Sure, let's solve the problem step-by-step.
You are given a model of the river bottom as:
[tex]\[ d = \frac{1}{5} |s - 250| - 50 \][/tex]
The harbormaster wants to place buoys where the river bottom is 20 feet below the surface, so let [tex]\( d = 20 \)[/tex].
Substitute [tex]\( d = 20 \)[/tex] into the equation and solve for [tex]\( s \)[/tex]:
[tex]\[ 20 = \frac{1}{5} |s - 250| - 50 \][/tex]
First, add 50 to both sides to isolate the absolute value term:
[tex]\[ 20 + 50 = \frac{1}{5} |s - 250| \][/tex]
[tex]\[ 70 = \frac{1}{5} |s - 250| \][/tex]
Next, multiply both sides by 5 to eliminate the fraction:
[tex]\[ 70 \times 5 = |s - 250| \][/tex]
[tex]\[ 350 = |s - 250| \][/tex]
This results in two possible equations, because the absolute value of a number [tex]\( |x| \)[/tex] equals x or -x:
[tex]\[ s - 250 = 350 \][/tex]
[tex]\[ s - 250 = -350 \][/tex]
Solve each equation separately:
For the first equation:
[tex]\[ s - 250 = 350 \][/tex]
[tex]\[ s = 350 + 250 \][/tex]
[tex]\[ s = 600 \][/tex]
For the second equation:
[tex]\[ s - 250 = -350 \][/tex]
[tex]\[ s = -350 + 250 \][/tex]
[tex]\[ s = -100 \][/tex]
Therefore, the horizontal distances from the left shore at which the buoys should be placed are:
[tex]\[ s = 600 \text{ feet} \][/tex]
[tex]\[ s = -100 \text{ feet} \][/tex]
Thus, the completed absolute value equation is:
[tex]\[ 20 = \frac{1}{5} |s - 250| - 50 \][/tex]
The horizontal distances at which the buoys should be placed are:
[tex]\[ s = 600 \][/tex]
[tex]\[ s = -100 \][/tex]
