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.
Since the circle is centered at the origin, we can use the equations
x² + y² = r²
x = r cos θ
y = r sin θ
We know y = 8 and r = 10. Solving for x
x² + 8² = 10²
x = 6
From this equation:
x = r cos θ
6 = 10 cos θ
cos θ = 6/10 = 0.6
Answer:
Step-by-step explanation:
Given is the shape of a trapezoid.
Therefore,
Area of the trapezoid

Answer:
-5/2(3x+4)<6-3x (multiply with -5/2)
-15/2x-10<6-3x (multiply with 2)
-15x-20<12-6x (change sides)
-15x+6x<12+20
-9x<12+20
-9x<32
x>-32/9
Hope this will help u :)