Answer:
The rate at which the distance from the plane to the station is increasing when it is 2 mi away from the station is 372 mi/h.
Step-by-step explanation:
Given information:
A plane flying horizontally at an altitude of "1" mi and a speed of "430" mi/h passes directly over a radar station.


We need to find the rate at which the distance from the plane to the station is increasing when it is 2 mi away from the station.

According to Pythagoras


.... (1)
Put z=1 and y=2, to find the value of x.




Taking square root both sides.

Differentiate equation (1) with respect to t.

Divide both sides by 2.

Put
, y=2,
in the above equation.

Divide both sides by 2.



Therefore the rate at which the distance from the plane to the station is increasing when it is 2 mi away from the station is 372 mi/h.
The point-slope form:

We have the point (4, -6) and the slope m = 3/5. Substitute:

Answer:
The second choice is the same so it is the last one
Answer:
11700 mm²
Step-by-step explanation:
A rectangle is a quadrilateral with two equal, opposite and parallel sides. Each of the angles in a rectangle is 90°.
The horizontal distance of the rectangle = r + r + r + r = 4r
The vertical distance = r + h + r = 2r + h
Where h is the distance between the midpoint of the 2 up circles and the midpoint of the down circle.
Using Pythagoras:
(2r)² = h² + r²
h² + r² = 4r²
h² = 3r²
h = √3r²
h = r√3
Vertical distance = 2r + h = 2r + r√3
Area of rectangle = vertical distance * horizontal distance
Area = 4r * (2r + r√3) = 8r² + 4r²√3 = 4r²(2 + √3)
Substituting:
Area = 4r²(2 + √3) = 4(28²)(2 + √3) = 11703.711 mm²
Area = 11700 mm² to 3 s.f