A lighthouse is built on an exposed reef, 4.5 miles off-shore. The shoreline is perfectly straight, and a town is located 6 miles downshore from the point on the shoreline nearest the lighthouse.
The lighthouse keeper needs to go from the lighthouse to the town to get fresh supplies. He can row a boat at 1.8 miles per hour, and run at 3.4 miles per hour.
How far downshore towards the town should he land, if he wants to get to the town as quickly as possible?

3.94 hours

Step-by-step explanation:

From the question, the following parameters are given:

Distance offshore = 4.5 miles

Distance downshore = 6 miles

Speed when running = 3.4 mph

Speed with boat = 1.8 mph

distance of boat rowing = sqrt(4.5^2 + x^2)

Where

Speed = distance/time

Time = distance/speed

Time = distance of boat rowing/1.8

distance of running = 6 - x

Time = (6-x)/3.4

total travel time

t = sqrt(4.5^2 + x^2)/1.8 + (6 - x)/3.4

dt/dx = x/(1.8×sqrt(4.5^2 + x^2) - 1/3.4

d^2t/dx^2 = +ve at any x

x is at its minimum when dt/dx=0

x/(1.8×sqrt(4.5^2 + x^2) = 1/3.4

x = 6/sqrt(253) × 4.5 = 1.697 miles

Substitute x in

t = sqrt(4.5^2 + x^2)/1.8 + (6 - x)/3.4

We obtaine

t = sqrt(4.5^2 + 1.697^2)/1.8 + (6 - 1.69)/3.4

t = 2.672+ 1.266 = 3.94 hours

according to the triangle sum theorem, all the angles in a triangle must add up to 180 degrees. if we know one angle is already 90, we do 180-90=90. the other 2 angles must add up to 90, exactly.

step-by-step explanation:

answer: c

(2, -4)

goes 7 to the

(9, 4)

easy