The depth of the submarine is 50 ft below sea level when it starts to descend at a rate of 10.5 ft/s. it dives at that rate for 5 s. what are the constraints on the x- and y-values?

The constraints on the x-values or independent variable – time are:

0 s < time < 5 s

The constraints on the y-values or dependent variable – Depth are:

50 ft < Depth (below sea level) < 102.5 ft

Step-by-step explanation:

Boundary restrictions that are placed upon the variables in an equation are called the Constraints.

The x-values are the independent variables in an equation and in our case it is time.

The y-values are the dependent variables in an equation and in our case it is depth (below the sea level).

The rate of descend provided in the question is constant and is equal to 10.5 ft/s.

We can write our equation of Depth as follows:

Depth (below sea level) = 50 + 10.5*(time)

The constraints on the x-values or independent variable – time are:

0 s < time < 5 s

We can put the boundary values of x variable constraints in the equation of Depth to find the constraint on y-values or dependent variable.

Depth (below sea level) = 50 + 10.5*(time)

Depth (below sea level) = 50 + 10.5*(0), at time 0 s

Depth (below sea level) = 50 ft

Depth (below sea level) = 50 + 10.5*(5), at time 5 s

Depth (below sea level) = 102.5 ft

So the constraints on the y-values or dependent variable – Depth are:

50 ft < Depth (below sea level) < 102.5 ft

distance covered by pamela in 9 l of fuel = 99km

let us apply unitary method here.

distance covered by pamela in 1 l of fuel =99/9 =11 km

now let us find distance covered by pamela if she has 12 l of fuel.

distance covered in 12 l = distance covered in 1 l of fuel * 12

we know that distance covered by here in 1 l is 11 km.

so distance covered in 12 l of fuel = 11*12 = 132 km.