7. The bottom of a river makes a V-shape that can be modeled with the absolute value function, d(h) = 1/6 |h – 300| - 50 where d is the depth of the river bottom (in feet) and h is the horizontal distance to the left-hand shore (in feet). A ship risks running aground if the bottom of its keel (its lowest point under the water) reaches down to the river bottom. Suppose you are the harbor master and you want to place buoys where the river bottom is 20 feet below the surface. If one buoy is to be placed 120 feet away from the left shore, how far should the second buoy be placed form the left shore?

(Note: A buoy is an anchored float serving as navigation mark, to show reefs or other hazards to guide ships.)

If one buoy is placed 120 feet away from the left shore, then the distance of the second buoy from the left shore = 480

Step-by-step explanation:

Given that:

The absolute value function:

To calculate the position of the buoy where the river bottom is 20 feet below the surface, we equate the absolute value function to the negative value of 20 feet and then solve it for h.

d(h) = -20

Let us add the sum of 50 on both sides, then:

By multiplying 6 on both sides, we have:

|h - 300| = 30 (6)

|h - 300| = 180

We can attempt the above expression by rewriting it as two equations:

h - 300 = 180   or     h  - 300  = - 180

Let us add 300 to both sides from the two expressions above:

h - 300 + 300 = 180 + 300     or   h - 300 + 300   = -180  + 300

h = 480     or  h = 120

Thus, if one buoy is placed 120 feet away from the left shore, then the distance of the second buoy from the left shore = 480

triangle and cube

step-by-step explanation:

the first one is x^2+2x+2

the second one is x^2+4

the last one is x^2-4x+8