Problem 85
Question
Evaluate \(\frac{a}{1-b}\) for the given values of \(a\) and \(b\) $$ a=0.9, b=-0.5 $$
Step-by-Step Solution
Verified Answer
The evaluated expression is 0.6.
1Step 1: Substitute the Given Values
First, substitute the given values of \(a\) and \(b\) into the expression \(\frac{a}{1-b}\). The expression becomes \(\frac{0.9}{1 - (-0.5)}\).
2Step 2: Simplify the Denominator
In the denominator, you have the term \(1 - (-0.5)\). This simplifies to \(1 + 0.5 = 1.5\).
3Step 3: Perform the Division
Now, divide the numerator by the simplified denominator: \(\frac{0.9}{1.5}\).
4Step 4: Simplify the Fraction
Convert \(\frac{0.9}{1.5}\) to a simpler fraction or decimal. By converting them both into fractions and finding a common term, \(\frac{0.9}{1.5} = \frac{3}{5} = 0.6\).
Key Concepts
Evaluating ExpressionsSubstitutionFractionsDecimal to Fraction Conversion
Evaluating Expressions
Evaluating expressions involves calculating the value of an expression by using specific values for the variables. In this exercise, our expression is \(\frac{a}{1-b}\), and we are given values \(a=0.9\) and \(b=-0.5\). To evaluate means to systematically substitute these values into the expression and simplify.
To begin, insert the given values into the expression so it becomes \(\frac{0.9}{1 - (-0.5)}\). The process doesn't end with substitution; you must continue by simplifying any arithmetic operations. This step ensures you arrive at a precise numerical value for the expression.
To begin, insert the given values into the expression so it becomes \(\frac{0.9}{1 - (-0.5)}\). The process doesn't end with substitution; you must continue by simplifying any arithmetic operations. This step ensures you arrive at a precise numerical value for the expression.
Substitution
Substitution is a key concept when evaluating expressions. It involves replacing variables with specific numbers. Here, the given values for \(a\) and \(b\) are substituted into the expression \(\frac{a}{1-b}\).
When substituting, the target expression should reflect these replacement values accurately. For our problem:
When substituting, the target expression should reflect these replacement values accurately. For our problem:
- The variable \(a\) is replaced with \(0.9\).
- The variable \(b\) is replaced with \(-0.5\), resulting in the modified expression \(\frac{0.9}{1 - (-0.5)}\).
Fractions
Understanding fractions is essential in solving expressions like \(\frac{0.9}{1-b}\). A fraction consists of a numerator and a denominator. In this context, \(0.9\) is the numerator and \(1 - b\) is the denominator.
Simplifying the denominator is often necessary before you perform the division. Here, simplifying \(1 - (-0.5)\) to \(1.5\) is a critical step. Once simplified, the expression becomes \(\frac{0.9}{1.5}\). Below, we'll convert this fraction into simpler terms through basic arithmetic or converting it to decimals.
Simplifying the denominator is often necessary before you perform the division. Here, simplifying \(1 - (-0.5)\) to \(1.5\) is a critical step. Once simplified, the expression becomes \(\frac{0.9}{1.5}\). Below, we'll convert this fraction into simpler terms through basic arithmetic or converting it to decimals.
Decimal to Fraction Conversion
Decimal to fraction conversion can simplify expressions further, making understanding easier. In our example, we expressed \(\frac{0.9}{1.5}\) as a fraction.
Both numbers, 0.9 and 1.5, can be rewritten in terms of fractions:
Both numbers, 0.9 and 1.5, can be rewritten in terms of fractions:
- 0.9 becomes \(\frac{9}{10}\)
- 1.5 becomes \(\frac{15}{10}\)
Other exercises in this chapter
Problem 84
Evaluate each expression for the given values of the variable. \(4 m+7 ; m=1,2,3,4,5\)
View solution Problem 84
Solve each system of equations by using either substitution or elimination. $$ \begin{array}{l}{10 x-y=13} \\ {3 x-4 y=15}\end{array} $$
View solution Problem 85
Evaluate each expression for the given values of the variable. \(4-2 k ; k=3,4,5,6,7\)
View solution Problem 85
Solve each system of equations by using either substitution or elimination. $$ \begin{array}{l}{3 c-7 d=-1} \\ {2 c-6 d=-6}\end{array} $$
View solution