- Analyze the expression as a division of coefficients and exponents: \(\frac{6.4 \times 10^{12}}{2.5 \times 10^4} = (\frac{6.4}{2.5}) \times 10^{12-4}\).
- Divide the coefficients: \(\frac{6.4}{2.5} = 2.56\).
- Simplify the exponents: \(10^{12} / 10^4 = 10^8\).
- Combine to find the final result: \(\boxed{2.56 \times 10^8}\).
Explanation
- Problem Analysis and Method
The goal is to calculate the quotient of two numbers given in scientific notation: \(6.4 \times 10^{12}\) and \(2.5 \times 10^4\). The division of numbers in scientific notation follows the rule:
\[\frac{a \times 10^n}{b \times 10^m} = \left(\frac{a}{b}\right) \times 10^{n-m}\]
where \(a\) and \(b\) are the coefficients, and \(n\) and \(m\) are the exponents of 10. Our task is to perform this operation and select the correct option from the provided list.
- Dividing the Coefficients
First, we divide the coefficients, which are 6.4 and 2.5:
\[\frac{6.4}{2.5} = 2.56\]
This value represents the coefficient of the final result.
- Calculating the Powers of 10
Next, we handle the exponents of 10. According to the rules of exponents for division, we subtract the exponent of the divisor from the exponent of the dividend:
\[\frac{10^{12}}{10^4} = 10^{12-4} = 10^8\]
This gives us the power of 10 for our final expression.
- Conclusion and Matching Option
Finally, combining the results from the previous steps, we get:
\[2.56 \times 10^8\]
Comparing this result to the given options:
- \(2.56 \times 10^{-8}\)
- \(3.9 \times 10^8\)
- \(2.56 \times 10^{-3}\)
- \(2.56 \times 10^3\)
- \(3.9 \times 10^{-8}\)
- \(2.56 \times 10^8\)
Our result matches the last option. Therefore, the correct answer is \(\boxed{2.56 \times 10^8}\).
Examples
Scientific notation is commonly used in fields like astronomy and physics to manage extremely large or small numbers. For example, if you know the total number of stars in a galaxy is roughly \(6.4 \times 10^{12}\) and you want to estimate how many stars each of the \(2.5 \times 10^4\) sectors in that galaxy contains, you would divide these two values. Mastering these calculations helps in simplifying complex scientific data into manageable units.