The figure shows two wires diverging from an intersection point at an angle of 20°. A vertical rod is located a distance x from the intersection point and is moving to the right at speed s. What is the rate of change of the area, dA/dt, formed by the two wires and the rod? A. dA/dt = (1/2(s)tan20)x
B. dA/dt = (1/2tan20)x
C. dA/dt = ((s)tan20)x
D. dA/dt = (tan20)x
E. dA/dt = (2(s)tan20)x

dA/dt = ( (S)tan20°)x

Option C) dA/dt = ((s)tan20)x is the correct Answer

Step-by-step explanation:

Given the data in the question and given figure in the image;

The second image mages a triangle ΔAOB

using SOH CAH TOA

such that;

tan = 0pp/adj

tan20° = AB / AO

tan20° = AB /x

AB = xtan20°

given; dx/dt = S

now, know that the area of a triangle is 1/2base×height

so Area of ΔAOB will be;

A = 1/2 × AO × AB

we substitute in AB = xtan20° and AO = x

A = 1/2 × x × xtan20°

A = 1/2 × x²tan20°

dA/dx = ( 1/2×2xtan20°)

dA/dx = ( xtan20°)

now

dA/dt = dA/dx × dx/dt

we substitute

dA/dt = ( xtan20°) × S

dA/dt = ( (S)tan20°)x

Therefore; Option C) dA/dt = ((s)tan20)x is the correct Answer

answer: 4 meters

step-by-step explanation: to you understand 2/3 better you can do 1 divided by 3 = . then multiply that by the 2 which is .66666666666666667 then you would do 6 x .66666666666666667 which is 4.      

hope this .