Solve for $d$.
$\begin{array}{l}
1.4 d=5.04 \\
d=
\end{array}$
We are given a simple linear equation with one variable, \(d\), which is represented by the equation \(1.4d = 5.04\). The goal is to isolate the variable \(d\) to determine its numerical value. This is a standard algebraic equation of the form \(kd = C\), where \(k = 1.4\) and \(C = 5.04\). To find \(d\), we need to divide the constant term \(C\) by the coefficient \(k\).
To isolate \(d\), we divide both sides of the equation \(1.4d = 5.04\) by \(1.4\):
\[d = \frac{5.04}{1.4}\]
To simplify this division, we can multiply both the numerator and the denominator by \(10\) to remove the decimal point from the denominator:
\[d = \frac{5.04 \times 10}{1.4 \times 10} = \frac{50.4}{14}\]
Now, we perform the long division of \(50.4\) by \(14\). First, how many times does \(14\) go into \(50\)? \(14 \times 3 = 42\). Subtracting \(42\) from \(50\) leaves a remainder of \(8\). Bringing down the decimal digit \(4\) gives us \(8.4\). Since \(14 \times 0.6 = 8.4\), the final result of the division is \(3 + 0.6 = 3.6\).
Having performed the necessary division, we have found that the value of \(d\) is \(3.6\). We can verify this by plugging the value back into the original equation: \(1.4 \times 3.6 = 5.04\). Since the result is correct, we conclude that the solution is \(\boxed{3.6}\).
This type of problem is common in everyday situations where you need to calculate unit rates or solve for an unknown quantity based on a total value. For example, if you know that the total cost of 1.4 kilograms of apples is \$5.04, you can use this equation to find the price per kilogram of the apples. By solving \(1.4d = 5.04\), you find that the cost per kilogram, \(d\), is \(\$3.60\). This skill is essential for budgeting and making informed purchasing decisions.