Chapter 1: Problem 41
Exer. 41-42: Approximate the real-number expression to four decimal places. (a) \(\left|3.2^{2}-\sqrt{3.15}\right|\) (b) \(\sqrt{(15.6-1.5)^{2}+(4.3-5.4)^{2}}\)
Short Answer
Expert verified
(a) 8.4652; (b) 14.1426
Step by step solution
01
Calculate the power in part (a)
First, compute the expression inside the absolute value. Start with calculating \(3.2^2\). This is \(3.2 \times 3.2 = 10.24\).
02
Calculate the square root in part (a)
Next, determine the value of \(\sqrt{3.15}\). Using a calculator, \(\sqrt{3.15} \approx 1.7748\).
03
Compute absolute value in part (a)
Subtract the square root found in Step 2 from the result in Step 1: \(10.24 - 1.7748 = 8.4652\). The absolute value is \(|8.4652| = 8.4652\), since it's already positive.
04
Calculate the expression inside the square root in part (b)
For part (b), first compute \((15.6 - 1.5)^2 = (14.1)^2 = 198.81\). Then calculate \((4.3 - 5.4)^2 = (-1.1)^2 = 1.21\).
05
Add the squared differences in part (b)
Add the results from the squared differences from Step 4: \(198.81 + 1.21 = 200.02\).
06
Calculate the square root in part (b)
Find the square root of the sum from Step 5: \(\sqrt{200.02} \approx 14.1426\).
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Absolute Value Calculation
The absolute value of a number is its distance from zero on the number line, regardless of direction. In mathematical terms, the absolute value of a number \( x \) is denoted as \( |x| \). The concept is easy to grasp because it essentially captures the magnitude of a number without considering its sign.
- If \( x \) is positive, \( |x| = x \).
- If \( x \) is negative, \( |x| = -x \), which turns it positive.
Square Root Approximation
Square root approximation is a method used to estimate the square root of a number that isn't a perfect square. When we can't find an exact whole number square root, we approximate it using a calculator or estimation techniques. This process is invaluable for solving various mathematical problems.
When you approximate the square root using a calculator, you input the number and receive a decimal value. For example, in the exercise, \( \sqrt{3.15} \approx 1.7748 \). This estimation :
When you approximate the square root using a calculator, you input the number and receive a decimal value. For example, in the exercise, \( \sqrt{3.15} \approx 1.7748 \). This estimation :
- helps in simplifying further calculations, and
- is used extensively in statistical and algebraic reasoning.
Power Calculation
Power calculation involves determining the result of multiplying a number by itself a certain number of times. This is a fundamental operation in mathematics. It is denoted as \( x^n \), where \( x \) is the base, and \( n \) is the exponent, indicating how many times \( x \) is used as a factor.
In the exercise, \( 3.2^2 \) means multiplying 3.2 by itself, which calculates to 10.24. Here is a simple breakdown:
In the exercise, \( 3.2^2 \) means multiplying 3.2 by itself, which calculates to 10.24. Here is a simple breakdown:
- Write down the base number twice.
- Multiply them together: \( 3.2 \times 3.2 = 10.24 \).
Distance Formula
The distance formula is used in mathematics to find the straight-line distance between two points in a plane. This formula is derived from the Pythagorean theorem and typically used in a Cartesian coordinate system.To apply the formula:
- Identify and define your points as \((x_1, y_1)\) and \((x_2, y_2)\).
- Plug these into the formula: \( \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \).