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allochka39001 [22]
2 years ago
13

I need help but not urgent but urgent!!

Mathematics
2 answers:
oksano4ka [1.4K]2 years ago
7 0

Answer:

8.6 miles (nearest tenth)

Step-by-step explanation:

d=1.2116\sqrt{h}

where:

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

<u>Captain</u>

Given the Captain is 15 ft above the ocean:

\implies h = 15

Substituting this into the equation:

\implies d_1=1.2116\sqrt{15}

<u>Lookout</u>

Given the lookout is 120ft above the ocean:

\implies h = 120

Substituting this into the equation:

\implies d_2=1.2116\sqrt{120}

<u>Solution</u>

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\sqrt{120}-1.2116\sqrt{15}

\implies 1.2116(\sqrt{120}-\sqrt{15})

\implies 8.579906391...

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

Mars2501 [29]2 years ago
6 0

Answer:

8.6 mile

Explanation:

\sf equation \ follows : \ \ \  d = 1.2116\sqrt{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\sqrt{15}

what the sailor sees:

\hookrightarrow \sf d_s = 1.2116\sqrt{120}

Difference between them:

\sf \rightarrow 1.2116\sqrt{120} \  - \   1.2116\sqrt{15}

\sf \rightarrow 8.58

\sf \rightarrow 8.6      (rounded to nearest tenth)

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