Problem 43 - Precalculus: Functions and Graphs

Question

Approximate the real-number expression. Express the answer in sclentific notation accurate to four significant figures. (a) \(\frac{1.2 \times 10^{3}}{3.1 \times 10^{2}+1.52 \times 10^{3}}\) (b) \(\left(1.23 \times 10^{-4}\right)+\sqrt{4.5 \times 10^{3}}\)

Step-by-Step Solution

Answer

(a) \(6.557 \times 10^{-1}\); (b) \(6.700 \times 10^{1}\).

Step 1: Evaluate the Denominator (a)

First, calculate the expression in the denominator: \( 3.1 \times 10^2 + 1.52 \times 10^3 \). Convert \(3.1 \times 10^2\) to \(0.31 \times 10^3\). Then add \(0.31 + 1.52 = 1.83\), so the denominator is \(1.83 \times 10^3\).

Step 2: Perform Division (a)

Now divide the numerator by the result from Step 1. The numerator is \(1.2 \times 10^3\). Divide by the denominator \(1.83 \times 10^3\): \[ \frac{1.2 \times 10^3}{1.83 \times 10^3} = \frac{1.2}{1.83} = 0.6557 \].In scientific notation, this is \(6.557 \times 10^{-1}\), rounded to four significant figures.

Step 3: Evaluate the Square Root (b)

Calculate the square root part of \( \left(1.23 \times 10^{-4}\right)+\sqrt{4.5 \times 10^{3}} \). The square root of \(4.5 \times 10^3\) is \(\sqrt{4.5} \times 10^{1.5}\). \(\sqrt{4.5} \approx 2.121\), so this becomes \(2.121 \times 10^{1.5} = 2.121 \times 31.62 \approx 67.0\).

Step 4: Add the Terms (b)

Sum \(1.23 \times 10^{-4}\) with the square root approximation. Since \(1.23 \times 10^{-4}\) is negligible compared to \(67.0\), the result is \[ 67.0 = 6.700 \times 10^{1} \] in scientific notation.

Key Concepts

Real-Number Expression

Real-number expressions are mathematical statements that consist of numbers, and they can be a combination of whole numbers, decimals, fractions, and irrational numbers. In the exercise provided, you encountered real-number expressions through both division and addition operations. When manipulating real-number expressions, particularly with scientific notation, you treat the numbers according to typical arithmetic rules but apply specific adjustments for the exponents. This allows for easier calculations and avoids handling very large or small numbers directly. Understanding real-number expressions is essential as it serves as the basis for more complex mathematical operations.

Significant Figures

Significant figures represent the numbers in a measurement that carry meaning contributing to its precision. Important for scientific and mathematical accuracy, significant figures tell you which digits are meaningful in expressing a measurement accurately. In the solution provided, numbers were rounded to four significant figures as requested in the problem.

• Start counting significant figures from the first non-zero digit.
• Include all following non-zero numbers and any zeros between them.
• Trailing zeros in a decimal point also count as significant.

For instance, the result of the initial division problem, \[6.557 \times 10^{-1}\] was expressed with four significant figures, making it precise.

Division in Scientific Notation

Conducting division using scientific notation involves dividing the coefficients and subtracting the exponents. This exercise illustrates this process succinctly. Consider the division:\[ \frac{1.2 \times 10^{3}}{1.83 \times 10^{3}} \]To perform this, follow these steps:

• Divide the numbers: \( \frac{1.2}{1.83} \approx 0.6557 \)
• Subtract the exponents: \(10^{3-3} = 10^{0} = 1\)

By combining these, you achieve the result \[0.6557 \times 10^{0}\] which simplifies to be presented in scientific notation as \[6.557 \times 10^{-1}\].Division in scientific notation streamlines complex calculations, especially with extreme numbers.

Square Roots

Square roots determine a number which, when multiplied by itself, gives the original number. In the context of this exercise, finding the square root of a value expressed in scientific notation requires special handling to account for both the decimal and the exponents. Consider the expression: \( \sqrt{4.5 \times 10^3} \). The steps are as follows:

• Take the square root of 4.5, approximately 2.121.
• For the exponent, halve it: \(10^{3} \rightarrow 10^{1.5} \).

So the complete result becomes \(2.121 \times 10^{1.5}\), which is further simplified in the solution to approximately 67.0. Square roots and other related operations with exponents necessitate attention to both the numerical and exponential components for accurate results.