The drag force F on a boat varies jointly with the wet surface area A f the boat and the square of the speed s of the boat. A boat with a wet surface area of 50 ft^2 7 mph traveling atexperiences a drag force of 98n Find the speed of a boat experiencing a drag force of 135 nand having a wet surface area of 60ft^2 .

S = 237.2 mph

Step-by-step explanation:

Given that the drag force F on a boat varies jointly with the wet surface area A of the boat and the square of the speed s of the boat. 

That is. F = KAS^2

Where

F = drag force

K = proportionality constant

S = speed

A boat with a wet surface area of 50 ft^2 traveling at 7 mph experiences a drag force of 98N

1 square foot = 3.587×10^-8 square mile

A = 50 × 3.587×10^-8

A = 1.7935×10^-6

Using the derived formula

F = KAS^2

98 = K × 1.7935×10^-6 × 7^2

98 = 8.788x10^-5K

K = 1115.14

To Find the speed of a boat experiencing a drag force of 135 N and having a wet surface area of 60ft^2 . We will use the same formula

A = 60 × 3.587×10^-8

A = 2.1522×10^-6

F = KAS^2

135 = 1115.14 × 2.1522×10^-6 × S^2

S^2 = 56249.9

S = 237.2 mph

256π units²   (answer b)

step-by-step explanation:

the formula for the surface area of a sphere of radius r

is             a = 4πr²

and so, if r = 8 units, then the surface area is:

a = 4π(8 units)² = 256π units²   (answer b)

step-by-step explanation:

∆abc has the points a(1, 7), b(-2, 2), and c(4, 2) as its vertices. formed with the point d(1, 2) as its third vertex, then ∆abd is triangle.