Solving the Mathematical Expression

The problem asks us to find the value of a given mathematical expression involving fractions and decimal numbers. The expression is:

\(\frac{1}{4} + \frac{{[{{(20.35)}^2} - {{(8.35)}^2}] \times 0.0175}}{{{{(1.05)}^2} + (1.05)(27.65)}}\)

We need to evaluate the fraction part first and then add \(\frac{1}{4}\) to the result.

Evaluating the Fraction Part

Let's analyze the numerator and the denominator of the fraction separately.

Simplifying the Numerator

The numerator is \({[{{(20.35)}^2} - {{(8.35)}^2}] \times 0.0175}\). The part inside the square brackets is in the form of \(a^2 - b^2\), where \(a = 20.35\) and \(b = 8.35\). We can use the algebraic identity \(a^2 - b^2 = (a+b)(a-b)\).

• Calculate \(a+b\): \(20.35 + 8.35 = 28.70\)
• Calculate \(a-b\): \(20.35 - 8.35 = 12.00\)

So, \({(20.35)}^2 - {{(8.35)}^2} = (28.70) \times (12.00)\).

Now, the numerator becomes \((28.70) \times (12.00) \times 0.0175\).

Simplifying the Denominator

The denominator is \({{{{(1.05)}^2} + (1.05)(27.65)}}\). We can see that \(1.05\) is a common factor in both terms. We can factor it out.

\({{(1.05)}^2} + (1.05)(27.65) = 1.05 \times (1.05 + 27.65)\)

• Calculate the sum inside the parenthesis: \(1.05 + 27.65 = 28.70\)

So, the denominator becomes \(1.05 \times 28.70\).

Putting the Fraction Back Together

Now the fraction is:

\(\frac{(28.70) \times (12.00) \times 0.0175}{1.05 \times (28.70)}\)

We can cancel out the common term \(28.70\) from both the numerator and the denominator.

\(\frac{\cancel{(28.70)} \times (12.00) \times 0.0175}{1.05 \times \cancel{(28.70)}} = \frac{12.00 \times 0.0175}{1.05}\)

Now, perform the multiplication in the numerator:

\(12.00 \times 0.0175 = 0.21\)

The fraction simplifies to:

\(\frac{0.21}{1.05}\)

To simplify this decimal fraction, we can multiply both the numerator and the denominator by 100 to remove the decimals:

\(\frac{0.21 \times 100}{1.05 \times 100} = \frac{21}{105}\)

Now, we simplify the fraction \(\frac{21}{105}\). Both 21 and 105 are divisible by 21.

• \(21 \div 21 = 1\)
• \(105 \div 21 = 5\)

So, the simplified fraction is \(\frac{1}{5}\).

Final Calculation

The original expression was \(\frac{1}{4} + \text{the fraction}\). We found that the fraction evaluates to \(\frac{1}{5}\). So, the expression becomes:

\(\frac{1}{4} + \frac{1}{5}\)

To add these fractions, we find a common denominator, which is the least common multiple of 4 and 5. The LCM of 4 and 5 is 20.

• Convert \(\frac{1}{4}\) to a fraction with denominator 20: \(\frac{1}{4} = \frac{1 \times 5}{4 \times 5} = \frac{5}{20}\)
• Convert \(\frac{1}{5}\) to a fraction with denominator 20: \(\frac{1}{5} = \frac{1 \times 4}{5 \times 4} = \frac{4}{20}\)

Now, add the fractions:

\(\frac{5}{20} + \frac{4}{20} = \frac{5+4}{20} = \frac{9}{20}\)

The value of the expression is \(\frac{9}{20}\).

Revision Table: Key Concepts for Expression Evaluation

Additional Information on Algebraic Simplification

Algebraic simplification is a fundamental skill in mathematics that helps in solving complex expressions and equations efficiently. Recognizing patterns like the difference of squares or common factors allows us to rewrite expressions in a simpler form, making calculations easier and less error-prone.

• The difference of squares identity \(a^2 - b^2 = (a+b)(a-b)\) is particularly useful when dealing with squares of numbers, especially decimals, as it converts squaring and subtracting into simpler addition and subtraction followed by multiplication.
• Factoring out common terms is essential for simplifying expressions and is often the first step in solving polynomial equations or simplifying rational expressions (fractions with algebraic terms).
• When dealing with fractions, always look for opportunities to simplify them before performing operations like addition or subtraction. This reduces the size of the numbers involved and makes the final calculation simpler. Converting decimal fractions to whole number fractions by multiplying by powers of 10 can also be a useful simplification technique.