Answer

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Hint: Keep a decimal point into one digit and convert others into terms of $10$.

Now in question it is asked to convert the number in standard form,

It is difficult to read numbers like$1202000004240$ or $0.0000042852512$. To make it easy

to read very large and small numbers, we write them in standard form.

So first of all what is standard form,

Standard form is a way of writing down very large or very small numbers

easily.${{10}^{3}}=1000$, so $4\times {{10}^{3}}=4000$ . So $4000$ can be written as$4\times {{10}^{3}}$ . This idea can be used to write even larger numbers down easily in

standard form. Any number that we can write as a decimal number, between $1.0$ and

$10.0$, multiplied by a power of $10$, is said to be in standard form.

Standard form in Math is mentioned for basically decimal numbers, equations, polynomials,

linear equations, etc. Its correct definition could be explained better in terms of decimal

numbers and following certain rules.

As we know, the decimal numbers are the simplified form of fractions. Some fractions give

decimal numbers which have numbers after decimal in thousandths, hundredths or tenths

place. But there are some fractions, which gives a big decimal number.

To represent such big numbers, we use such simpler forms, which is also stated as Scientific

notation.

Let us take an example $43333.21$

So we can write $43333.21=4.333321\times {{10}^{4}}$……..(1)

In the (1) we can see it, So $4.333321\times {{10}^{4}}$is standard form of $43333.21$.

So in this way we can write the digits in standard form.

So generally Standard form is a way of writing down very large or very small numbers easily.

Now for us given number is $3908.78$ ,

So we have to convert $3908.78$ in standard form,

Multiplying and dividing by 100 to $3908.78$, We get

$=\dfrac{3908.78\times 100}{100}=\dfrac{390878}{100}$

Now simplifying it We get

$=\dfrac{3.90878\times {{10}^{5}}}{100}=\dfrac{3.90878\times {{10}^{5}}}{{{10}^{2}}}$…… (2)

Now we know the property that $\dfrac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}}$,

So using the above property in (2) We get,

$\begin{align}

& =\dfrac{3.90878\times {{10}^{5}}}{{{10}^{2}}}=3.90878\times \dfrac{{{10}^{5}}}{{{10}^{2}}}=3.90878\times {{10}^{5-2}} \\

& =3.90878\times {{10}^{3}} \\

\end{align}$

So $3.90878\times {{10}^{3}}$ is standard form of $3908.78$.

So we have got the standard form of $3908.78$ which is $3.90878\times {{10}^{3}}$.

Hence proved.

$3.90878\times {{10}^{3}}$ is standard form of $3908.78$ is the answer.

Note: So basically don’t jumble while converting the number. Such as $3908.78$ to $3.90878\times {{10}^{3}}$. Use the property accordingly Such as I had used this property

$\dfrac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}}$. Remember that the properties are used according

to the problem.

Now in question it is asked to convert the number in standard form,

It is difficult to read numbers like$1202000004240$ or $0.0000042852512$. To make it easy

to read very large and small numbers, we write them in standard form.

So first of all what is standard form,

Standard form is a way of writing down very large or very small numbers

easily.${{10}^{3}}=1000$, so $4\times {{10}^{3}}=4000$ . So $4000$ can be written as$4\times {{10}^{3}}$ . This idea can be used to write even larger numbers down easily in

standard form. Any number that we can write as a decimal number, between $1.0$ and

$10.0$, multiplied by a power of $10$, is said to be in standard form.

Standard form in Math is mentioned for basically decimal numbers, equations, polynomials,

linear equations, etc. Its correct definition could be explained better in terms of decimal

numbers and following certain rules.

As we know, the decimal numbers are the simplified form of fractions. Some fractions give

decimal numbers which have numbers after decimal in thousandths, hundredths or tenths

place. But there are some fractions, which gives a big decimal number.

To represent such big numbers, we use such simpler forms, which is also stated as Scientific

notation.

Let us take an example $43333.21$

So we can write $43333.21=4.333321\times {{10}^{4}}$……..(1)

In the (1) we can see it, So $4.333321\times {{10}^{4}}$is standard form of $43333.21$.

So in this way we can write the digits in standard form.

So generally Standard form is a way of writing down very large or very small numbers easily.

Now for us given number is $3908.78$ ,

So we have to convert $3908.78$ in standard form,

Multiplying and dividing by 100 to $3908.78$, We get

$=\dfrac{3908.78\times 100}{100}=\dfrac{390878}{100}$

Now simplifying it We get

$=\dfrac{3.90878\times {{10}^{5}}}{100}=\dfrac{3.90878\times {{10}^{5}}}{{{10}^{2}}}$…… (2)

Now we know the property that $\dfrac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}}$,

So using the above property in (2) We get,

$\begin{align}

& =\dfrac{3.90878\times {{10}^{5}}}{{{10}^{2}}}=3.90878\times \dfrac{{{10}^{5}}}{{{10}^{2}}}=3.90878\times {{10}^{5-2}} \\

& =3.90878\times {{10}^{3}} \\

\end{align}$

So $3.90878\times {{10}^{3}}$ is standard form of $3908.78$.

So we have got the standard form of $3908.78$ which is $3.90878\times {{10}^{3}}$.

Hence proved.

$3.90878\times {{10}^{3}}$ is standard form of $3908.78$ is the answer.

Note: So basically don’t jumble while converting the number. Such as $3908.78$ to $3.90878\times {{10}^{3}}$. Use the property accordingly Such as I had used this property

$\dfrac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}}$. Remember that the properties are used according

to the problem.

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