Answer:
Yes its a quadrilateral
Step-by-step explanation:
It has 4 sides
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.
18 x 3 + 12 - 1 x 34 + 13 - 2 x 56 = 67
<=Work=>
18 x 3 = 54
1 x 34 = 34
2 x 56 = 112
(54) + 12 - (34) + 13 - (112)
66 - 34 + 13 - 112
32 + 13 - 112
45 - 112
= 67
Let`s assume that points M, N and P are the touching points of those 3 circles:Then:Y M + M Z = 14,Z N + N X = 20X P + P Y = 18And also: M Z = ZN, Y M = P Y and N X = X P.Now we have a system of 3 equations ( Y M, M Z and X P are the radii of each circle ):Y M + M Z = 14M Z + X P = 20X P + Y M = 18 Y M - M Z = - 14+X P + Y M = 18 X P - M Z = 4Y M - M Z = - 14+M Z + X P = 20 X P - Y M = 6 /* ( - 1 )X P - M Z = 4 X P + Y M = - 6 X P - M Z = 4 Y M - M Z = - 2 Y M + M Z = 14 2 Y M = 12 => Y M = 6M Z - 6 = 2 => M Z = 8X P + 6 = 18
X P = 12
Radii of the circles are: 12, 8 and 6.
1.45 meters is your answer