I think B should be the top one
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.
The answer would be p + s = q + s ...so it would be B.
Well, first, the equation that you are looking for is Y = MX + B. To find B, look at the first dot on the Y-axis. It starts at positive 3, so, you know your equation will end with +3. Next, to find MX, start at the either dot, and find the Rise/Run. In this particular equation, the rise is 1, and, the run is -2, because it's going backwards (negative line). Meaning, the line's equation would be, f( x )= -1/2x + 3.
Answer:
3ab
-------------------
(b+a)
Step-by-step explanation:
3/a - 3/b
-------------------
1/a^2 - 1/b^2
Multiply the top and bottom by a^2 b^2/ a^2/b^2 to clear the fractions
(3/a - 3/b) a^2 b^2
-------------------
(1/a^2 - 1/b^2) a^2b^2
3ab^2 - 3 a^2 b
-------------------
b^2 - a^2
Factor out 3ab on the top
3ab( b-a)
-------------------
b^2 - a^2
The bottom is the difference of squares
3ab( b-a)
-------------------
(b-a) (b+a)
Cancel like terms from the top and bottom
3ab
-------------------
(b+a)