Question

the movement of the progress bar may be uneven because questions can be worth more or less (including zero) depending on your answer. the pressure applied to a leverage bar varies inversely as the distance from the object. if 150 pounds is required for a distance of 10 inches from the object, how much pressure is needed for a distance of 3 inches? 5 pounds 500 pounds 450 pounds 45 pounds

Explanation:

Step1: Define inverse variation formula

For inverse variation, \( P = \frac{k}{d} \), where \( P \) is pressure, \( d \) is distance, and \( k \) is constant.

Step2: Find constant \( k \)

Given \( P = 150 \), \( d = 10 \). Substitute into formula: \( 150 = \frac{k}{10} \). Solve for \( k \): \( k = 150 \times 10 = 1500 \).

Step3: Calculate pressure for \( d = 3 \)

Use \( k = 1500 \), \( d = 3 \) in \( P = \frac{k}{d} \): \( P = \frac{1500}{3} = 500 \)? Wait, no—wait, inverse variation: as distance decreases, pressure increases. Wait, 15010 = 1500, then 1500/3 = 500? Wait, but let's check again. Wait, inverse variation: \( P_1d_1 = P_2d_2 \). So \( 150 \times 10 = P_2 \times 3 \). Then \( P_2 = \frac{150 \times 10}{3} = \frac{1500}{3} = 500 \)? Wait, but the options have 500. Wait, but let's recalculate. 15010=1500, 1500/3=500. So yes.

Answer:

500 pounds (the option with "500 pounds")