Question

Perform the following calculations. Round the answers. Challenge \(\left(1.32 \times 10^{3} \mathrm{g}\right) \div\left(2.5 \times 10^{2} \mathrm{cm}^{3}\right)\)

Short answer

The result of the calculation is approximately \(5.28\).

Step-by-step solution

Divide coefficients

Begin by dividing the coefficients 1.32 and 2.5: \(\dfrac{1.32}{2.5} = 0.528\)

Subtract exponents

Next, we will subtract the exponents: \((10^{3})/(10^{2}) = 10^{(3-2)} = 10^{1}\)

Combine the results

Now, we will multiply the result from Step 1 with the result from Step 2: \(0.528 \times 10^{1} = 5.28\)

Round the answer

Finally, we will round our answer to a reasonable number of decimal places (in this case, two decimal places): \(5.28\) The result of the calculation is approximately \(5.28\).

Key concepts

Coefficients

In scientific notation, we often deal with numbers expressed in the form of a coefficient and an exponent connected to a base of 10. The coefficient is a number, often between 1 and 10, which is the more detailed part of the number. For instance, in the scientific notation \(1.32 \times 10^3\), the coefficient is 1.32. Coefficients are the real, manageable pieces we work with in calculations.

• When performing operations like multiplication or division in scientific notation, it's crucial first to handle the coefficients as you would in any basic arithmetic operation.
• In our example exercise, the original coefficients are 1.32 and 2.5, which we divide as the first step \(\frac{1.32}{2.5} = 0.528\).
• This division extracts the core numerical relationship of the two values at hand, excluding the influence of the powers of 10.

Understanding coefficients in scientific notation helps simplify complex calculations by removing the complexity of large or very small numbers.

Exponents

Exponents in scientific notation indicate how many times the base, which is often 10, is multiplied by itself. In a number like \(10^3\), the exponent is 3, telling us that 10 is used as a factor three times: \(10 \times 10 \times 10 = 1000\).

• When dividing numbers in scientific notation, you subtract the exponent of the denominator from the exponent of the numerator.
• For example, in the operation \((10^3)/(10^2)\), you subtract the exponents: \(3 - 2 = 1\), resulting in \(10^1\).
• This subtraction helps in effectively scaling the numbers, allowing the simplification of complex calculations.

Recognizing the role of exponents is critical because they determine the scale and size of the number in scientific notation, simplifying large numerical operations significantly.

Rounding

Rounding is an important step when you're dealing with numbers that result from complex calculations, especially in scientific contexts where precision is key but practicality in reporting numbers is essential. The basic idea is to limit the number of decimal places to make a number easier to use or interpret.

• In our example, the number 5.28 is the result after rounding from 5.280, where we opted for two decimal places.
• Rounding helps avoid small errors in measurements from impacting the overall results significantly.
• Many times, the decision about how many decimal places to keep depends on the context of the problem or standards within a particular field.

Understanding when and how to round numbers ensures the data remains accessible, without sacrificing necessary precision.

Division of Numbers in Scientific Notation

Dividing numbers in scientific notation often involves both the coefficients and the exponents, simplifying large numerical calculations containing powers of ten.

• The division process involves two primary steps: first, divide the coefficients; second, subtract the exponents.
• From our example, we divided the coefficients \(1.32\) by \(2.5\) to get \(0.528\), and adjusted the exponents: \(10^3\) divided by \(10^2\) gives us \(10^1\).
• The result, \(0.528 \times 10^1\), turns into a more manageable form through further simplification or rounding.

Mastering this division process allows us to handle operations involving very large or small numbers more easily, providing clearer insights and making complex calculations feasible.