Problem 74 - Elementary and Intermediate Algebra

Question

Use scientific notation to perform the calculations. Give all answers in scientific notation and standard notation. \(\frac{(0.48)(14,400,000)}{96,000,000}\)

Step-by-Step Solution

Answer

The result is \(7.2 \times 10^{-2}\) in scientific notation and 0.072 in standard notation.

Step 1: Express in Scientific Notation

First, express each number in scientific notation. 0.48 becomes \(4.8 \times 10^{-1}\), 14,400,000 becomes \(1.44 \times 10^7\), and 96,000,000 becomes \(9.6 \times 10^7\).

Step 2: Multiply the Numerator

Multiply the numbers in the numerator: \((4.8 \times 10^{-1}) \times (1.44 \times 10^7) = (4.8 \times 1.44) \times (10^{-1} \times 10^7)\). Calculating the multiplication, \(4.8 \times 1.44 = 6.912\). For the powers of 10, add the exponents: \(-1 + 7 = 6\). The result is \(6.912 \times 10^6\).

Step 3: Divide by the Denominator

Now, divide the result from Step 2 by the number in the denominator:\(\frac{6.912 \times 10^6}{9.6 \times 10^7} = \frac{6.912}{9.6} \times \frac{10^6}{10^7}\).Calculate the numerical fraction: \(\frac{6.912}{9.6} = 0.72\). For the powers of 10, subtract the exponents: \(6 - 7 = -1\). So the answer is \(0.72 \times 10^{-1}\).

Step 4: Express in Conventional Scientific Notation

Convert \(0.72 \times 10^{-1}\) into scientific notation and standard notation.Scientific notation: Move the decimal point one place to the right: \(7.2 \times 10^{-2}\). Standard notation: This equals 0.072.

Key Concepts

Multiplication in Scientific Notation

To multiply numbers using scientific notation, start by multiplying their coefficients, which are the numerical parts before the powers of ten. This makes the process more manageable, especially with very large or small numbers. For example, if you're multiplying \((4.8 \times 10^{-1})\) and \((1.44 \times 10^7)\), focus first on multiplying 4.8 and 1.44.

Here's a step-by-step on how to do it:

• Multiply the coefficients: \((4.8) \times (1.44) = 6.912\).
• Next, deal with the powers of ten: For \(10^{-1} \times 10^7\), simply add the exponents (because of the property of exponents \(a^m \times a^n = a^{m+n}\)) to get \(10^{6}\).

So, the result of the multiplication is \(6.912 \times 10^6\).
This method simplifies multiplication and avoids excessive use of zeros.

Division in Scientific Notation

When dividing using scientific notation, the process is just as straightforward as multiplication but involves a couple of different rules – particularly, subtracting the exponents. Say you have to divide \((6.912 \times 10^6)\) by \((9.6 \times 10^7)\). Start by dividing the coefficients.

• Calculate the division of coefficients: \(\frac{6.912}{9.6} = 0.72\).
• Then, divide the powers of ten by subtracting the exponents: \(10^6 / 10^7 = 10^{-1}\).

The division gives you \(0.72 \times 10^{-1}\).
This result is not yet in standard scientific notation, though. Adjust the coefficient to be between 1 and 10: \(7.2 \times 10^{-2}\). In standard notation, this is 0.072.
Dividing in scientific notation helps maintain clarity, especially with numbers that have many zeros, sparing you from potential errors.

Step-by-step Calculations in Algebra

Performing step-by-step calculations in algebra is about breaking down complex problems into simpler, more manageable steps. This method enhances understanding and ensures accuracy. Look at our original problem:

• Step 1: Convert numbers into scientific notation to simplify multiplication and division. For example, convert 0.48 to \(4.8 \times 10^{-1}\), ensuring each number fits into a base between 1 and 10.
• Step 2: Address the numerator first by multiplying. Whether multiplying or dividing, deal with numbers and powers of ten separately.
• Step 3: Finally, divide the result from the multiplication by the number in the denominator, again tackling the coefficients and exponents separately.
• Step 4: Convert your final result into standard scientific notation if needed, like turning \(0.72 \times 10^{-1}\) into \(7.2 \times 10^{-2}\).

Understanding and practicing these methodical steps improves problem-solving skills and boosts confidence in algebra. This approach is particularly effective since it organizes complex operations "into bite-sized chunks."