A boat can travel 10 miles against a current in the same time that it can travel 40 miles with the current. The rate of the current is 3 mph. Find the rate of the boat in still water.

x = 5 mph

the rate of the boat in still water is 5 mph.

Step-by-step explanation:

Let x represent the speed of the boat in still water;

Given that the rate of the current is 3 mph.

When it is traveling against current, its relative speed is;

v1 = x - 3

When it is traveling with current, its relative speed is;

v1 = x + 3

Given;

A boat can travel 10 miles against a current in the same time that it can travel 40 miles with the current.

d1 = 10 miles

d2 = 40 miles

Distance = speed × time

time = distance/speed

The time taken for both cases are the same t;

t = d1/v1 = d2/v2

d1/v1 = d2/v2

Substituting the values;

10/(x-3) = 40/(x+3)

Cross multiply;

10(x+3)= 40(x-3)

10x + 30 = 40x - 120

Collecting the like terms;

40x -10x = 120+30

30x = 150

x = 150/30

x = 5 mph

the rate of the boat in still water is 5 mph.

just do 5.2 + 2 ( 7.2) and 1/3 x 3 (.9) and youv'e got your equation. 0.9( 9/10) + 7.2 ^2 = 16.4 hope i could !

step-by-step explanation: