A person who is 6 feet tall walks away from a flagpole toward the tip of the shadow of the flagpole. When the person is 12 feet from the flagpole, the tips of the person's shadow and the shadow cast by the flagpole coincide at a point 3 feet in front of the person.

h = 29.4 f

Step-by-step explanation: (see annex)

I conclude the question of the problem is the height " h " of the flagpole

From the draft in annex, is clear the angle ABC is common to the triangles

ABC and BDE

In triangle BDE  

BE = √ (3)² + (6)²         ⇒ BE = √9 + 36       ⇒BE = √45

BE = 6,71 f

then      sin ∠ ABC  =  6/6.71     ⇒   sin ∠ ABC  = 0,89

then arcsin 0,89      ∠ABC  =  63°

Now in triangle ABC

tan  ∠ ABC  = h/( 12 + 3 )        ⇒   tan  ∠ ABC  = h/15

tan ∠ 63°   =  1.96

Then       1.96  =  h/15

h  =  1.96 * 15

h = 29.4 f

the first dot plot best represents the data

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i dont know

step-by-step explanation: