Lost at Sea You are part of the rescue team in a ship at sea. One of your divers is 250 feet below sea level, and she injured herself. She only has a 7-minute supply of air in her tank, and can only rise towards the surface at a rate of 10 feet per minute. You are sending down a rescue sub. The sub can descend at a rate of 30 feet per minute.

Secret Question to the problem If you were to graph this situation, in which quadrant should you place the scuba diver? Why? Lost at Sea You are part of the rescue team in a ship at sea. One of your divers is 250 feet below sea level, and she injured herself. She only has a 7-minute supply of air in her tank, and can only rise towards the surface at a rate of 10 feet per minute. You are sending down a rescue sub. The sub can descend at a rate of 30 feet per minute. What would it mean if you placed your scuba diver in quadrant 1?

Time (minutes) This graph must use the 4th quadrant. The time counts up in minutes making it a positive number on the x-axis. The diver and submarine must have negative y-values because they are at a depth below sea level. The x-axis would represent zero. Depth (feet)

Time (minutes) The submarine has a depth of 0 feet below zero at a time of 0 minutes. The y-intercept is zero. The blue line represents the ascent of the diver. The problem states she can ascend at a rate of 10 feet per minute. The slope is 10. The red line represents the descent of the submarine. The problem states it can descend at a rate of 30 feet per minute. The slope is -30. Depth (feet) The diver has a depth of 250 feet below zero at a time of 0 minutes. The y-intercept is -250. y = -30x y = 10x – 250 or y = -250 +10x

= = = -30x = -250 + 10x 1x = 6.25 y = -30x y = -250 +10x -40x = -250 Time (minutes) -30x = -250 + 10x -10x -10x -40x = -250 ÷(-40) ÷(-40) 1x = 6.25 The sub will reach the diver after 6.25 minutes. Depth (feet) y = -30x y = -30(6.25) y = -187.5 ≈ The sub will save the diver at -188 ft below sea level