Problem 57
Question
Evaluate each expression. $$ \begin{array}{ll}{\text { (a) }|(-2) \cdot 6|} & {\text { (b) }\left|\left(-\frac{1}{3}\right)(-15)\right|}\end{array} $$
Step-by-Step Solution
Verified Answer
(a) 12, (b) 5.
1Step 1: Evaluate Expression (a)
For expression (a), you need to calculate \(|(-2) \cdot 6|\). First, perform the multiplication within the absolute value function: \((-2) \times 6 = -12\).Next, apply the absolute value which makes any negative value positive: \|-12| = 12\.
2Step 2: Evaluate Expression (b)
For expression (b), calculate \(|\left(-\frac{1}{3}\right)(-15)|\). Start with the multiplication: \[-\frac{1}{3} \times -15 = \frac{15}{3} = 5\].Here, since there are two negatives, the result is positive. The absolute value just confirms this result, \(|5| = 5\).
Key Concepts
MultiplicationEvaluating ExpressionsNegative Numbers
Multiplication
Multiplication is a fundamental operation in mathematics, involving two numbers known as factors. The product is the result of this operation. Sometimes students find multiplication tricky when it involves negative numbers or fractions. But don't worry! It's not as hard as it seems.
First, let's understand the basic rule: **When you multiply two numbers**:
First, let's understand the basic rule: **When you multiply two numbers**:
- If both are positive, the product is positive.
- If both are negative, the product is also positive.
- If one is negative and the other is positive, the product is negative.
Evaluating Expressions
Evaluating expressions involves a series of calculations to find the value of an expression. The key is to follow the right order of operations, sometimes remembered by the acronym PEMDAS (Parentheses, Exponents, Multiplication and Division (from left to right), Addition and Subtraction (from left to right)).
Let's apply this to our exercise. In expression (a), the key steps included multiplying \(-2\) by \(6\) before applying the absolute value. For expression (b), the steps are similar: multiply \(-\frac{1}{3}\) by \(-15\) before evaluating the absolute value.
A useful tip: always handle operations inside the absolute value brackets first before applying the absolute value. This ensures the accuracy of the evaluation.
Let's apply this to our exercise. In expression (a), the key steps included multiplying \(-2\) by \(6\) before applying the absolute value. For expression (b), the steps are similar: multiply \(-\frac{1}{3}\) by \(-15\) before evaluating the absolute value.
A useful tip: always handle operations inside the absolute value brackets first before applying the absolute value. This ensures the accuracy of the evaluation.
Negative Numbers
Negative numbers represent values less than zero and are crucial in understanding mathematics. When dealing with them, especially in operations like multiplication, they have specific rules. Here, both expressions in this task involve negative numbers.
Let's break it down:
Let's break it down:
- **Multiplying a negative by a positive**: The product is negative. For instance, in \(-2 \times 6\), the result is \(-12\).
- **Multiplying two negatives**: The product is positive. From \(-\frac{1}{3} \times -15\), we get \(5\). This happens because two negatives "cancel out" each other.
Other exercises in this chapter
Problem 57
\(47-72\) . Simplify the expression, and eliminate any negative exponent(s). $$ \frac{6 y^{3} z}{2 y z^{2}} $$
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Factor the expression completely. $$ (a+b)^{2}-(a-b)^{2} $$
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\(49-68=\) Simplify the expression and eliminate any negative exponent(s). Assume that all letters denote positive numbers. $$ \left(8 y^{3}\right)^{-2 / 3} $$
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Multiply the algebraic expressions using a Special Product Formula, and simplify. \(\left(x^{2}+1\right)^{2}\)
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