What is the product?
$\frac{5 k}{6} \cdot \frac{3}{2 k^3}$
To simplify the expression \(\frac{5k}{6} \cdot \frac{3}{2k^3}\), we start by applying the fundamental rule for multiplying two fractions: \(\frac{a}{b} \cdot \frac{c}{d} = \frac{ac}{bd}\). This tells us to multiply the numerators together and the denominators together.
First, we calculate the numerator:
\[5k \times 3 = 15k\]
Next, we calculate the denominator:
\[6 \times 2k^3 = 12k^3\]
Combining these, we get the fraction:
\[\frac{15k}{12k^3}\]
Now we simplify the expression \(\frac{15k}{12k^3}\) by breaking it down into its numerical and variable components.
For the numerical part, we have the fraction \(\frac{15}{12}\). Both 15 and 12 are divisible by 3:
\[\frac{15 \div 3}{12 \div 3} = \frac{5}{4}\]
For the variable part, we have \(\frac{k}{k^3}\). Using the exponent property \(\frac{x^a}{x^b} = x^{a-b}\), we get:
\[\frac{k^1}{k^3} = k^{1-3} = k^{-2} = \frac{1}{k^2}\]
Putting these two parts together gives:
\[\frac{5}{4} \cdot \frac{1}{k^2} = \frac{5}{4k^2}\]
Comparing our result \(\frac{5}{4k^2}\) to the provided options, we see that it matches the first option. Therefore, the product of the two fractions is \(\boxed{\frac{5}{4k^2}}\).
In real-world applications, simplifying algebraic expressions is essential in fields like physics and engineering. For example, if you are calculating the force between two objects or the rate of work done as variables change, you often end up with complex fractional expressions. Simplifying these expressions allows you to quickly understand how the resulting value changes relative to the input variable \(k\), making it easier to predict outcomes in experiments or mechanical designs.