Question

I need help but not urgent but urgent!!

Answer 1

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)

Captain

Given the Captain is 15 ft above the ocean:

$\implies h = 15$

Substituting this into the equation:

$\implies d_1=1.2116\sqrt{15}$

Lookout

Given the lookout is 120ft above the ocean:

$\implies h = 120$

Substituting this into the equation:

$\implies d_2=1.2116\sqrt{120}$

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:

$\implies d_2-d_1$

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

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

$\implies 8.579906391...$

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

Answer 2

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)