146k views
11 votes
I need help but not urgent but urgent!!

I need help but not urgent but urgent!!-example-1

2 Answers

11 votes

Answer:

8.6 miles (nearest tenth)

Step-by-step explanation:


d=1.2116√(h)

where:

  • d = approximate distance a person can see (in miles)
  • h = person's height above the ocean (in feet)

Captain

Given the Captain is 15 ft above the ocean:


\implies h = 15

Substituting this into the equation:


\implies d_1=1.2116√(15)

Lookout

Given the lookout is 120ft above the ocean:


\implies h = 120

Substituting this into the equation:


\implies d_2=1.2116√(120)

Solution

To find how much farther the lookout can see than the captain, subtract the distance the captain can see from the distance the lookout can see:


\implies d_2-d_1


\implies 1.2116√(120)-1.2116√(15)


\implies 1.2116(√(120)-√(15))


\implies 8.579906391...


\implies 8.6\: \sf miles\:(nearest\:tenth)

User Khurshid Ansari
by
6.6k points
12 votes

Answer:

8.6 mile

Step-by-step explanation:


\sf equation \ follows : \ \ \ d = 1.2116√(h)

  • d represents the distance a person can see.
  • h represents the height above the ocean.

what the captain sees:


\hookrightarrow \sf d_c = 1.2116√(15)

what the sailor sees:


\hookrightarrow \sf d_s = 1.2116√(120)

Difference between them:


\sf \rightarrow 1.2116√(120) \ - \ 1.2116√(15)


\sf \rightarrow 8.58


\sf \rightarrow 8.6 (rounded to nearest tenth)

User Josh Morrison
by
6.5k points
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