Problem 58
Question
Find each value. Check each result with a calculator. \(\frac{2^{3}-7}{5^{2}}\)
Step-by-Step Solution
Verified Answer
The value is \(\frac{1}{25}\).
1Step 1: Simplify the Numerator
Start by simplifying the expression in the numerator. You need to solve \(2^3\). Calculate \(2^3\) which is \(2 \times 2 \times 2 = 8\). Next, subtract 7 from 8: \(8 - 7 = 1\). Therefore, the simplified numerator is 1.
2Step 2: Simplify the Denominator
Now, simplify the denominator \(5^2\). Calculate \(5^2\) which is \(5 \times 5 = 25\). Thus, the simplified denominator is 25.
3Step 3: Combine and Simplify the Fraction
Combine the simplified numerator and denominator into the fraction: \(\frac{1}{25}\). This means the value of the expression \(\frac{2^3 - 7}{5^2}\) is \(\frac{1}{25}\).
4Step 4: Verify the Result with a Calculator
Using a calculator, compute each part of the expression: \(2^3 = 8\) and \(5^2 = 25\). Then calculate \(\frac{2^3 - 7}{5^2} = \frac{8 - 7}{25} = \frac{1}{25}\). The calculator should confirm the value \(\frac{1}{25}\).
Key Concepts
Understanding the NumeratorDefining the DenominatorExploring ExponentiationThe Process of Simplification
Understanding the Numerator
In any fraction, the numerator is the top part. It shows how many parts of a whole are being considered. In our exercise, the numerator is a small mathematical expression: \(2^3 - 7\).
First, let's handle \(2^3\), which involves exponentiation. This is the repeated multiplication of 2 three times: \(2 \times 2 \times 2\).
This equals 8. From 8, we subtract 7, resulting in a simplified numerator of 1.
First, let's handle \(2^3\), which involves exponentiation. This is the repeated multiplication of 2 three times: \(2 \times 2 \times 2\).
This equals 8. From 8, we subtract 7, resulting in a simplified numerator of 1.
- The numerator tells us the number of parts taken.
- In our problem, the numerator shows the result of \(2^3\) minus 7.
Defining the Denominator
The denominator in a fraction is the bottom number. It signifies the total number of equal parts in the whole. For our exercise, the denominator is expressed as \(5^2\).
This involves exponentiation, similar to the numerator, meaning we multiply 5 by itself: \(5 \times 5\).
The result is 25, which is our simplified denominator.
This involves exponentiation, similar to the numerator, meaning we multiply 5 by itself: \(5 \times 5\).
The result is 25, which is our simplified denominator.
- The denominator indicates the number of equal parts the whole divide into.
- Here, it shows that the whole is divided into 25 equal parts.
Exploring Exponentiation
Exponentiation is a fundamental operation in mathematics. It means multiplying a number by itself a specific number of times, as indicated by the exponent. In the given problem, both the numerator and the denominator undergo this process.
For the numerator, the calculation \(2^3\) means multiplying 2 by itself three times: \(2 \times 2 \times 2 = 8\).
For the denominator, \(5^2\) means multiplying 5 by itself twice: \(5 \times 5 = 25\).
For the numerator, the calculation \(2^3\) means multiplying 2 by itself three times: \(2 \times 2 \times 2 = 8\).
For the denominator, \(5^2\) means multiplying 5 by itself twice: \(5 \times 5 = 25\).
- Exponents show the power to which the base number is raised.
- This operation reduces complexity by converting repeated multiplication into a simple expression.
The Process of Simplification
Simplification is about making expressions easier to understand and solve. It involves breaking down complex problems into simple, solvable parts. In our example, simplification begins by solving both the numerator and denominator individually.
The numerator \(2^3 - 7\) simplifies to 1 after performing the operations.
The denominator \(5^2\) simplifies to 25. The fractional expression then becomes \(\frac{1}{25}\).
The numerator \(2^3 - 7\) simplifies to 1 after performing the operations.
The denominator \(5^2\) simplifies to 25. The fractional expression then becomes \(\frac{1}{25}\).
- Simplification ensures expressions are easier to compute and verify.
- It minimizes the chance for errors in complex calculations.
Other exercises in this chapter
Problem 58
Find the least common multiple of the numbers. 8 and 8
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Determine which of the whole numbers are prime and which are composite. 339
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Determine the value of each of the powers. Use a calculator to check each result. \(2^{20}\)
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Write each number as a product of prime factors. 700
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