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.

1     sepan

2     encuentren

3     tenga

4     hable

5   parezca

6   llegue

the correct functions are

h(x) = 3x + 2  

g(x) = 2^x

step-by-step explanation:

i proceed to calculate the different values ​​of k (x)

case 1)  k(x) = g(x) ∘ h(x)

g(x) ∘ h(x)=g(h(x))

k(x) = [2]^[3x + > is the option  4. k(x) = 2^(3x + 2 )

case 2)  k(x) = g(x) + h(x)

k(x) = [2^x]+[3x + 2]=2^x+3x+> is the option  1.) k(x) = 2^x + 3x + 2

case 3)k(x) = h(x) ÷ g(x)

k(x) =  [3x + 2]/[2^x]=3x/(2^x)+2/(2^x)=3x*(2^-x)+2^(1-x)

is the option  3.)   k(x) = (3x)2^(-x) + 2^(-x + 1 )

case 4)k(x) = g(x) - h(x)

k(x) =  [2^x]-[3x + 2]=2^x-3x-> is the option  2.) k(x) = 2^x - 3x - 2 

case 5)  k(x) = g(x) × h(x)

k(x) = [2^x]*[3x + > is the option  5.) k(x) = 2^x(3x + 2)

case 6)k(x) = h(x) ∘ g(x)

h(x) ∘ g(x)=h(g(x))

k(x)=3*[2^x]+> is the option  6. k(x) = 3(2^x) +2