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Chapter 1: Problem 14

Use the distributive property to rewrite each expression without using parentheses. 2\(a\left(0.5 z+z^{2}\right)\)

Short Answer

Expert verified
\(az + 2az^2\)

Step by step solution

01

Apply the distributive property

The distributive property states that for any numbers or variables a, b, and c, the equation \( a(b + c) = ab + ac \) holds. Apply this property to the expression given: \(2a(0.5z + z^2)\).
02

Distribute the first term

Multiply \(2a\) by the first term inside the parentheses (\(0.5z\)). This gives: \(2a \times 0.5z = a \times z\).
03

Distribute the second term

Multiply \(2a\) by the second term inside the parentheses (\(z^2\)). This gives: \(2a \times z^2 = 2az^2\).
04

Combine the results

Now, combine the results from Steps 2 and 3 to get the final expression: \(az + 2az^2\).

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Key Concepts

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

Algebraic Expressions
Algebraic expressions are a combination of numbers, variables, and mathematical operations like addition, subtraction, multiplication, and division. In the problem given, the expression is presented as:
2a(0.5z+z^{2}).
An algebraic expression can appear complex, but breaking it down into simpler steps helps to better understand and work with it. The variables here are 'a' and 'z,' and we use the distributive property to simplify the expression further. Whenever you see an expression with parentheses, it often signals that you may need to use the distributive property, especially if you want to eliminate those parentheses.
Mathematical Operations
Mathematical operations include basic processes like addition, subtraction, multiplication, and division. For the expression 2a(0.5z+z^{2}), we primarily focus on multiplication.
Here's a quick refresher on how multiplication is handled here:
  • First, you multiply 2a by each term inside the parentheses individually.
  • Then, you combine the results from each multiplication step.

This simple approach allows you to systematically approach each operation. For instance, multiplying 2a by 0.5z gives us az. Following the same rule, multiplying 2a by z^{2} gives us 2az^{2}. Combining the results of these operations results in the simplified expression az + 2az^{2}.
Distributive Property
The distributive property is a fundamental concept in algebra that simplifies expressions and solves equations. The property is defined as:
. a(b + c) = ab + ac. In the problem given: 2a(0.5z+z^{2}), applying the distributive property means you distribute 2a to both terms inside the parentheses:
  • First, distribute 2a to 0.5z. This gives you az.
  • Next, distribute 2a to z^{2}, giving you 2az^{2}.
Combining these, the expression becomes az + 2az^{2}.
The distributive property helps you break down complex algebraic expressions into simpler, manageable pieces. The end goal is to eliminate parentheses, making the expression easier to work with in subsequent calculations.

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Most popular questions from this chapter

Each table describes a linear relationship. For each relationship, find the slope of the line and the \(y\) -intercept. Then write an equation for the relationship in the form \(y=m x+b .\) $$\begin{array}{|c|c|c|c|c|c|}\hline x & {2} & {4} & {6} & {8} & {10} \\\ \hline y & {8} & {12} & {16} & {20} & {24} \\ \hline\end{array}$$

One day Lydia walked from Allentown to Brassville at a constant rate of 4 kilometers per hour. The towns are 30 kilometers apart. a. Write an equation for the relationship between the distance Lydia traveled, \(d,\) and the hours she walked, \(h\). b. Graph your equation to show the relationship between hours walked and distance traveled. Put distance traveled on the vertical axis. c. How many hours did it take Lydia to reach Brassville? d. Now write an equation for the relationship between the hours walked, \(h\), and the distance remaining to complete the trip, \(r\). e. Graph the equation you wrote for Part d on the same set of axes you used for Part b. Label the vertical axis for both \(d\) and \(r\). f. How can you use your graph from Part e to determine how many hours it took Lydia to reach Brassville?

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Below is world population data for the years 1950 through 1990. $$\begin{array}{|c|c|}\hline & {\text { Population }} \\ {\text { Year }} & {\text { (billions) }} \\ \hline 1950 & {2.52} \\ {1960} & {3.02} \\ {1970} & {3.70} \\ {1980} & {4.45} \\ {1990} & {5.29} \\ \hline\end{array}$$ a. Plot the points on a graph with 鈥淵ears since 1900鈥� on the horizontal axis and 鈥淧opulation (billions)鈥� on the vertical axis. Try to fit a line to the data. b. Write an equation to fit your line. c. Use your equation to project the world population for the year 2010, which is 110 years after 1900. d. What does your equation tell you about world population in 1900? Does this make sense? Explain. e. According to United Nations figures, the world population in 1900 was 1.65 billion. The UN has predicted that world population in the year 2010 will be 6.79 billion. Are the 1900 data and the prediction for 2010 different from your predictions? How do you explain your answer?

For each equation, identify the slope and the y-intercept. Graph the line to check your answer. $$ y=-2 x $$
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