Problem 88
Question
Evaluate the expression. \(\frac{1}{2} n^{3}\) when \(n=-2\)
Step-by-Step Solution
Verified Answer
The evaluated expression is \(-4\).
1Step 1: Substitution
First, substitute the given value of \(n = -2\) into the expression \(\frac{1}{2} n^{3}\). This gives us \(\frac{1}{2} (-2)^{3}\).
2Step 2: Exponentiation
Next, compute the exponentiation part \((-2)^{3}\). This results in \(-8\). So the expression now transforms to \(\frac{1}{2} * -8\).
3Step 3: Multiplication
Finally, calculate the product \(\frac{1}{2} * -8\). The answer is \(-4\).
Key Concepts
ExponentiationSubstitution in AlgebraMultiplication of Fractions
Exponentiation
In mathematics, exponentiation is an operation that involves raising a number, known as the base, to the power of an exponent. The exponent signifies how many times the base is multiplied by itself. For example, in the expression \(b^n\), \(b\) is the base, and \(n\) is the exponent or power.
Let's apply this to the exercise where we have \(n^3\). This means \(n\) is the base, and it is multiplied by itself three times: \(n \times n \times n\). When \(n=-2\), as in the given expression, \(n^3\) becomes \((-2) \times (-2) \times (-2)\) which equals \(-8\), because a negative number multiplied by itself an odd number of times results in a negative product.
Remember that the sign of the result depends on the exponent: if it's even, the result is positive, but if it's odd—as in our case—the result is negative.
Let's apply this to the exercise where we have \(n^3\). This means \(n\) is the base, and it is multiplied by itself three times: \(n \times n \times n\). When \(n=-2\), as in the given expression, \(n^3\) becomes \((-2) \times (-2) \times (-2)\) which equals \(-8\), because a negative number multiplied by itself an odd number of times results in a negative product.
Remember that the sign of the result depends on the exponent: if it's even, the result is positive, but if it's odd—as in our case—the result is negative.
Substitution in Algebra
The technique of substitution in algebra is essentially replacing a variable with its value. This helps in simplifying expressions or solving equations. When we have an expression with a variable, like \(\frac{1}{2} n^{3}\), and we know the value of that variable, we can substitute the variable with its value for further calculation.
In our exercise, we substitute \(n\) with \(n=-2\). With substitution, we go from the abstract \(\frac{1}{2} n^{3}\) to the more concrete \(\frac{1}{2} (-2)^{3}\), which we can compute. Proper substitution is crucial as the next steps of the problem-solving process completely depend on it. Substitution allows for a clearer understanding and often simplifies the steps needed to find a solution.
In our exercise, we substitute \(n\) with \(n=-2\). With substitution, we go from the abstract \(\frac{1}{2} n^{3}\) to the more concrete \(\frac{1}{2} (-2)^{3}\), which we can compute. Proper substitution is crucial as the next steps of the problem-solving process completely depend on it. Substitution allows for a clearer understanding and often simplifies the steps needed to find a solution.
Multiplication of Fractions
The process of multiplying fractions is straightforward: Simply multiply the numerators (top numbers) together, and then the denominators (bottom numbers) together. The rule is \(\frac{a}{b} \cdot \frac{c}{d} = \frac{a \cdot c}{b \cdot d}\). Unlike when adding or subtracting fractions, you don't need a common denominator.
In the context of our problem, we multiply the fraction \(\frac{1}{2}\) by \(\frac{-8}{1}\)—since any integer can be written as a fraction with 1 as the denominator—giving us \(-4\) because \(\frac{1}{2} \cdot \frac{-8}{1} = \frac{1 \cdot -8}{2 \cdot 1} = \frac{-8}{2}\), which simplifies to \(\frac{-8}{2} = -4\). Multiplication of fractions is often used together with other operations, such as exponentiation and substitution, to evaluate expressions like the one in this exercise.
In the context of our problem, we multiply the fraction \(\frac{1}{2}\) by \(\frac{-8}{1}\)—since any integer can be written as a fraction with 1 as the denominator—giving us \(-4\) because \(\frac{1}{2} \cdot \frac{-8}{1} = \frac{1 \cdot -8}{2 \cdot 1} = \frac{-8}{2}\), which simplifies to \(\frac{-8}{2} = -4\). Multiplication of fractions is often used together with other operations, such as exponentiation and substitution, to evaluate expressions like the one in this exercise.
Other exercises in this chapter
Problem 87
Evaluate the expression. \((5 y)^{4}\) when \(y=2\)
View solution Problem 87
Use substitution to solve the system. $$\begin{aligned}&x+15 y=6\\\&-x-5 y=84\end{aligned}$$
View solution Problem 88
Use substitution to solve the system. $$\begin{aligned}&4 x-y=5\\\&2 x+4 y=15\end{aligned}$$
View solution Problem 89
Evaluate the expression. \(\frac{1}{y^{2}}\) when \(y=5\)
View solution