A snoll is at the bottom of a well that is 16.5 m deep. Starting from it climbs up 4.5 m in one hour and then rest for one hour. When it rests, it slips back 3 m . If the pattern of climbing and resting continues, at wha tme will the snail get out of the well?

To solve this problem, we need to determine the time it takes for the snail to climb out of the well. Given: - Depth of the well: 16.5 m - Climbing rate: 4.5 m/h - Slipping rate: 3 m/h Let's denote the time it takes for the snail to climb out of the well as \( t \) hours. The snail climbs 4.5 m in one hour and then slips back 3 m in the next hour. This pattern continues until the snail climbs out of the well. The total distance the snail climbs in one cycle (climbing and resting) is: \[ 4.5 \, \text{m} - 3 \, \text{m} = 1.5 \, \text{m} \] Since the snail climbs 1.5 m in one cycle, it will take \( \frac{16.5}{1.5} \) cycles to climb out of the well. The time it takes for the snail to climb out of the well is the number of cycles multiplied by the time it takes for one cycle: \[ t = \frac{16.5}{1.5} \times 2 \] Now, let's calculate the time it takes for the snail to climb out of the well. Calculate the value by following steps: - step0: Calculate: \(\frac{16.5}{1.5}\times 2\) - step1: Divide the terms: \(11\times 2\) - step2: Multiply the numbers: \(22\)