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Chapter 5: Problem 57

Multiply. $$ (a w+0.1)(-a w+0.1) $$

Short Answer

Expert verified
a^2 w^2 - 0.01

Step by step solution

01

- Identify the pattern

Notice that the given expression (a w + 0.1)(-a w + 0.1) resembles the difference of squares formula. The formula is (a + b)(a - b) = a^2 - b^2.
02

- Apply the difference of squares

In the expression (a w + 0.1)(-a w + 0.1), identify the terms: For this expression, 'a' is equivalent to 'aw' and 'b' is equivalent to '0.1'.
03

- Plug into the formula

Substitute 'aw' for 'a' and '0.1' for 'b' in the formula (a + b)(a - b) = a^2 - b^2: (aw)^2 - (0.1)^2.
04

- Simplify

Simplify the expression: (aw)^2 = a^2 w^2 and (0.1)^2 = 0.01.So, a^2 w^2 - 0.01.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Algebra
Algebra is a branch of mathematics that uses symbols and letters to represent numbers and quantities in formulas and equations. It is fundamental for developing problem-solving skills.

Here are the building blocks of algebra:
  • **Variables**: Letters such as 'a', 'w', and 'x' used to represent numbers.
  • **Constants**: Numbers on their own, like 0.1 in the exercise.
  • **Expressions**: Combinations of variables and constants, such as 'aw + 0.1'.

These components come together to form equations and inequalities, which are used to solve various mathematical problems.
Exploring Polynomials
Polynomials are algebraic expressions that consist of variables and coefficients. They are constructed using addition, subtraction, multiplication, and non-negative integer exponents of variables.

Key elements of polynomials include:
  • **Terms**: Each part of the polynomial separated by plus or minus signs (e.g., 'aw' and '0.1').
  • **Coefficients**: Numbers in front of variables (e.g., 'a' in 'aw').
  • **Degree**: The highest power of the variable in the polynomial.

For instance, the polynomial 'aw + 0.1' has two terms: 'aw' and '0.1', where 'aw' has a coefficient 'a' and 'w' is the variable.
Expression Simplification using Difference of Squares
Expression simplification makes complex mathematical expressions more manageable by breaking them down into simpler forms.

The difference of squares is a useful technique in this process. It states that: \(a^2 - b^2 = (a + b)(a - b)\).

Steps to use the difference of squares:
  • **Identify**: Recognize structures in the form \( (a + b)(a - b) \).
  • **Substitute**: For the given expression \( (aw + 0.1)(-aw + 0.1)\), identify 'a' as 'aw' and 'b' as '0.1'.
  • **Simplify**: Apply the formula \( a^2 - b^2 \). This gives \( (aw)^2 - (0.1)^2 = a^2w^2 - 0.01 \).

This method simplifies the multiplication problem into a straightforward subtraction, showing the power of algebraic techniques.

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Most popular questions from this chapter

The average number of accidents per day involving drivers of age \(r\) can be approximated by the polynomial $$0.4 r^{2}-40 r+1039$$ For what age is the number of daily accidents smallest?

Computer spreadsheet applications allow values for cells in a spreadsheet to be calculated from values in other cells. For example, if the cell Cl contains the formula $$ =\mathrm{A} 1+2 * \mathrm{B} 1 $$ the value in Cl will be the sum of the value in Al and twice the value in B1. This formula is a polynomial in the two variables Al and B1. The cell D6 contains the formula $$ =\mathrm{Al}-0.2 * \mathrm{B} 1+0.3^{*} \mathrm{Cl} $$ What is the value in \(\mathrm{D} 6\) if the value in \(\mathrm{Al}\) is \(10,\) the value in \(\mathrm{B} 1\) is \(-3,\) and the value in \(\mathrm{Cl}\) is \(30 ?\)

For each pair of functions fand \(g\), determine the domain of the sum, the difference, and the product of the two functions. $$ \begin{aligned} &f(x)=9-x^{2}\\\ &g(x)=\frac{3}{x-6}+2 x \end{aligned} $$

Evaluate. $$ 4+x^{3}, \text { for } x=10 $$

Simplify. \(\frac{9}{2} x^{8}+\frac{1}{9} x^{2}+\frac{1}{2} x^{9}+\frac{9}{2} x+\frac{9}{2} x^{9}+\frac{8}{9} x^{2}+\frac{1}{2} x-\frac{1}{2} x^{8}\)
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