The formula for the distance from a point (x, y, z) to the plane ax +by +cz +d = 0 is this:
.. distance = |ax +by +cz +d|/√(a^2 +b^2 +c^2)
For your point and plane, that is
.. distance = |-8 -2*7 -4*6 -8|/√(1^2 +2^2 +4^2)
.. = 54/√21
.. = (18/7)√21
Answer:
1, 13
-1, -13
here are 2 coefficient factors that multipy to get 13
brain list please?
I believe the answer is 2 3/7 I hope I could help
Answer: the rate of the jet in still air is 685 mph and the rate of the jetstream is 75 mph
Step-by-step explanation:
Let x represent the rate of the jet in still air.
Let y represent the rate of the jet jetstream.
Distance = speed × time
Flying against the jetstream, a jet travels 3050 miles in 5 hours. This means that the total speed of the jet would be (x - y) mph. Therefore,
3050 = 5(x - y)
3050/5 = x - y
x - y = 610 - - - - - - - - - - - - - -1
Flying with the jetstream, the same jet travels 3800 miles in 4 hours. This means that the total speed of the jet would be (x + y) mph. Therefore,
3800 = 4(x + y)
3800/5 = x + y
x + y = 760 - - - - - - - - - - - - - -2
Adding equation 1 to equation 2, it becomes
2x = 1370
x = 1370/2 = 685mph
Substituting x = 685 into equation 1, it becomes
685 - y = 610
y = 685 - 610 = 75mph
Answer: 7
Step-by-step explanation:
