Answer:
<u>Answer</u><u>:</u><u> </u><u>A</u><u>.</u><u> </u><u>(</u><u>6</u><u>,</u><u> </u><u>4</u><u>)</u>
Equation of the line:
when x is 6:
They are traveling at right angles to each other so we can say one is traveling north to south and the other west to east. Then we can say that there positions, y and x are:
y=150-600t x=200-800t
By using the Pythagorean Theorem we can find the distance between these two planes as a function of time:
d^2=y^2+x^2, using y and x from above
d^2=(150-600t)^2+(200-800t)^2
d^2=22500-180000t+360000t^2+40000-320000t+640000t^2
d^2=1000000t^2-500000t+62500
d=√(1000000t^2-500000t+6250)
So the rate of change is the derivative of d
dd/dt=(1/2)(2000000t-500000)/√(1000000t^2-500000t+6250)
dd/dt=(1000000t-250000)/√(1000000t^2-500000t+6250)
So the rate depends upon t and is not a constant, so for the instantaneous rate you would plug in a specific value of t...
...
To find how much time the controller has to change the airplanes flight path, we only need to solve for when d=0, or even d^2=0...
1000000t^2-500000t+62500=0
6250(16t^2-8t+1)=0
6250(16^2-4t-4t+1)=0
6250(4t(4t-1)-1(4t-1))=0
6250(4t-1)(4t-1)=0
6250(4t-1)^2=0
4t-1=0
4t=1
t=1/4 hr
Well technically, the controller has t<1/4 because at t=1/4 impact will occur :)
Answer:
1
Step-by-step explanation:
There are about 110,000 days in 300 years, so the expected number of failures is about 10/11 ≈ 1.
_____
This assumes the launch conditions are identical for each of the launches, or that whatever variation there might be has no effect on the probability. These are bad assumptions.
Answer:
a. 8x^3y^3
b. x^5
Step-by-step explanation:
Answer:
hey brother answer for your tricky question is point d286 cubic meter
hope it may help you brother
