xy}}{{100}} \\
   = \dfrac{{{x^2}}}{{100}} + \dfrac{{16{y^2}}}{{100}} + \dfrac{{8xy}}{{100}} \\
 \]
This can be further simplified to
\[ = \dfrac{{{x^2}}}{{100}} + \dfrac{{4{y^2}}}{{25}} + \dfrac{{2xy}}{{25}}\].
This can be further simplified to
Hence, ${(0.1x + 0.4y)^2} = \dfrac{{{x^2}}}{{100}} + \dfrac{{4{y^2}}}{{25}} + \dfrac{{2xy}}{{25}}$

Note: While converting a decimal number into a fraction, we calculate the number of digits after the decimal point and we multiply the numerator and the denominator with \[1\] with that many zeros in it. For example, for the decimal number $0.85$to be converted into a fraction, the numerator $0.85$ and the denominator $1$ will be multiplied with $100$. The formula of ${(a + b)^2} = {a^2} + {b^2} + 2ab$ should be carefully used when the terms are in fractions or when there in variables in fractions.