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

47.28 or 47.2857143

step-by-step explanation:

you just divide 331 by 7 and you get that answer. ( i think so tho)

< xcy ≅ < xby

step-by-step explanation:

in δcax and δbax,

so, δcax ≅ δbax (side-angle-side or sas)

since δcax ≅ δbax, we can conclude < acx ≅ < abx

in δacy and δaby,

so, δacy ≅ δaby (side-angle-side or sas)

since δacy ≅ δaby, we can conclude < acy ≅ < aby

now,

< acy ≅ < aby

=> < acx + < xcy ≅ < abx + < xby

=> < acx + < xcy ≅ < acx + < xby   (since < acx ≅ < abx already proved above)

subtracting < acx from both sides, we get

< acx + < xcy -< acx ≅ < acx + < xby -< acx

cancelling out < acx and -< acx from both sides, we get

< xcy ≅ < xby (proved)