Answer:
Both flights approach each other at a speed of 624.70 Knots. The FAA minimum separation is not violated as both airplanes are 7.26 Nautical miles away from each other at the time when one of the flights( flight AA) passes through Frada Heights.
Step-by-step explanation:
To solve this kind of problem, the knowledge of concept of relative velocity is needed as the first question requested how fast the flights were approaching each other. To find the minimum distance between both flights, the closest point of approach between both flights should be taken into consideration which was Frada heights. Flight AA passes through Frada Heights in a shorter time of 0.079 hours. This is the time at which both flights approach each other the closest and so the minimum distance (separation) between them. This was calculated to be 7.26NM which is greater than the FAA's minimum this requirement for flight was not violated.
Detailed calculation steps can be found in the attachment below.

Let's simplify ~




Therefore, A is the Correct choice !
Hello,
(t*s)(x)=t(x)*s(x)=(4x²-x+3)*(x-7)
Answer D
Answer:
m<PTR = 140°
Step-by-step explanation:
First, find the value of x. To find the value of x, derive an equation which you'd use in solving for x.
m<PTQ = (x + 28)°
m<RTS = (2x + 16)°
m<PTQ = m<RTS (vertical opposite angles are congruent)
Therefore:
x + 28 = 2x + 16
Solve for x. Combine like terms
28 - 16 = 2x - x
12 = x
x = 12
Find m<PTQ
m<PTQ = (x + 28)
plug in the value of x
m<PTQ = 12 + 28 = 40°
m<PTR + m<PTQ = 180° (supplementary angles)
m<PTR + 40° = 180° (substitution)
m<PTR = 180 - 40 (subtracting 40 from each side)
m<PTR = 140°
It is B because angle DBA is supplementary to angle DBE because together they create a straight line which equals 180 degrees.