Verified Answer :
- For problem 9, the constant of proportionality is \(k = \frac{1}{3}\), and for \(x = 24\), \(y = \boxed{8}\).
- For problem 10, the constant of proportionality is \(k = \frac{1}{4}\), and for \(y = -6\), \(x = \boxed{-24}\).
- For problem 11, the constant of proportionality is \(k = 3.5\), and for \(x = 8\), \(y = \boxed{28}\).
- For problem 12, the constant of proportionality is \(k = -2\), and for \(y = 24\), \(x = \boxed{-12}\).
- For problem 13, additional information is required to define the constant \(k\), so the wages are expressed as \(W = 50k\).
Explanation
- Understanding Direct Variation
Direct variation implies that two variables \(y\) and \(x\) have a relationship defined by the equation \(y = kx\), where \(k\) is a non-zero constant known as the constant of proportionality. To solve for unknown values in these problems, we first identify the constant \(k\) using a given pair of \((x, y)\) values, then apply that \(k\) to the new condition. For problems where we need to find \(x\) given \(y\), we can either solve \(y = kx\) for \(x\) or use the equivalent form \(x = \frac{1}{k}y\). We will address each sub-problem individually based on these principles.
- Solving Sub-problems 9 and 10
For problem 9, we are given \(y = 2\) when \(x = 6\). The constant of proportionality is \(k = \frac{y}{x} = \frac{2}{6} = \frac{1}{3}\). We want to find \(y\) when \(x = 24\). Using our formula \(y = kx\), we calculate \(y = \frac{1}{3} \times 24 = 8\).
For problem 10, we are given \(x = -12\) when \(y = -3\). We find the constant \(k\) using \(y = kx\), so \(-3 = k(-12)\), which gives \(k = \frac{-3}{-12} = \frac{1}{4}\). We need to find \(x\) when \(y = -6\). Substituting into \(y = kx\), we have \(-6 = \frac{1}{4}x\). Multiplying both sides by 4, we get \(x = -24\).
- Solving Sub-problems 11 and 12
For problem 11, we are given \(y = 7\) when \(x = 2\). The constant \(k\) is \(\frac{7}{2} = 3.5\). Finding \(y\) when \(x = 8\), we calculate \(y = 3.5 \times 8 = 28\).
For problem 12, we are given \(y = 6\) when \(x = -3\). The constant \(k\) is \(\frac{6}{-3} = -2\). To find \(x\) when \(y = 24\), we use \(y = kx\), so \(24 = -2x\). Dividing by -2, we find \(x = -12\).
- Analyzing Sub-problem 13 and Final Conclusion
For problem 13, the wage \(W\) varies directly with the time \(T\), which means \(W = kT\). To find the wage for 50 hours, we would need to know the wage rate \(k\) (the amount earned per hour) or a specific reference point where a wage was earned for a known time. Since no such information is provided, we can only express the answer as the function \(W = k \times 50\), where \(k\) is the unknown hourly wage. Without additional data, the specific numerical value for the wages cannot be determined.
Examples
Direct variation is a foundational concept in economics and physics. For example, if you are paid a constant hourly rate, your total wages are directly proportional to the number of hours you work. If you earn \(\$20\) per hour, the relationship is \(W = 20T\). Knowing this, if you work for \(40\) hours, you can predict your total earnings: \(W = 20 \times 40 = \$800\). Similarly, when traveling at a constant speed, the distance you cover is directly proportional to the time spent driving. This concept allows us to make accurate predictions based on limited data points.